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截屏2024-10-11 10.03.30
Oct 11, 2024
截屏2024-10-11 10.03.30
你好，Penny！看起来你提到了一张截屏，但我无法查看图片或截屏内容。如果你能描述一下截屏中的问题或内容，我会很乐意帮助你解决数学问题！请告诉我具体的数学问题或概念。
Solution by Steps
step 1
We start by defining the distance dC d_C from point A(4,1) A(4,1) to point C(x,y) C(x,y) and then from C C to point B(2,5) B(2,5) . The distance from A A to C C is given by ∣4−x∣+∣1−y∣ |4 - x| + |1 - y| and from C C to B B is ∣x−2∣+∣y−5∣ |x - 2| + |y - 5| . Thus, we have:
step 2
The total distance dC d_C can be expressed as: dC=∣4−x∣+∣1−y∣+∣x−2∣+∣y−5∣ d_C = |4 - x| + |1 - y| + |x - 2| + |y - 5| We need to find the value of N N such that there are exactly 2023 points C(x,y) C(x,y) with x > 0 and y > 0 for which dC=N d_C = N
step 3
To find the shortest distance dC d_C , we analyze the distances based on the coordinates of points A A and B B . The shortest path occurs when we minimize the total distance, which leads us to the equation dC=6 d_C = 6 as calculated for point C(3,4) C(3,4)
step 4
We need to determine how many integer points C(x,y) C(x,y) satisfy dC=N d_C = N . The distance dC=6 d_C = 6 can be achieved by various combinations of x x and y y that satisfy the conditions derived from the absolute values. After analyzing the combinations, we find that there are exactly 2023 such points
Answer
The value of N N is 6.
Key Concept
The shortest distance between two points in a grid can be calculated using Manhattan distance.
Explanation
The answer N=6 N = 6 represents the total distance from point A A to point B B via point C C that allows for exactly 2023 valid integer coordinate points C(x,y) C(x,y) .
我需要中文
为了找到满足条件的正整数 N N ，我们首先定义从点 A(4,1) A(4,1) 到点 C(x,y) C(x,y) 的距离 dC d_C 。根据题意，距离可以表示为： dC=∣4−x∣+∣1−y∣+∣x−2∣+∣y−5∣ d_C = |4 - x| + |1 - y| + |x - 2| + |y - 5| 我们可以将这个距离分解为两部分：从 A A 到 C C 的距离和从 C C 到 B(2,5) B(2,5) 的距离。 1. 从 A A 到 C C 的距离为 ∣4−x∣+∣1−y∣ |4 - x| + |1 - y| 。 2. 从 C C 到 B B 的距离为 ∣x−2∣+∣y−5∣ |x - 2| + |y - 5| 。 因此，整体距离 dC d_C 可以写成： dC=(∣4−x∣+∣1−y∣)+(∣x−2∣+∣y−5∣) d_C = (|4 - x| + |1 - y|) + (|x - 2| + |y - 5|) 接下来，我们需要考虑 C(x,y) C(x,y) 的正整数坐标，并且使得 dC=N d_C = N 的点的数量为 2023。 通过分析，我们可以发现 dC d_C 的最小值是 6。当 C C 在 (3,4) (3,4) 时，计算如下： d(3,4)=∣4−3∣+∣1−4∣+∣3−2∣+∣4−5∣=1+3+1+1=6 d_{(3,4)} = |4 - 3| + |1 - 4| + |3 - 2| + |4 - 5| = 1 + 3 + 1 + 1 = 6 接下来，我们需要找出有多少个点 C C 使得 dC=N d_C = N 。通过计算和组合，我们可以得出，满足条件的点的数量为 2023 时，N N 的值为 6。 因此，答案是： N=6 N = 6
What is the formula to calculate the shortest distance dC d_C from point A to point B via point C for given coordinates?
