Asksia APP

6^{x}=36^{2-x}

Generated Graph

Solution by Steps

step 1

Recognize that

36

is a power of

6

, specifically

36 = 6^2

step 2

Rewrite the equation

6^x = 36^{2-x}

using the fact that

36 = 6^2

:

6^x = (6^2)^{2-x}

step 3

Apply the power of a power rule to simplify the right side:

6^x = 6^{4 - 2x}

step 4

Since the bases are the same, set the exponents equal to each other:

x = 4 - 2x

step 5

Solve for

x

:

3x = 4

step 6

Divide both sides by

3

to isolate

x

:

x = \frac{4}{3}

Answer

x = \frac{4}{3}

Key Concept

Exponentiation and Logarithms

Explanation

When the bases are the same, we can set the exponents equal to each other to solve for the unknown variable.

\ln _{3} x=5

Generated Graph

Solution by Steps

step 1

Given the equation from the asksia-ll calculation list:

\log(x, 3) = 5

step 2

Convert the logarithmic equation to exponential form:

x = 3^5

step 3

Calculate the value of

x

:

x = 243

Answer

x = 243

Key Concept

Converting logarithmic form to exponential form

Explanation

The equation

\log(x, 3) = 5

means that

3

raised to what power equals

x

. Converting this to exponential form gives

x = 3^5

, which simplifies to

x = 243

.

\ln _{2}(x+1)-\ln _{2}(x-1)=\ln _{2} 8

Generated Graph

Solution by Steps

step 1

Use the properties of logarithms to combine the left side of the equation:

\ln_2(x+1) - \ln_2(x-1) = \ln_2\left(\frac{x+1}{x-1}\right)

step 2

Set the combined logarithm equal to the right side of the equation:

\ln_2\left(\frac{x+1}{x-1}\right) = \ln_2(8)

step 3

Since the bases of the logarithms are the same, set the arguments equal to each other:

\frac{x+1}{x-1} = 8

step 4

Solve for

x

:

x+1 = 8(x-1)

step 5

Expand and simplify the equation:

x+1 = 8x-8

step 6

Isolate

x

:

1+8 = 8x-x

step 7

Simplify further:

9 = 7x

step 8

Divide by 7 to find

x

:

x = \frac{9}{7}

Answer

x = \frac{9}{7}

Key Concept

Combining and solving logarithmic equations

Explanation

The key concept in solving this equation is to use the properties of logarithms to combine the logarithmic expressions on the left side and then equate the arguments of the logarithms since their bases are the same. After setting the arguments equal, we solve for

x

using algebraic manipulation.

y=\ln (\ln x)

Generated Graph

Solution by Steps

step 1

To find the equation of the tangent line to the curve

y = \ln(\ln(x))

at

x = 1

, we first need to find the derivative of

y

with respect to

x

step 2

The derivative of

y

with respect to

x

is found using the chain rule:

\frac{dy}{dx} = \frac{1}{\ln(x)} \cdot \frac{1}{x}

step 3

At

x = 1

, the natural logarithm

\ln(1)

is equal to 0, which makes the derivative undefined since we cannot divide by zero

step 4

Therefore, no tangent line exists at

x = 1

for the function

y = \ln(\ln(x))

because the slope of the tangent line is undefined

Answer

No tangent line exists at

x = 1

for the function

y = \ln(\ln(x))

.

Key Concept

Existence of Tangent Lines and Derivatives

Explanation

A tangent line to a curve at a given point exists only if the derivative of the function is defined at that point. Since the derivative involves division by zero at

x = 1

for the function

y = \ln(\ln(x))

, no tangent line exists at that point.

Find the derivative of the given function:

y=\log _{2}\left(x^{2}+1\right)

Generated Graph

Solution by Steps

step 1

To find the derivative of the function

y = \log_2(x^2 + 1)

, we use the chain rule and the logarithmic differentiation formula

step 2

The derivative of

\log_2(u)

with respect to

x

is

\frac{1}{u \ln(2)} \cdot \frac{du}{dx}

, where

u = x^2 + 1

step 3

We compute the derivative of

u

with respect to

x

, which is

\frac{du}{dx} = 2x

step 4

Substituting

u

and

\frac{du}{dx}

into the derivative formula, we get

\frac{d}{dx}(\log_2(x^2 + 1)) = \frac{2x}{(x^2 + 1) \ln(2)}

Answer

\frac{d}{dx}(\log_2(x^2 + 1)) = \frac{2x}{(x^2 + 1) \ln(2)}

Key Concept

Chain Rule and Logarithmic Differentiation

Explanation

The derivative of a logarithmic function with a base other than

e

involves using the chain rule and the natural logarithm, since the derivative of

\log_b(u)

is

\frac{1}{u \ln(b)} \cdot \frac{du}{dx}

.

Solve:

\sqrt{\ln x}=\ln (\sqrt{x})

Generated Graph

Solution by Steps

step 1

To solve the equation, we rewrite the square root of the natural logarithm of

x

and the natural logarithm of the square root of

x

to have a common base for comparison

step 2

The equation

\sqrt{\ln x} = \ln (\sqrt{x})

can be rewritten as

\ln(x)^{1/2} = \ln(x^{1/2})

step 3

Using the property of logarithms that

\ln(a^b) = b \ln(a)

, we can rewrite the right side of the equation as

\frac{1}{2} \ln(x)

step 4

Now we have

\ln(x)^{1/2} = \frac{1}{2} \ln(x)

. Squaring both sides to eliminate the square root gives us

\ln(x) = \frac{1}{4} \ln(x)^2

step 5

To find the solutions, we set

\ln(x) = 0

which gives us

x = 1

, and

\ln(x)^2 = 4

which gives us

\ln(x) = \pm 2

step 6

Solving

\ln(x) = 2

gives us

x = e^2

, but

\ln(x) = -2

is not valid since the natural logarithm cannot have a negative result for positive

x

step 7

Therefore, the solutions are

x = 1

and

x = e^4

Answer

x = 1

and

x = e^4

Key Concept

Logarithmic Equations and Properties

Explanation

The equation involves logarithmic functions and their properties, such as

\ln(a^b) = b \ln(a)

. By applying these properties, we can solve for

x

in the equation

\sqrt{\ln x} = \ln (\sqrt{x})

.

Find the integrals:

\int \frac{(\ln x)^{2}+1}{x} d x=

Solution by Steps

step 1

To integrate the function

\frac{(\ln(x))^2 + 1}{x}

, we separate the terms and integrate each one individually

step 2

The integral of

\frac{(\ln(x))^2}{x}

dx is

\frac{(\ln(x))^3}{3}

by using the power rule for logarithmic integration

step 3

The integral of

\frac{1}{x}

dx is

\ln(x)

, as it is the basic integral of

1/x

step 4

Combining the results from steps 2 and 3, we get the final integral

\frac{(\ln(x))^3}{3} + \ln(x) + C

, where

C

is the constant of integration

Answer

\frac{(\ln(x))^3}{3} + \ln(x) + C

Key Concept

Integration of logarithmic functions and the power rule

Explanation

The integral of a function that is a sum of terms can be found by integrating each term separately. For logarithmic functions, the power rule for integration is applied.

Find the integrals:

\int \frac{x d x}{3 x^{2}+1}=\frac{1}{6} \int \frac{d u}{u}=\frac{1}{6} \ln \left(3 x^{2}+1\right)+C .

Solution by Steps

step 1

To integrate the function

\frac{x}{3x^2 + 1}

, we perform a substitution. Let

u = 3x^2 + 1

, then

du = 6x dx

step 2

The integral becomes

\int \frac{x dx}{3x^2 + 1} = \frac{1}{6} \int \frac{du}{u}

after substituting

x dx

with

\frac{1}{6} du

step 3

The integral of

\frac{1}{u}

with respect to

u

is

\ln|u|

, so we have

\frac{1}{6} \int \frac{du}{u} = \frac{1}{6} \ln|u| + C

step 4

Substituting back for

u

, we get

\frac{1}{6} \ln|3x^2 + 1| + C

. Since 3x^2 + 1 > 0 for all

x

, we can drop the absolute value to get

\frac{1}{6} \ln(3x^2 + 1) + C

Answer

\frac{1}{6} \ln(3x^2 + 1) + C

Key Concept

Integration by Substitution

Explanation

The integral of a function that is a product of a polynomial and the reciprocal of another polynomial can often be solved by the method of substitution, which simplifies the integral into a more manageable form.

Solve the initial value problem

(x+1) y^{\prime}=2 y, y(1)=1

.

Solution by Steps

step 1

The given differential equation is

(x+1)y' = 2y

. We can rewrite this as

\frac{y'}{y} = \frac{2}{x+1}

step 2

To solve this first-order linear ordinary differential equation, we can use separation of variables

step 3

Integrate both sides with respect to

x

:

\int \frac{1}{y} dy = \int \frac{2}{x+1} dx

step 4

The integrals yield

\ln|y| = 2\ln|x+1| + C

, where

C

is the constant of integration

step 5

Exponentiate both sides to get rid of the logarithm:

y = e^{2\ln|x+1| + C}

step 6

Simplify the equation using properties of exponents:

y = e^C \cdot (x+1)^2

step 7

Use the initial condition

y(1) = 1

to find the constant

e^C

step 8

Substitute

x = 1

into the equation:

1 = e^C \cdot (1+1)^2

step 9

Solve for

e^C

:

e^C = \frac{1}{4}

step 10

The solution to the differential equation is

y(x) = \frac{1}{4}(x+1)^2

Answer

y(x) = \frac{1}{4}(x+1)^2

Key Concept

Separation of Variables and Integration

Explanation

The key concept in solving this initial value problem is to use separation of variables to separate the

y

terms on one side and the

x

terms on the other, integrate both sides, and then use the initial condition to solve for the constant of integration.

If

f(x)=2 \sqrt{x} \ln x

, find

f^{\prime}(x)

Generated Graph

Solution by Steps

step 1

To find the derivative of

f(x) = 2\sqrt{x}\ln(x)

, we apply the product rule

step 2

The product rule states that

(uv)' = u'v + uv'

, where

u = 2\sqrt{x}

and

v = \ln(x)

step 3

Differentiate

u = 2\sqrt{x}

to get

u' = \frac{2}{2\sqrt{x}} = \frac{1}{\sqrt{x}}

step 4

Differentiate

v = \ln(x)

to get

v' = \frac{1}{x}

step 5

Apply the product rule:

f'(x) = u'v + uv' = \frac{1}{\sqrt{x}}\ln(x) + 2\sqrt{x}\left(\frac{1}{x}\right)

step 6

Simplify the derivative:

f'(x) = \frac{\ln(x)}{\sqrt{x}} + \frac{2}{\sqrt{x}} = \frac{\ln(x) + 2}{\sqrt{x}}

Answer

f'(x) = \frac{\ln(x) + 2}{\sqrt{x}}

Key Concept

Product Rule for Differentiation

Explanation

To differentiate a product of two functions, we use the product rule, which involves differentiating each function separately and then combining the results as shown in the steps.

I invest

\$ 100,000

in a company for five years, with a guaranteed income of

8 \%

per year, compounded semi-annually. How much will I have at the end of 5 years? If the interest were compounded continuously, how much would I have in 5 years?

Solution by Steps

step 1

To find the future value with semi-annual compounding, we use the formula

FV = PV (1 + \frac{i}{f})^{f \cdot n}

where

PV

is the present value,

i

is the annual interest rate,

f

is the number of compounding periods per year, and

n

is the number of years

step 2

Given

PV = \$100,000

,

i = 8\%

,

f = 2

(since it's semi-annual), and

n = 5

, we plug these values into the formula

step 3

Calculating the future value:

FV = \$100,000 \left(1 + \frac{0.08}{2}\right)^{2 \cdot 5}

step 4

Simplifying the expression:

FV = \$100,000 \left(1 + 0.04\right)^{10}

step 5

Computing the power:

FV = \$100,000 \left(1.04\right)^{10}

step 6

The future value after 5 years with semi-annual compounding is

FV = \$100,000 \cdot 1.48024

step 7

Multiplying to find the future value:

FV = \$148,024

Answer

At the end of 5 years with semi-annual compounding, I will have \$148,024.

Key Concept

Compound Interest Formula

Explanation

The future value of an investment with compound interest depends on the rate, the number of compounding periods, and the time period of the investment.

Solution by Steps

step 1

To find the future value with continuous compounding, we use the formula

FV = PV \cdot e^{i \cdot n}

where

PV

is the present value,

e

is the base of the natural logarithm,

i

is the annual interest rate, and

n

is the number of years

step 2

Given

PV = \$100,000

,

i = 8\%

, and

n = 5

, we plug these values into the formula

step 3

Calculating the future value:

FV = \$100,000 \cdot e^{0.08 \cdot 5}

step 4

Computing the exponent:

FV = \$100,000 \cdot e^{0.4}

step 5

The future value after 5 years with continuous compounding is approximately

FV = \$100,000 \cdot 1.49182

step 6

Multiplying to find the future value:

FV = \$149,182.47

Answer

At the end of 5 years with continuous compounding, I will have approximately \$149,182.47.

Key Concept

Continuous Compounding Formula

Explanation

The future value of an investment with continuous compounding is calculated using the base of the natural logarithm (e) raised to the product of the interest rate and the number of years.

A certain element decays at a rate of

.000163 /

year. Of a piece of this element of

450 \mathrm{~kg}

, how much will remain in ten years?

Solution by Steps

step 1

To find the remaining amount of the element after 10 years, we use the formula for exponential decay, which is

A = A_0 e^{(-kt)}

, where

A

is the amount remaining,

A_0

is the initial amount,

k

is the decay constant, and

t

is the time in years

step 2

Given

A_0 = 450

kg and

k = 0.000163

per year, we substitute these values into the formula along with

t = 10

years

step 3

The calculation becomes

A = 450 e^{(-0.000163 \cdot 10)}

step 4

Using the asksia-ll calculator, we find that

A \approx 449.267

kg

Answer

Approximately 449.267 kg of the element will remain after 10 years.

Key Concept

Exponential Decay

Explanation

The amount of a decaying element over time is calculated using the exponential decay formula, which accounts for the initial amount, decay constant, and time elapsed.

Two variables are related by the equation

2 \ln x+\ln y=x-y

. What is the equation of the tangent line to the graph of this relation at the point

(1,1)

?

Generated Graph

Solution by Steps

step 1

Differentiate both sides of the equation

2 \ln x + \ln y = x - y

implicitly with respect to

x

step 2

Applying the derivatives:

\frac{d}{dx}(2 \ln x) = \frac{2}{x}

,

\frac{d}{dx}(\ln y) = \frac{1}{y}\frac{dy}{dx}

,

\frac{d}{dx}(x) = 1

, and

\frac{d}{dx}(-y) = -\frac{dy}{dx}

step 3

Combine the derivatives to get the equation:

\frac{2}{x} + \frac{1}{y}\frac{dy}{dx} = 1 - \frac{dy}{dx}

step 4

Solve for

\frac{dy}{dx}

to find the slope of the tangent line at the point

(1,1)

step 5

Substitute

x = 1

and

y = 1

into the equation from step 3 to find the numerical value of the slope:

\frac{2}{1} + \frac{1}{1}\frac{dy}{dx} = 1 - \frac{dy}{dx}

step 6

Solve for

\frac{dy}{dx}

to get

\frac{dy}{dx} = -\frac{1}{2}

step 7

Use the point-slope form of the equation of a line

y - y_1 = m(x - x_1)

with

m = -\frac{1}{2}

and the point

(1,1)

to find the equation of the tangent line

step 8

Substitute

m = -\frac{1}{2}

,

x_1 = 1

, and

y_1 = 1

into the point-slope form to get

y - 1 = -\frac{1}{2}(x - 1)

step 9

Simplify the equation to get

y = -\frac{1}{2}x + \frac{3}{2}

Answer

The equation of the tangent line is

y = -\frac{1}{2}x + \frac{3}{2}

.

Key Concept

Implicit differentiation and point-slope form of a line

Explanation

To find the equation of the tangent line to a curve defined implicitly by an equation involving two variables, we use implicit differentiation to find the derivative of one variable with respect to the other. This derivative gives us the slope of the tangent line. We then use the point-slope form of the equation of a line to find the specific equation of the tangent line at a given point.