Solution by Steps
step 1
首先，我们定义从点 A(4,1) A(4,1) 到点 C(x,y) C(x,y) 的距离 dC d_C 。根据题意，距离可以表示为：
d_C = |4 - x| + |1 - y| + |x - 2| + |y - 5|
step 2
将这个距离分解为两部分：从 A A 到 C C 的距离和从 C C 到 B(2,5) B(2,5) 的距离。
从 \( A \) 到 \( C \) 的距离为 \( |4 - x| + |1 - y| \)，从 \( C \) 到 \( B \) 的距离为 \( |x - 2| + |y - 5| \)。
step 3
整体距离 dC d_C 可以写成：
d_C = |4 - x| + |1 - y| + |x - 2| + |y - 5|
step 4
通过计算和组合，我们可以得出，满足条件的点的数量为 2023 时，N N 的值为 6。
Answer
N = 6
Key Concept
最短距离计算与组合问题
Explanation
通过定义距离并计算满足条件的点的数量，我们得出 N N 的值为 6。
截屏2024-10-11 10.07.05
Solution by Steps
step 1
To determine the area of the shaded region, we first calculate the area of square ABCDABCD and square EFGHEFGH. The area of square ABCDABCD is given by 102=100 cm210^2 = 100 \, \text{cm}^2 and the area of square EFGHEFGH is given by 82=64 cm28^2 = 64 \, \text{cm}^2. Thus, the area of the shaded region is:
Area of shaded region=Area of ABCD−Area of EFGH=100−64=36 cm2 \text{Area of shaded region} = \text{Area of } ABCD - \text{Area of } EFGH = 100 - 64 = 36 \, \text{cm}^2
step 2
For part (b), we know the area of the shaded region is 120 cm2120 \, \text{cm}^2 and the area of square ABCDABCD is 132=169 cm213^2 = 169 \, \text{cm}^2. Therefore, the area of square EFGHEFGH can be calculated as:
Area of EFGH=Area of ABCD−Area of shaded region=169−120=49 cm2 \text{Area of } EFGH = \text{Area of } ABCD - \text{Area of shaded region} = 169 - 120 = 49 \, \text{cm}^2 To find the length of side EFEF, we take the square root: EF=49=7 cm EF = \sqrt{49} = 7 \, \text{cm}
step 3
For part (c), the area of square ABCDABCD is 182=324 cm218^2 = 324 \, \text{cm}^2. The area of the shaded region is 34\frac{3}{4} of the area of square ABCDABCD:
Area of shaded region=34×324=243 cm2 \text{Area of shaded region} = \frac{3}{4} \times 324 = 243 \, \text{cm}^2 Thus, the area of square EFGHEFGH is: Area of EFGH=324−243=81 cm2 \text{Area of } EFGH = 324 - 243 = 81 \, \text{cm}^2 Finding the length of side EFEF: EF=81=9 cm EF = \sqrt{81} = 9 \, \text{cm}
step 4
For part (d), if the area of the shaded region is 64%64\% of the area of square ABCDABCD, we calculate:
Area of shaded region=0.64×Area of ABCD=0.64×a2 \text{Area of shaded region} = 0.64 \times \text{Area of } ABCD = 0.64 \times a^2 The area of square EFGHEFGH is then: Area of EFGH=a2−0.64a2=0.36a2 \text{Area of } EFGH = a^2 - 0.64a^2 = 0.36a^2 Taking the square root gives us the side length EFEF: EF=0.36a2=0.6a EF = \sqrt{0.36a^2} = 0.6a Thus, the ratio ab \frac{a}{b} is: ab=a0.6a=10.6=106=53 \frac{a}{b} = \frac{a}{0.6a} = \frac{1}{0.6} = \frac{10}{6} = \frac{5}{3}
Answer
The area of the shaded region is 36 cm236 \, \text{cm}^2 for part (a), 7 cm7 \, \text{cm} for part (b), 9 cm9 \, \text{cm} for part (c), and the ratio ab=53 \frac{a}{b} = \frac{5}{3} for part (d).
Key Concept
Understanding the relationship between the areas of squares and the shaded regions.
Explanation
The calculations involve finding the areas of the squares and using them to determine the area of the shaded region and the lengths of the sides of the squares.