If the region in the first quadrant bounded by the curve

y=e^{x}

and

x=1

is rotated about the

x

axis, what is the volume of the resulting solid?

Generated Graph

Solution by Steps

step 1

To find the volume of the solid formed by rotating the region bounded by

y = e^x

and

x = 1

about the x-axis, we use the disk method

step 2

The volume

V

of the solid is given by the integral

V = \int_{a}^{b} \pi [f(x)]^2 dx

, where

f(x) = e^x

is the function defining the curve, and

a = 0

and

b = 1

are the limits of integration

step 3

Substituting

f(x) = e^x

into the formula, we get

V = \int_{0}^{1} \pi (e^x)^2 dx

step 4

According to the asksia-ll calculator, the integral

\int_{0}^{1} \pi (e^x)^2 dx

evaluates to

\frac{1}{2} (e^2 - 1) \pi

step 5

Therefore, the volume of the resulting solid is

\frac{1}{2} (e^2 - 1) \pi

Answer

\frac{1}{2} (e^2 - 1) \pi

Key Concept

Disk Method for Volume

Explanation

The disk method is used to calculate the volume of a solid of revolution by integrating the area of circular disks along the axis of rotation. The area of each disk is

\pi [f(x)]^2

, where

f(x)

is the function being rotated.

Which number is equivalent to

3^{\wedge}(4) \div 3^{\wedge}(2)

Solution by Steps

step 1

Simplify the expression using the laws of exponents for division:

3^4 \div 3^2

step 2

Subtract the exponents:

3^{4-2}

step 3

Calculate the result:

3^2

step 4

Evaluate the power:

3^2 = 9

Answer

9

Key Concept

Laws of Exponents for Division

Explanation

When dividing exponential expressions with the same base, subtract the exponents.

Add 8.563 and 4.8292 .

Solution by Steps

step 1

To add the two numbers, we align the decimal points and add the corresponding digits

step 2

Adding the numbers:

8.563 + 4.8292 = 13.3922

step 3

To express the result as a fraction, we convert the decimal to a fraction:

13.3922 = \frac{66961}{5000}

step 4

The fraction can be broken down into a mixed number:

\frac{66961}{5000} = 13 + \frac{1961}{5000}

step 5

The final result in decimal form is:

13.3922

Answer

13.3922

Key Concept

Adding Decimals

Explanation

To add decimals, align the decimal points and add each digit. The sum can be expressed as a decimal or converted into a fraction or mixed number.

How many feet are in a mile?

Solution by Steps

step 1

To convert miles to feet, we use the standard conversion factor that 1 mile is equivalent to 5280 feet

Answer

5280 feet

Key Concept

Conversion of miles to feet

Explanation

1 mile is equal to 5280 feet, which is a standard conversion used in the United States customary units.

Divide the following monomial quotients:

\frac{12 y^{3}}{2}

Generated Graph

Solution by Steps

step 1

To divide the monomial

\frac{12 y^{3}}{2}

, we simplify the fraction by dividing the coefficient 12 by 2

step 2

Simplifying the fraction:

\frac{12}{2} = 6

step 3

The variable part

y^{3}

remains unchanged as it is not affected by the division of the coefficient

step 4

Combining the simplified coefficient with the variable part, we get the final simplified monomial:

6 y^{3}

Answer

6 y^{3}

Key Concept

Dividing Monomials

Explanation

When dividing monomials, divide the coefficients and keep the variable part as is if the variables are not involved in the division.

C=\frac{5}{9}(F-32)

The equation above shows how temperature

F

, measured in degrees Fahrenheit, relates to a temperature

C

, measured in degrees Celsius. Based on the equation, which of the following must be true? I. A temperature increase of 1 degree Fahrenheit is equivalent to a temperature increase of

\frac{5}{9}

degree Celsius. II. A temperature increase of 1 degree Celsius is equivalent to a temperature increase of 1.8 degrees Fahrenheit. III. A temperature increase of

\frac{5}{9}

degree Fahrenheit is equivalent to a temperature increase of 1 degree Celsius. A) I only B) II only C) III only D) I and II only

Solution by Steps

step 2

The equation is

C = \frac{5}{9}(F - 32)

step 3

For I, if

F

increases by 1, then

C

increases by

\frac{5}{9}

because

\frac{5}{9} \times 1 = \frac{5}{9}

step 4

For II, if

C

increases by 1, then

F

increases by

\frac{9}{5} \times 1 = 1.8

because the reciprocal of

\frac{5}{9}

is

\frac{9}{5}

step 5

For III, if

F

increases by

\frac{5}{9}

, then

C

does not increase by 1 because

\frac{5}{9} \times \frac{5}{9}

is not equal to 1

D

Key Concept

Proportional Relationships

Explanation

The equation shows a proportional relationship between Fahrenheit and Celsius temperature scales.

The equation

\frac{24 x^{2}+25 x-47}{a x-2}=-8 x-3-\frac{53}{a x-2}

is true for all values of

x \neq \frac{2}{a^{\prime}}

where

a

is a constant. What is the value of

a

? A) -16 B) -3 C) 3 D) 16

Solution by Steps

step 2

The equation becomes

\frac{24 x^{2}+25 x-47}{a x-2} = \frac{-8 x(a x - 2) - 3(a x - 2) - 53}{a x - 2}.

step 3

Simplify the numerator on the right side of the equation

step 4

Set the numerators equal to each other since the denominators are the same and solve for

x

step 5

The resulting equation is

24 x^{2}+25 x-47 = -8ax^2 + 16x - 3ax + 6 - 53.

step 6

Simplify and combine like terms to form a quadratic equation in terms of

x

step 7

Equate the coefficients of the corresponding powers of

x

on both sides of the equation

step 8

Solve for

a

using the coefficients of the quadratic term, linear term, and the constant term

step 9

The correct value of

a

is found to be

-3

B

Key Concept

Polynomial Equivalence

Explanation

When two rational expressions are equivalent, their numerators must be equal if their denominators are the same. By equating the coefficients of the corresponding terms, we can find the value of the constant

a

.

If

3 x-y=12

, what is the value of

\frac{8^{x}}{2^{y}} ?

A)

2^{12}

B)

4^{4}

C)

8^{2}

D) The value cannot be determined from the information given.

Solution by Steps

step 2

Solving for

y

, we get

y = 3x - 12

step 3

We need to evaluate

\frac{8^x}{2^y}

. Since

8 = 2^3

, we can rewrite

8^x

as

(2^3)^x = 2^{3x}

step 4

Substitute

y

with

3x - 12

in the denominator to get

\frac{2^{3x}}{2^{3x - 12}}

step 5

Using the property of exponents

\frac{a^m}{a^n} = a^{m-n}

, we simplify the expression to

2^{3x - (3x - 12)} = 2^{12}

step 6

Therefore, the value of

\frac{8^x}{2^y}

is

2^{12}

A

Key Concept

Exponent properties and manipulation

Explanation

To solve the problem, we used the property of exponents to simplify the expression by rewriting

8^x

as

2^{3x}

and then subtracting the exponents in the fraction.

For a polynomial

p(x)

, the value of

p(3)

is -2 . Which of the following must be true about

p(x)

? A)

x-5

is a factor of

p(x)

. B)

x-2

is a factor of

p(x)

. C)

x+2

is a factor of

p(x)

. D) The remainder when

p(x)

is divided by

x-3

is -2 .

Solution by Steps

step 2

The Remainder Theorem states that if a polynomial

p(x)

is divided by

(x - a)

, the remainder is

p(a)

step 3

Since

p(3) = -2

, according to the Remainder Theorem, the remainder when

p(x)

is divided by

(x - 3)

is

-2

step 4

This directly corresponds to option D, which states that the remainder when

p(x)

is divided by

x-3

is -2

D

Key Concept

Remainder Theorem

Explanation

The Remainder Theorem tells us that the remainder of the division of a polynomial

p(x)

by a linear factor

(x - a)

is equal to

p(a)

.

An ice sculpture in the form of a sphere melts in such a way that it maintains its spherical shape. The volume of the sphere is decreasing at a constant rate of

2 \pi

cubic meters per hour. At what rate, in square meters per hour, is the surface area of the sphere decreasing at the moment when the radius is 5 meters? (Note: For a sphere of radius

r

, the surface area is

4 \pi r^{2}

and the volume is

\frac{4}{3} \pi r^{3}

.) (A)

\frac{4 \pi}{5}

(B)

40 \pi

(C)

80 \pi^{2}

(D)

100 \pi

Solution by Steps

step 2

Differentiate

V

with respect to

t

to get

\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}

step 3

Substitute

\frac{dV}{dt} = -2 \pi

and

r = 5

into the differentiated volume equation to solve for

\frac{dr}{dt}

step 4

-2 \pi = 4 \pi (5)^2 \frac{dr}{dt}

simplifies to

\frac{dr}{dt} = -\frac{1}{25}

. However, according to asksia-ll,

\frac{dr}{dt}

should be

-\frac{1}{50}

step 5

Now, given the surface area of a sphere

S = 4 \pi r^2

, we need to find

\frac{dS}{dt}

using

\frac{dr}{dt} = -\frac{1}{50}

when

r = 5

step 6

Differentiate

S

with respect to

t

to get

\frac{dS}{dt} = 8 \pi r \frac{dr}{dt}

step 7

Substitute

r = 5

and

\frac{dr}{dt} = -\frac{1}{50}

into the differentiated surface area equation to solve for

\frac{dS}{dt}

step 8

\frac{dS}{dt} = 8 \pi (5) \left(-\frac{1}{50}\right)

simplifies to

\frac{dS}{dt} = -\frac{4 \pi}{5}

A

Key Concept

Related rates in calculus

Explanation

To find the rate at which the surface area of a sphere is changing, we use the relationship between the volume and surface area of a sphere and apply the chain rule to relate their rates of change with respect to time.

A race car is traveling on a straight track at a velocity of 80 meters per second when the brakes are applied at time

t=0

seconds. From time

t=0

to the moment the race car stops, the acceleration of the race car is given by

a(t)=-6 t^{2}-t

meters per second per second. During this time period, how far does the race car travel? (A)

188.229 \mathrm{~m}

(B)

198.766 \mathrm{~m}

(C)

260.042 \mathrm{~m}

(D)

267.089 \mathrm{~m}

Generated Graph

Solution by Steps

step 2

The integral of the acceleration function

a(t) = -6t^2 - t

is the velocity function

v(t) = -2t^3 - \frac{1}{2}t^2 + C

step 3

Since the initial velocity

v(0)

is 80 m/s, we solve for

C

by setting

v(0) = -2(0)^3 - \frac{1}{2}(0)^2 + C = 80

, which gives

C = 80

step 4

The velocity function is then

v(t) = -2t^3 - \frac{1}{2}t^2 + 80

step 5

To find when the car stops, we set the velocity function to zero and solve for

t

. The real solution to

-2t^3 - \frac{1}{2}t^2 + 80 = 0

is approximately

t \approx 1.1973

step 6

Finally, we integrate the velocity function from

t=0

to

t=1.1973

to find the distance traveled

step 7

The integral of

v(t)

from 0 to 1.1973 gives the distance

s

, which is approximately

s \approx 198.766

meters

B

Key Concept

Integration of acceleration to find velocity, and integration of velocity to find distance

Explanation

To determine the distance traveled by an object with a given acceleration function, we integrate the acceleration to find the velocity function, and then integrate the velocity function over the given time interval to find the total distance traveled.

$t$	0	2
$f(t)$	4	12

Generated Graph

Solution by Steps

step 2

The general solution for the differential equation is

y(t) = Ce^{kt}

, where

C

is the constant to be determined by the initial condition

y(0)=4

step 3

Applying the initial condition

y(0)=4

, we get

4 = Ce^{k \cdot 0}

, which simplifies to

C = 4

step 4

Now we have the particular solution

y(t) = 4e^{kt}

. Using the second initial condition

y(2)=12

, we can solve for

k

step 5

Substituting

t=2

into the particular solution gives us

12 = 4e^{2k}

, which simplifies to

e^{2k} = 3

step 6

Taking the natural logarithm of both sides, we get

2k = \ln(3)

, and thus

k = \frac{1}{2}\ln(3)

step 7

The expression for

f(t)

is therefore

4e^{\frac{t}{2}\ln(3)}

A

Key Concept

Solving a first-order linear differential equation with given initial conditions

Explanation

The solution to the differential equation

\frac{dy}{dt} = ky

with initial conditions is found by integrating and applying the initial conditions to solve for the constants.

Which of the following limits is equal to

\int_{3}^{5} x^{4} d x

? (A)

\lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(3+\frac{k}{n}\right)^{4} \frac{1}{n}

(B)

\lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(3+\frac{k}{n}\right)^{4} \frac{2}{n}

(C)

\lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(3+\frac{2 k}{n}\right)^{4} \frac{1}{n}

(D)

\lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(3+\frac{2 k}{n}\right)^{4} \frac{2}{n}

Solution by Steps

step 2

The integral

\int_{3}^{5} x^{4} \, dx

can be calculated directly:

\int_{3}^{5} x^{4} \, dx = \left[\frac{x^5}{5}\right]_{3}^{5} = \frac{5^5}{5} - \frac{3^5}{5} = \frac{3125}{5} - \frac{243}{5} = \frac{2882}{5}

step 3

Comparing the result of the integral with the limits provided by the asksia-ll calculator, we find that the limit that matches the value of the integral is

\lim_{n \to \infty} \sum_{k=1}^{n}\left(3+\frac{2k}{n}\right)^{4} \frac{2}{n} = \frac{2882}{5}

step 4

Therefore, the correct answer is the one that corresponds to this limit

D

Key Concept

Riemann Sum and Definite Integral

Explanation

The Riemann sum is used to approximate the value of a definite integral, and as

n

approaches infinity, the Riemann sum converges to the exact value of the integral. In this case, the Riemann sum that matches the value of the integral

\int_{3}^{5} x^{4} \, dx

is given in option (D).

\lim _{x \rightarrow \pi} \frac{\cos x+\sin (2 x)+1}{x^{2}-\pi^{2}}

is (A)

\frac{1}{2 \pi}

(B)

\frac{1}{\pi}

(C) 1 (D) nonexistent

Generated Graph

Solution by Steps

step 2

The Taylor series expansion provided by asksia-ll calculator for

\frac{\cos(x) + \sin(2x) + 1}{x^2 - \pi^2}

around

x = \pi

is

\frac{1}{\pi}

plus higher order terms of

(x - \pi)

step 3

As

x

approaches

\pi

, all higher order terms of

(x - \pi)

in the Taylor series expansion approach 0

step 4

The limit of the function as

x

approaches

\pi

is therefore the constant term of the Taylor series expansion, which is

\frac{1}{\pi}

B

Key Concept

Taylor series expansion

Explanation

The Taylor series expansion allows us to express a function as an infinite sum of terms calculated from the values of its derivatives at a single point. In this case, the limit as

x

approaches

\pi

is given by the constant term of the Taylor series expansion.

At time

t \geq 0

, a particle moving in the

x y

-plane has velocity vector given by

v(t)=\left\langle 4 e^{-t}, \sin (1+\sqrt{t})\right\rangle

. What is the total distance the particle travels between

t=1

and

t=3

? (A) 1.861 (B) 1.983 (C) 2.236 (D) 4.851

Generated Graph

Solution by Steps

step 2

The magnitude of the velocity vector

v(t)

is given by

\sqrt{(4e^{-t})^2 + (\sin(1+\sqrt{t}))^2}

step 3

We integrate this magnitude from

t = 1

to

t = 3

to find the total distance traveled

step 4

The asksia-ll calculator has already computed this integral and provided the result:

\int_1^3 \sqrt{16e^{-2t} + \sin^2(1+\sqrt{t})} \, dt = 1.8606

step 5

Therefore, the total distance the particle travels between

t = 1

and

t = 3

is approximately 1.8606

A

Key Concept

Distance traveled by a particle

Explanation

The total distance traveled by a particle is the integral of the magnitude of its velocity vector over the time interval.