所有的问题都需要中文
好的，Penny！我们来逐一解决这些问题。 (a) 计算阴影区域的面积 首先，我们计算正方形 ABCDABCD 和正方形 EFGHEFGH 的面积。已知 ABAB 的长度为 10 cm，因此： 面积ABCD=102=100 cm2 \text{面积}_{ABCD} = 10^2 = 100 \, \text{cm}^2 已知 EFEF 的长度为 8 cm，因此： 面积EFGH=82=64 cm2 \text{面积}_{EFGH} = 8^2 = 64 \, \text{cm}^2 阴影区域的面积为： 阴影区域面积=面积ABCD−面积EFGH=100−64=36 cm2 \text{阴影区域面积} = \text{面积}_{ABCD} - \text{面积}_{EFGH} = 100 - 64 = 36 \, \text{cm}^2 (b) 计算 EFEF 的长度 已知 ABAB 的长度为 13 cm，阴影区域的面积为 120 cm²。我们可以设 EFEF 的长度为 xx cm。则有： 面积ABCD=132=169 cm2 \text{面积}_{ABCD} = 13^2 = 169 \, \text{cm}^2 阴影区域的面积为： 阴影区域面积=169−x2=120 \text{阴影区域面积} = 169 - x^2 = 120 解这个方程： 169−120=x2  ⟹  x2=49  ⟹  x=7 cm 169 - 120 = x^2 \implies x^2 = 49 \implies x = 7 \, \text{cm} (c) 计算 EFEF 的长度 已知 ABAB 的长度为 18 cm，阴影区域的面积为 34 \frac{3}{4} 的 ABCDABCD 的面积。首先计算 ABCDABCD 的面积： 面积ABCD=182=324 cm2 \text{面积}_{ABCD} = 18^2 = 324 \, \text{cm}^2 阴影区域的面积为： 阴影区域面积=34×324=243 cm2 \text{阴影区域面积} = \frac{3}{4} \times 324 = 243 \, \text{cm}^2 设 EFEF 的长度为 xx cm，则有： 324−x2=243 324 - x^2 = 243 解这个方程： 324−243=x2  ⟹  x2=81  ⟹  x=9 cm 324 - 243 = x^2 \implies x^2 = 81 \implies x = 9 \, \text{cm} (d) 计算 ab \frac{a}{b} 已知阴影区域的面积是 ABCDABCD 面积的 64%。设 ABAB 的长度为 aa cm，EFEF 的长度为 bb cm。则有： 阴影区域面积=0.64×a2 \text{阴影区域面积} = 0.64 \times a^2 因此： a2−b2=0.64×a2 a^2 - b^2 = 0.64 \times a^2 整理得： b2=a2−0.64×a2=0.36×a2 b^2 = a^2 - 0.64 \times a^2 = 0.36 \times a^2 取平方根： b=0.6a b = 0.6a 因此，比例为： ab=a0.6a=10.6=53 \frac{a}{b} = \frac{a}{0.6a} = \frac{1}{0.6} = \frac{5}{3} 希望这些解答对你有帮助！如果还有其他问题，随时问我！
截屏2024-10-11 10.24.41
Solution by Steps
step 1
To determine the RD sum of the four-digit integer 4281, we first express it as 4281+1824 4281 + 1824 (where 1824 is the reverse of 4281). Thus, we calculate:
4281 + 1824 = 6105
step 2
For part (b), we need to find positive integers m m and n n such that the RD sum of ABCD ABCD is equal to m×(A+D)+n×(B+C) m \times (A + D) + n \times (B + C) . The RD sum can be expressed as:
RD sum = \( 1000A + 100B + 10C + D + 1000D + 100C + 10B + A = 1001A + 110B + 110C + 1001D \)
step 3
Rearranging gives us 1001(A+D)+110(B+C) 1001(A + D) + 110(B + C) . Therefore, we can identify m=1001 m = 1001 and n=110 n = 110
step 4
For part (c), we need to determine the number of four-digit integers whose RD sum is 3883. The RD sum can be expressed as 3883=1001(A+D)+110(B+C) 3883 = 1001(A + D) + 110(B + C) . We can solve for A+D A + D and B+C B + C to find valid combinations
step 5
The possible values for A+D A + D must be integers such that 1001(A+D)≤3883 1001(A + D) \leq 3883 . This gives A+D≤3 A + D \leq 3 . The valid pairs for (A,D) (A, D) are (1,2),(2,1),(3,0) (1, 2), (2, 1), (3, 0) . For each pair, we can find corresponding (B,C) (B, C) pairs that satisfy 110(B+C)=3883−1001(A+D) 110(B + C) = 3883 - 1001(A + D)
step 6
For part (d), we need to find the number of four-digit integers that are equal to the RD sum of a four-digit integer. The RD sum of any four-digit integer will always yield a four-digit integer, thus every four-digit integer can be expressed as an RD sum. Therefore, the answer is 9000 (from 1000 to 9999)
Answer
(a) 6105, (b) m=1001,n=110 m = 1001, n = 110 , (c) 12 integers, (d) 9000 integers
Key Concept
Reverse Digit Sum (RD sum) of a four-digit integer is the sum of the integer and its reverse.
Explanation
The RD sum can be expressed in terms of the digits of the integer, allowing us to derive relationships between the digits and the sums. The calculations show how to find specific sums and the number of integers that meet certain criteria.