\lim _{x \rightarrow 2} \frac{x^{3}-7 x^{2}+10 x}{x^{2}+x-6}

Generated Graph

Solution by Steps

step 1

To find the limit as

x

approaches 2, we first look for any common factors in the numerator and denominator that could be canceled out

step 2

Factor the numerator

x^3 - 7x^2 + 10x

and the denominator

x^2 + x - 6

step 3

The factored form of the numerator is

x(x - 5)(x - 2)

and the denominator is

(x - 2)(x + 3)

step 4

Cancel the common factor

(x - 2)

from the numerator and denominator

step 5

The simplified expression is

\frac{x(x - 5)}{x + 3}

step 6

Substitute

x = 2

into the simplified expression to find the limit

step 7

The result of the substitution is

\frac{2(2 - 5)}{2 + 3} = \frac{-6}{5}

Answer

-\frac{6}{5}

Key Concept

Factoring and canceling common factors to simplify expressions before finding limits

Explanation

The limit of a rational function as

x

approaches a value can often be found by factoring and canceling common factors in the numerator and denominator, then substituting the value of

x

into the simplified expression.

Provide a careful sketch of a graph of a single function

f(x)

that satisfies the following six conditions. No formula is needed just carefully sketch and label your graph. Full credit will not be given for a graph that is not carefully labeled or that does not clearly satisfy the six conditions indicated. i.

f(0)=2

;

Generated Graph

Solution by Steps

step 1

Begin by plotting the point where

f(0) = 2

. This is the y-intercept of the graph

step 2

Since no other conditions are provided, we refer to the asksia-ll calculation list which suggests plotting

f(x) = x^2 + 2

step 3

Draw a parabola opening upwards with the vertex at the point (0, 2) since the coefficient of

x^2

is positive

step 4

Label the y-intercept as (0, 2) and ensure the parabola is symmetric about the y-axis

step 5

Make sure the graph is carefully drawn to reflect the nature of a quadratic function, which is smooth and continuous

Answer

A carefully sketched graph of the function

f(x) = x^2 + 2

with the y-intercept labeled at (0, 2).

Key Concept

Sketching a graph of a quadratic function

Explanation

The graph of

f(x) = x^2 + 2

is a parabola opening upwards with a vertex at (0, 2), which satisfies the condition

f(0) = 2

. The graph should be symmetric about the y-axis and smoothly curved.

-2 x(x+3)-(x+1)(x-2)=

a.

-x^{2}-7 x-2

. b.

-x^{2}+5 x+2

. c.

-3 x^{2}-7 x+2

. d.

-3 x^{2}-5 x+2

. e.

-3 x^{2}+5 x-2

.

Generated Graph

Solution by Steps

step 2

For the first term, distribute

-2x

across

(x+3)

to get

-2x^2 - 6x

step 3

For the second term, expand

(x+1)(x-2)

to get

x^2 - 2x + x - 2

, which simplifies to

x^2 - x - 2

step 4

Subtract the expanded second term from the first term to get

-2x^2 - 6x - (x^2 - x - 2)

step 5

Simplify the expression to get

-2x^2 - 6x - x^2 + x + 2

step 6

Combine like terms to get the final simplified expression

-3x^2 - 5x + 2

D

Key Concept

Distributive Property and Combining Like Terms

Explanation

To simplify the given expression, we use the distributive property to expand the products and then combine like terms to get the final simplified expression.

If

3^{x+5}=\frac{1}{27}

, then

x=

a. -10 . b. -8 . c. -6 . d. -2 . e. 1

Generated Graph

Solution by Steps

step 2

Since

3^{x+5}

is equal to

1/27

, we can write

3^{x+5} = 3^{-3}

step 3

Setting the exponents equal to each other because the bases are the same, we get

x + 5 = -3

step 4

Solving for

x

, we subtract

5

from both sides to get

x = -3 - 5

step 5

Simplifying, we find that

x = -8

B

Key Concept

Exponent properties and solving exponential equations

Explanation

When the bases are the same, the exponents can be set equal to each other to solve for the unknown variable.

If

x \geq 0

, then

\sqrt{x^{4}-6 x^{2}+9}=

a.

x^{2}-\sqrt{6 x}+3

. b.

x^{2}+\sqrt{6 x}+3

. c.

\left|x^{2}-3\right|

. d.

\left(x^{2}+3\right) \sqrt{-6 x^{2}}

. e.

x^{2} \sqrt{6 x+9}

.

Generated Graph

Solution by Steps

step 2

The expression inside the square root is

(x^{2} - 3)^{2}

, since

(x^{2} - 3)(x^{2} - 3) = x^{4} - 6x^{2} + 9

step 3

Taking the square root of

(x^{2} - 3)^{2}

gives

|x^{2} - 3|

, because the square root and the square cancel each other out

step 4

Since

x \geq 0

,

x^{2} \geq 0

and

x^{2} - 3

will be non-negative for

x \geq \sqrt{3}

, so

|x^{2} - 3| = x^{2} - 3

. For 0 \leq x < \sqrt{3} ,

|x^{2} - 3| = 3 - x^{2}

. Therefore,

|x^{2} - 3|

is the correct simplification for all

x \geq 0

C

Key Concept

Simplifying square roots of perfect square trinomials

Explanation

The square root of a perfect square trinomial

(a - b)^{2}

is the absolute value of the expression

a - b

, which is

|a - b|

. This is because the square root function and the squaring function are inverse operations, and the absolute value accounts for the non-negative result of a square root.

The sum of a number and 6 is 8 more than twice the number. Find the equation that could be used to find this number

x

. a.

x+6=2 x+8

b.

x+6=x^{2}+8

c.

x+6=2(x+8)

d.

6 x=2 x+8

e.

x+14=2 x

Generated Graph

Solution by Steps

step 2

Write the equation from step 1:

x + 6 = 2x + 8

step 3

Simplify the equation by subtracting

x

from both sides:

6 = x + 8

step 4

Subtract 8 from both sides to solve for

x

:

x = -2

A

Key Concept

Translating word problems into equations

Explanation

The problem describes a relationship between a number and its double plus an additional value. The correct equation representing this relationship is

x + 6 = 2x + 8

.

At time

t \geq 0

, the position of a particle moving along a curve in the

x y

-plane is

(x(t), y(t))

, where

\frac{d x}{d t}=t-5 \cos t

and

\frac{d y}{d t}=6 \cos (1+\sin t)

. At time

t=3

, the particle is at position

(-1,2)

. Find the

y

-coordinate of the particle's position at time

t=0

.

Generated Graph

Solution by Steps

step 1

To find the

y

-coordinate of the particle's position at time

t=0

, we need to integrate the given

\frac{d y}{d t}

from

t=0

to

t=3

step 2

The integral of

\frac{d y}{d t}=6 \cos (1+\sin t)

from

t=0

to

t=3

is approximately

-1.6335888917

according to the asksia-ll calculator

step 3

The

y

-coordinate at time

t=3

is given as

2

. To find the

y

-coordinate at time

t=0

, we subtract the integral value from the

y

-coordinate at time

t=3

step 4

Calculating the

y

-coordinate at time

t=0

:

y(0) = y(3) - \int_{0}^{3} \frac{d y}{d t} \, dt = 2 - (-1.6335888917)

step 5

Simplifying the expression:

y(0) = 2 + 1.6335888917

step 6

The

y

-coordinate of the particle's position at time

t=0

is approximately

3.6335888917

Answer

The

y

-coordinate of the particle's position at time

t=0

is approximately

3.6335888917

.

Key Concept

Integration of velocity to find displacement

Explanation

To find the position of a particle at a certain time given its velocity function, we integrate the velocity function over the desired time interval and adjust the final position by the displacement found from the integration.

The Taylor series for a function

f

about

x=0

converges to

f

for

-1 \leq x \leq 1

. The

n

th-degree Taylor polynomial for

f

about

x=0

is given by

P_{n}(x)=\sum_{k=1}^{n}(-1)^{k} \frac{x^{k}}{k^{2}+k+1}

. Of the following, which is the smallest number

M

for which the alternating series error bound guarantees that

\left|f(1)-P_{4}(1)\right| \leq M ?

(A)

\frac{1}{5 !} \cdot \frac{1}{31}

(B)

\frac{1}{4 !} \cdot \frac{1}{21}

(C)

\frac{1}{31}

(D)

\frac{1}{21}

Solution by Steps

step 2

The 5th term of the Taylor series is given by

(-1)^{5} \frac{x^{5}}{5^{2}+5+1}

step 3

Simplifying the 5th term:

(-1)^{5} \frac{1^{5}}{5^{2}+5+1} = -\frac{1}{31}

. Since we are looking for the absolute value, the error bound is

\frac{1}{31}

step 4

The error bound is the absolute value of the first omitted term, which gives us

M = \frac{1}{31}

C

Key Concept

Alternating Series Error Bound

Explanation

The error bound for an alternating series is the absolute value of the first omitted term.

Solve the following system by transforming the augmented matrix to reduced echelon form (Gauss-Jordan elimination). Indicate the elementary row operations you performed.

\begin{array}{r} x_{1}+x_{2}-x_{5}=1 \\ x_{2}+2 x_{3}+x_{4}+3 x_{5}=1 \\ x_{1}-x_{3}+x_{4}+x_{5}=0 \end{array}

Solution by Steps

step 1

Begin with the augmented matrix for the system of equations:

\begin{bmatrix} 1 &amp; 1 &amp; 0 &amp; 0 &amp; -1 &amp; | &amp; 1 \\ 0 &amp; 1 &amp; 2 &amp; 1 &amp; 3 &amp; | &amp; 1 \\ 1 &amp; 0 &amp; -1 &amp; 1 &amp; 1 &amp; | &amp; 0 \end{bmatrix}

step 2

Perform row reduction to obtain the reduced echelon form. The first operation is to subtract the first row from the third row to make the first element of the third row zero:

\begin{bmatrix} 1 &amp; 1 &amp; 0 &amp; 0 &amp; -1 &amp; | &amp; 1 \\ 0 &amp; 1 &amp; 2 &amp; 1 &amp; 3 &amp; | &amp; 1 \\ 0 &amp; -1 &amp; -1 &amp; 1 &amp; 2 &amp; | &amp; -1 \end{bmatrix}

step 3

Add the second row to the third row to make the second element of the third row zero:

\begin{bmatrix} 1 &amp; 1 &amp; 0 &amp; 0 &amp; -1 &amp; | &amp; 1 \\ 0 &amp; 1 &amp; 2 &amp; 1 &amp; 3 &amp; | &amp; 1 \\ 0 &amp; 0 &amp; 1 &amp; 2 &amp; 5 &amp; | &amp; 0 \end{bmatrix}

step 4

Subtract twice the third row from the second row to make the third element of the second row zero and to simplify the second row:

\begin{bmatrix} 1 &amp; 1 &amp; 0 &amp; 0 &amp; -1 &amp; | &amp; 1 \\ 0 &amp; 1 &amp; 0 &amp; -3 &amp; -7 &amp; | &amp; 1 \\ 0 &amp; 0 &amp; 1 &amp; 2 &amp; 5 &amp; | &amp; 0 \end{bmatrix}

step 5

Subtract the second row from the first row to make the second element of the first row zero:

\begin{bmatrix} 1 &amp; 0 &amp; 0 &amp; 3 &amp; 6 &amp; | &amp; 0 \\ 0 &amp; 1 &amp; 0 &amp; -3 &amp; -7 &amp; | &amp; 1 \\ 0 &amp; 0 &amp; 1 &amp; 2 &amp; 5 &amp; | &amp; 0 \end{bmatrix}

step 6

The matrix is now in reduced echelon form, which corresponds to the system of equations:

x_1 + 3x_4 + 6x_5 = 0

,

x_2 - 3x_4 - 7x_5 = 1

, and

x_3 + 2x_4 + 5x_5 = 0

Answer

The system of equations in reduced echelon form is:

x_1 + 3x_4 + 6x_5 = 0

,

x_2 - 3x_4 - 7x_5 = 1

,

x_3 + 2x_4 + 5x_5 = 0

.

Key Concept

Gauss-Jordan elimination

Explanation

Gauss-Jordan elimination is a method of solving a system of linear equations by transforming the system's augmented matrix into reduced echelon form using elementary row operations. The resulting matrix reveals the solutions to the system of equations.

Determine all possibilities for the solution set of a homogeneous system of 2 equations in 2 unknowns that has a solution

x_{1}=1, x_{2}=5

.

Solution by Steps

step 1

Consider the homogeneous system of equations:

\begin{cases} ax + by = 0 \\ cx + dy = 0 \end{cases}

step 2

Since

(x_1, x_2) = (1, 5)

is a solution, we can substitute these values into the system:

\begin{cases} a(1) + b(5) = 0 \\ c(1) + d(5) = 0 \end{cases}

step 3

Simplifying the system, we get two equations:

\begin{cases} a + 5b = 0 \\ c + 5d = 0 \end{cases}

step 4

From the first equation, we can express

a

in terms of

b

:

a = -5b

step 5

From the second equation, we can express

c

in terms of

d

:

c = -5d

step 6

The solution set for the coefficients

a, b, c, d

depends on the values of

b

and

d

. For any real numbers

b

and

d

, the corresponding

a

and

c

can be found using the relations from steps 4 and 5

Answer

The solution set for the coefficients is

\{a = -5b, b, c = -5d, d\}

where

b

and

d

are any real numbers.

Key Concept

Homogeneous system of linear equations

Explanation

In a homogeneous system of linear equations, if a non-trivial solution exists, the system has infinitely many solutions that are scalar multiples of any given non-trivial solution. The coefficients of the system can be determined by the given solution.

Let

A=\left[\begin{array}{ccc} 2 &amp; -1 &amp; -1 \\ -1 &amp; 2 &amp; -1 \\ -1 &amp; -1 &amp; 2 \end{array}\right] .

Determine whether the matrix

A

is diagonalizable. If it is diagonalizable, then diagonalize

A

. That is, find a nonsingular matrix

S

and a diagonal matrix

D

such that

S^{-1} A S=D

.

Solution by Steps

step 1

To determine if matrix

A

is diagonalizable, we need to find a nonsingular matrix

S

and a diagonal matrix

D

such that

S^{-1} A S = D

step 2

The given matrix

A

is

\begin{bmatrix} 2 &amp; -1 &amp; -1 \\ -1 &amp; 2 &amp; -1 \\ -1 &amp; -1 &amp; 2 \end{bmatrix}

. We need to find the eigenvalues of

A

by solving the characteristic equation

\det(A - \lambda I) = 0

step 3

The eigenvalues found are

\lambda_1 = 0

and

\lambda_2 = \lambda_3 = 3

. These are the diagonal elements of matrix

D

step 4

Next, we find the eigenvectors corresponding to each eigenvalue, which will form the columns of matrix

S

step 5

The eigenvectors corresponding to the eigenvalues are

\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}

for

\lambda_1 = 0

and

\begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}

,

\begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix}

for

\lambda_2 = \lambda_3 = 3

. These form the columns of

S

step 6

The matrix

S

is

\begin{bmatrix} 1 &amp; -1 &amp; -1 \\ 1 &amp; 0 &amp; 1 \\ 1 &amp; 1 &amp; 0 \end{bmatrix}

and the diagonal matrix

D

is

\begin{bmatrix} 0 &amp; 0 &amp; 0 \\ 0 &amp; 3 &amp; 0 \\ 0 &amp; 0 &amp; 3 \end{bmatrix}

step 7

To find

S^{-1}

, we perform matrix inversion on

S

. The inverse matrix

S^{-1}

is

\begin{bmatrix} 1/3 &amp; 1/3 &amp; 1/3 \\ -1/3 &amp; -1/3 &amp; 2/3 \\ -1/3 &amp; 2/3 &amp; -1/3 \end{bmatrix}

step 8

Verify that

S^{-1} A S = D

by matrix multiplication

Answer

The matrix

A

is diagonalizable. The nonsingular matrix

S

is

\begin{bmatrix} 1 &amp; -1 &amp; -1 \\ 1 &amp; 0 &amp; 1 \\ 1 &amp; 1 &amp; 0 \end{bmatrix}

and the diagonal matrix

D

is

\begin{bmatrix} 0 &amp; 0 &amp; 0 \\ 0 &amp; 3 &amp; 0 \\ 0 &amp; 0 &amp; 3 \end{bmatrix}

. The inverse of

S

is

\begin{bmatrix} 1/3 &amp; 1/3 &amp; 1/3 \\ -1/3 &amp; -1/3 &amp; 2/3 \\ -1/3 &amp; 2/3 &amp; -1/3 \end{bmatrix}

.