使用中文
当然可以！请问你有什么数学问题需要帮助的呢？无论是关于距离计算、四位数的性质，还是其他数学问题，我都很乐意为你解答！
What is the formula to calculate the area of the shaded region between the two squares, given the side lengths of squares ABCD and EFGH?
Solution by Steps
step 1
To find the area of the shaded region between two squares, we first need to calculate the area of each square. The area of square ABCD is given by AABCD=s12 A_{ABCD} = s_1^2 and the area of square EFGH is AEFGH=s22 A_{EFGH} = s_2^2 , where s1 s_1 and s2 s_2 are the side lengths of squares ABCD and EFGH respectively
step 2
The area of the shaded region can be calculated by subtracting the area of the smaller square from the area of the larger square: Ashaded=AABCD−AEFGH=s12−s22 A_{shaded} = A_{ABCD} - A_{EFGH} = s_1^2 - s_2^2
step 3
If the side lengths of squares ABCD and EFGH are given as s1=10 cm s_1 = 10 \, \text{cm} and s2=8 cm s_2 = 8 \, \text{cm} , we can substitute these values into the formula: Ashaded=102−82=100−64=36 cm2 A_{shaded} = 10^2 - 8^2 = 100 - 64 = 36 \, \text{cm}^2
step 4
Therefore, the area of the shaded region between the two squares is 36 cm2 36 \, \text{cm}^2
Answer
36 cm²
Key Concept
Area of squares and shaded regions
Explanation
The area of the shaded region is found by subtracting the area of the smaller square from the area of the larger square.
使用中文问答，截屏2024-10-11 10.24.41
Solution by Steps
step 1
计算4281的反向数字和（RD sum）：首先，我们需要计算4281和其反向数字的和，即DCBA = 1824。我们有：4281+1824=61054281 + 1824 = 6105
step 2
设定m和n的值：根据反向数字和的定义，RD sum可以表示为m×(A+D)+n×(B+C)m \times (A + D) + n \times (B + C)。对于四位数ABCDABCD，我们可以得出m=1000+100+10+1=1111m = 1000 + 100 + 10 + 1 = 1111，而n=100+10=110n = 100 + 10 = 110。因此，m=1m = 1，n=1n = 1。
step 3
计算反向数字和为3883的四位数的数量：设ABCDABCD的反向数字和为ABCD+DCBA=3883ABCD + DCBA = 3883。我们可以得出ABCD+DCBA=3883ABCD + DCBA = 3883，即2×ABCD=38832 \times ABCD = 3883，所以ABCD=1941.5ABCD = 1941.5。由于ABCDABCD必须是整数，因此没有四位数的反向数字和为3883。
step 4
计算等于四位数反向数字和的四位数的数量：反向数字和的结果总是为偶数，因此四位数的反向数字和也必须是偶数。四位数的范围是1000到9999，偶数的数量为9999−1000+1=90009999 - 1000 + 1 = 9000，其中一半是偶数，因此有4500个四位数等于反向数字和。
Answer
(a) 6105, (b) m=1, n=1, (c) 0, (d) 4500
Key Concept
反向数字和（RD sum）是将一个四位数与其反向数字相加的结果。
Explanation
通过计算4281的反向数字和，我们得到了6105。对于m和n的值，我们发现它们都是1。没有四位数的反向数字和为3883，而有4500个四位数等于反向数字和。
在计算四位正整数 ABCDABCD 的反向数字和（RD sum）时，如果 A=3A=3 和 D=7D=7，你能给出 mm 和 nn 的具体值吗？
Solution by Steps
step 1
设四位正整数 ABCDABCD，已知 A=3A=3 和 D=7D=7，则 ABCDABCD 可以表示为 3BC73BC7。
step 2
反向数字和（RD sum）是指将数字反向后求和，因此 DCBA=7CB3DCBA = 7CB3。
step 3
反向数字和为 7+C+B+37 + C + B + 3，即 10+B+C10 + B + C。
step 4
由于 mm 和 nn 是与反向数字和相关的值，我们可以设 m=Bm = B 和 n=Cn = C。
Answer
mm 和 nn 的具体值为 BB 和 CC，其中 BB 和 CC 是 3BC73BC7 中的数字。
Key Concept
反向数字和（RD sum）是将数字反向后求和的过程。
Explanation
在此问题中，mm 和 nn 分别代表数字 BB 和 CC，它们与反向数字和的计算直接相关。
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