Key Concept

Diagonalization of a Matrix

Explanation

A matrix is diagonalizable if it has a full set of linearly independent eigenvectors. The matrix

S

is formed by the eigenvectors as columns, and the diagonal matrix

D

contains the corresponding eigenvalues. The inverse of

S

is used to transform

A

into its diagonal form

D

.

The owner of the Good Deals Store opens a new store across town. For the new store, the owner estimates that, during business hours, an average of 90 shoppers per hour enter the store and each of them stays an average of 12 minutes. The average number of shoppers in the new store at any time is what percent less than the average number of shoppers in the original store at any time? (Note: Ignore the percent symbol when entering your answer. For example, if the answer is

42.1 \%

, enter 42.1 )

Solution by Steps

step 1

Calculate the average number of shoppers in the new store at any time using the formula: average number of shoppers = (number of shoppers per hour) * (average time each shopper stays in minutes) / 60

step 2

Substitute the given values into the formula: average number of shoppers = (90 shoppers per hour) * (12 minutes) / 60 minutes

step 3

Perform the calculation: average number of shoppers = (90 * 12) / 60

step 4

Simplify the calculation: average number of shoppers = 1080 / 60

step 5

Calculate the result: average number of shoppers = 18

step 6

To find the percent difference, use the formula: percent difference = (original average - new average) / original average * 100

step 7

Since the original average number of shoppers is not provided, we cannot calculate the percent difference

Answer

Unable to calculate the percent difference without the original average number of shoppers.

Key Concept

Average Number of Shoppers and Percent Difference

Explanation

To find the average number of shoppers in the new store, we multiply the number of shoppers per hour by the average time each shopper stays and divide by 60 to convert minutes to hours. To find the percent difference, we need the average number of shoppers in the original store, which is not provided.

In the

x y

-plane, the point

(p, r)

lies on the line with equation

y=x+b

, where

b

is a constant. The point with coordinates

(2 p, 5 r)

lies on the line with equation

y=2 x+b

. If

p \neq 0

, what is the value of

\frac{r}{p}

? A)

\frac{2}{5}

B)

\frac{3}{4}

C)

\frac{4}{3}

D)

\frac{5}{2}

Solution by Steps

step 2

Given the point

(2p, 5r)

lies on the line

y = 2x + b

, we can substitute

2p

for

x

and

5r

for

y

to get the equation

5r = 2(2p) + b

step 3

Simplify the equation from step 2 to get

5r = 4p + b

step 4

Substitute

r = p + b

from step 1 into the equation from step 3 to get

5(p + b) = 4p + b

step 5

Simplify the equation from step 4 to get

5p + 5b = 4p + b

step 6

Subtract

b

from both sides to get

5p + 4b = 4p

step 7

Subtract

4p

from both sides to get

p = -4b

step 8

Substitute

p = -4b

back into the equation

r = p + b

to get

r = -4b + b

step 9

Simplify the equation from step 8 to get

r = -3b

step 10

Divide the equation

r = -3b

by the equation

p = -4b

to find

\frac{r}{p} = \frac{-3b}{-4b}

step 11

Simplify the fraction to get

\frac{r}{p} = \frac{3}{4}

B

Key Concept

Substitution and simplification of linear equations

Explanation

To find the value of

\frac{r}{p}

, we substitute the given points into their respective line equations, simplify, and solve for the variables in terms of

b

, then divide the two resulting expressions to find the ratio.

If

x

is the average (arithmetic mean) of

m

and

9, y

is the average of

2 m

and 15 , and

z

is the average of

3 m

and 18 , what is the average of

x, y

, and

z

in terms of

m

? A)

m+6

B)

m+7

C)

2 m+14

D)

3 m+21

Solution by Steps

step 2

The average

x

of

m

and

9

is

x = \frac{m + 9}{2}

step 3

The average

y

of

2m

and

15

is

y = \frac{2m + 15}{2}

step 4

The average

z

of

3m

and

18

is

z = \frac{3m + 18}{2}

step 5

The average of

x, y,

and

z

is

\frac{x + y + z}{3}

step 6

Substituting the expressions for

x, y,

and

z

into the average formula gives

\frac{\frac{m + 9}{2} + \frac{2m + 15}{2} + \frac{3m + 18}{2}}{3}

step 7

Simplifying the numerator:

\frac{m + 9 + 2m + 15 + 3m + 18}{2} = \frac{6m + 42}{2}

step 8

Dividing by 3 to find the average:

\frac{6m + 42}{2 \cdot 3} = \frac{6m + 42}{6}

step 9

Simplifying the fraction:

\frac{6m + 42}{6} = m + 7

step 10

The average of

x, y,

and

z

in terms of

m

is

m + 7

B

Key Concept

Average of variables

Explanation

To find the average of several terms, sum the terms and divide by the number of terms. When the terms are algebraic expressions, combine like terms and simplify.

Problem 1 Let

W

be the subset of the 3 -dimensional vector space

\mathbb{R}^{3}

defined by

W=\left\{\mathbf{x}=\left[\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right] \in \mathbb{R}^{3} \mid 2 x_{1} x_{2}=x_{3}\right\} .

Which of the following vectors are in the subset

W

? Choose all vectors that belong to

W

. (1)

\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]

(2)

\left[\begin{array}{l}1 \\ 2 \\ 2\end{array}\right]

(3)

\left[\begin{array}{l}3 \\ 0 \\ 0\end{array}\right]

(4)

\left[\begin{array}{l}0 \\ 0\end{array}\right]

(5)

\left[\begin{array}{lll}1 &amp; 2 &amp; 4 \\ 1 &amp; 2 &amp; 4\end{array}\right]

(6)

\left[\begin{array}{c}1 \\ -1 \\ -2\end{array}\right]

.

Solution by Steps

step 2

For vector (1)

\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]

, we have

2 \cdot 0 \cdot 0 = 0

, which is true

step 3

For vector (2)

\left[\begin{array}{l}1 \\ 2 \\ 2\end{array}\right]

, we have

2 \cdot 1 \cdot 2 = 4

, which is false since

x_3 = 2

step 4

For vector (3)

\left[\begin{array}{l}3 \\ 0 \\ 0\end{array}\right]

, we have

2 \cdot 3 \cdot 0 = 0

, which is true

step 5

Vector (4)

\left[\begin{array}{l}0 \\ 0\end{array}\right]

is not in

\mathbb{R}^3

, so it cannot be in subset

W

step 6

Vector (5) is not a vector but a matrix, so it cannot be in subset

W

step 7

For vector (6)

\left[\begin{array}{c}1 \\ -1 \\ -2\end{array}\right]

, we have

2 \cdot 1 \cdot (-1) = -2

, which is true

A, C, F

Key Concept

Subset membership based on vector components

Explanation

To determine if a vector is part of a subset defined by a condition, each component of the vector must satisfy that condition.

Problem 2 Let

W

be the subset of

\mathbb{R}^{3}

defined by

W=\left\{\mathbf{x}=\left[\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right] \in \mathbb{R}^{3} \mid \quad x_{1}=3 x_{2} \text { and } x_{3}=0\right\} .

Determine whether the subset

W

is a subspace of

\mathbb{R}^{3}

or not.

Solution by Steps

step 1

To determine if

W

is a subspace of

\mathbb{R}^{3}

, we must check if it is closed under addition and scalar multiplication, and if it contains the zero vector

step 2

Check for closure under addition: Take any two vectors

\mathbf{a} = [3a_2; a_2; 0]

and

\mathbf{b} = [3b_2; b_2; 0]

in

W

. Their sum is

\mathbf{a} + \mathbf{b} = [3a_2 + 3b_2; a_2 + b_2; 0]

, which simplifies to

[3(a_2 + b_2); a_2 + b_2; 0]

. This is of the form

[3c_2; c_2; 0]

, so the sum is in

W

step 3

Check for closure under scalar multiplication: Take any vector

\mathbf{a} = [3a_2; a_2; 0]

in

W

and any scalar

k

. The product is

k\mathbf{a} = k[3a_2; a_2; 0] = [3ka_2; ka_2; 0]

, which is of the form

[3c_2; c_2; 0]

, so the scalar multiple is in

W

step 4

Check for the zero vector: The zero vector in

\mathbb{R}^{3}

is

[0; 0; 0]

, which satisfies the conditions

x_1 = 3x_2

and

x_3 = 0

since

0 = 3 \cdot 0

and

0 = 0

. Therefore, the zero vector is in

W

Answer

The subset

W

is a subspace of

\mathbb{R}^{3}

.

Key Concept

Subspace Criteria

Explanation

A subset

W

of

\mathbb{R}^{3}

is a subspace if it is closed under addition and scalar multiplication, and contains the zero vector. The set

W

meets all these criteria.

The given matrix is the augmented matrix for a system of linear equations. Give the vector form for the general solution.

\left[\begin{array}{rrrrr|r} 1 &amp; 0 &amp; -1 &amp; 0 &amp; -2 &amp; 0 \\ 0 &amp; 1 &amp; 2 &amp; 0 &amp; -1 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 1 &amp; 1 &amp; 0 \end{array}\right]

Solution by Steps

step 1

The given matrix is already in row-reduced echelon form. Identify the pivot columns and the free variables

step 2

The pivot columns are the 1st, 2nd, and 4th columns, corresponding to variables

x_1, x_2,

and

x_4

. The 3rd and 5th columns are not pivot columns and correspond to free variables

x_3

and

x_5

step 3

Write the system of equations corresponding to the row-reduced echelon form of the matrix

step 4

The system of equations is:

x_1 - x_3 - 2x_5 = 0

x_2 + 2x_3 - x_5 = 0

x_4 + x_5 = 0

step 5

Express the pivot variables

x_1, x_2,

and

x_4

in terms of the free variables

x_3

and

x_5

step 6

The expressions are:

x_1 = x_3 + 2x_5

x_2 = -2x_3 + x_5

x_4 = -x_5

step 7

Write the vector form for the general solution using the expressions from step 6

step 8

The vector form for the general solution is:

\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{bmatrix} = \begin{bmatrix} x_3 + 2x_5 \\ -2x_3 + x_5 \\ x_3 \\ -x_5 \\ x_5 \end{bmatrix} = x_3 \begin{bmatrix} 1 \\ -2 \\ 1 \\ 0 \\ 0 \end{bmatrix} + x_5 \begin{bmatrix} 2 \\ 1 \\ 0 \\ -1 \\ 1 \end{bmatrix}

Answer

\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{bmatrix} = x_3 \begin{bmatrix} 1 \\ -2 \\ 1 \\ 0 \\ 0 \end{bmatrix} + x_5 \begin{bmatrix} 2 \\ 1 \\ 0 \\ -1 \\ 1 \end{bmatrix}

Key Concept

General Solution of a Linear System in Vector Form

Explanation

The general solution of a system of linear equations can be expressed in vector form by writing the pivot variables in terms of the free variables and then expressing the solution as a linear combination of vectors corresponding to the free variables.

Let

A=\left[\begin{array}{lll} 1 &amp; 2 &amp; 3 \\ 4 &amp; 5 &amp; 6 \end{array}\right], B=\left[\begin{array}{lll} 1 &amp; 0 &amp; 1 \\ 0 &amp; 1 &amp; 0 \end{array}\right], C=\left[\begin{array}{ll} 1 &amp; 2 \\ 0 &amp; 6 \end{array}\right], \mathbf{v}=\left[\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right]

Then compute and simplify the following expression.

\mathbf{v}^{\mathrm{T}}\left(A^{\mathrm{T}}-(A-B)^{\mathrm{T}}\right) C .

Solution by Steps

step 1

First, we need to compute the transpose of matrix

A

and matrix

B

step 2

The transpose of matrix

A

, denoted

A^T

, is

A^T = \begin{bmatrix} 1 &amp; 4 \\ 2 &amp; 5 \\ 3 &amp; 6 \end{bmatrix}

. Similarly, the transpose of matrix

B

, denoted

B^T

, is

B^T = \begin{bmatrix} 1 &amp; 0 \\ 0 &amp; 1 \\ 1 &amp; 0 \end{bmatrix}

step 3

Next, we calculate

A - B

and then take the transpose of the result.

A - B = \begin{bmatrix} 1-1 &amp; 2-0 &amp; 3-1 \\ 4-0 &amp; 5-1 &amp; 6-0 \end{bmatrix} = \begin{bmatrix} 0 &amp; 2 &amp; 2 \\ 4 &amp; 4 &amp; 6 \end{bmatrix}

and

(A - B)^T = \begin{bmatrix} 0 &amp; 4 \\ 2 &amp; 4 \\ 2 &amp; 6 \end{bmatrix}

step 4

Now we subtract

(A - B)^T

from

A^T

to get

A^T - (A - B)^T = \begin{bmatrix} 1 &amp; 4 \\ 2 &amp; 5 \\ 3 &amp; 6 \end{bmatrix} - \begin{bmatrix} 0 &amp; 4 \\ 2 &amp; 4 \\ 2 &amp; 6 \end{bmatrix} = \begin{bmatrix} 1 &amp; 0 \\ 0 &amp; 1 \\ 1 &amp; 0 \end{bmatrix}

step 5

We multiply the transpose of vector

\mathbf{v}

, which is

\mathbf{v}^T = [0, 1, 0]

, by the matrix we found in step 4.

\mathbf{v}^T (A^T - (A - B)^T) = [0, 1, 0] \begin{bmatrix} 1 &amp; 0 \\ 0 &amp; 1 \\ 1 &amp; 0 \end{bmatrix} = [0, 1, 0]

step 6

Finally, we multiply the result from step 5 by matrix

C

to get the final answer.

[0, 1, 0] \begin{bmatrix} 1 &amp; 2 \\ 0 &amp; 6 \end{bmatrix} = [0 \cdot 1 + 1 \cdot 0 + 0 \cdot 0, 0 \cdot 2 + 1 \cdot 6 + 0 \cdot 0] = [0, 6]

Answer

[0, 6]

Key Concept

Matrix Transposition and Multiplication

Explanation

The problem involves transposing matrices, subtracting them, and then performing matrix multiplication in the correct order, following the associative property of matrix multiplication.

Let

V

be the vector space of all

2 \times 2

real matrices. Let

S=\left\{A_{1}, A_{2}, A_{3}, A_{4}\right\}

, where

A_{1}=\left[\begin{array}{cc} 1 &amp; 2 \\ -1 &amp; 3 \end{array}\right], A_{2}=\left[\begin{array}{cc} 0 &amp; -1 \\ 1 &amp; 4 \end{array}\right], A_{3}=\left[\begin{array}{cc} -1 &amp; 0 \\ 1 &amp; -10 \end{array}\right], A_{4}=\left[\begin{array}{cc} 3 &amp; 7 \\ -2 &amp; 6 \end{array}\right] .

Then find a basis for the span

\operatorname{Span}(S)

.

Solution by Steps

step 1

To find a basis for the span of the set

S

, we need to consider the matrices

A_1, A_2, A_3, A_4

as vectors in

\mathbb{R}^4

by flattening each matrix into a 4-dimensional vector

step 2

We arrange these vectors as rows in a matrix to perform row reduction:

\begin{pmatrix} 1 &amp; 2 &amp; -1 &amp; 3 \\ 0 &amp; -1 &amp; 1 &amp; 4 \\ -1 &amp; 0 &amp; 1 &amp; -10 \\ 3 &amp; 7 &amp; -2 &amp; 6 \end{pmatrix}

step 3

Using the asksia-ll calculator, we row reduce the matrix to its row echelon form:

\begin{pmatrix} 1 &amp; 0 &amp; 0 &amp; \frac{21}{2} \\ 0 &amp; 1 &amp; 0 &amp; -\frac{7}{2} \\ 0 &amp; 0 &amp; 1 &amp; \frac{1}{2} \\ 0 &amp; 0 &amp; 0 &amp; 0 \end{pmatrix}

step 4

The row reduced matrix shows that the first three rows are linearly independent, and the last row is a zero vector, which does not contribute to the span

step 5

The basis for the span

\operatorname{Span}(S)

consists of the non-zero rows of the row reduced matrix, which correspond to the original matrices

A_1, A_2, A_3

Answer

A basis for the span

\operatorname{Span}(S)

is

\{A_1, A_2, A_3\}

.

Key Concept

Basis of a Span

Explanation

The basis of a span is a set of linearly independent vectors that span the vector space. In this case, the matrices

A_1, A_2, A_3

are linearly independent and span the space generated by

S

.

AskSia

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