Sixth grade volume I mathematics Yangtze River exercise book side 38 I will think about the answer

Sixth grade volume I mathematics Yangtze River exercise book side 38 I will think about the answer


32+(1/5+1/5)=32 2/5



There is a pool. The water surface is a square with a side length of 10 feet. In the center of the pool, there is a new reed, which is 1 foot higher than the water surface. If the reed is pulled vertically to the bank, its top just reaches the water surface of the bank. What is the depth of the pool and the length of the reed?


Let the depth of the pool be x feet, then the length of the reed is (x + 1) feet. According to the meaning of the title, we get: (102) & nbsp; 2 + x2 = (x + 1) 2, and the solution is: x = 12, so the length of the reed is: 12 + 1 = 13 (feet). A: the depth of the pool is 12 feet, and the length of the reed is 13 feet



There is a problem in the ancient mathematics monograph nine chapters arithmetic
Buy 13 pigs with the money of 2 cows and 5 sheep, and the remaining money is 1000; buy 9 sheep with the money of 3 cows and 3 pigs, and the money is just right; buy 5 cows with the money of 6 sheep and 8 pigs, and the difference is 600. How much is the price of each cow, pig and sheep without a head?


Let cattle, sheep and pigs be XYZ
2x+5y=13z+1000
3x+3z=9y
6y+8Z=5x-600
x=1200
y=500
z=300



19999.9 + 1999.9 + 199.9 + 19.9 + 1.9


19999.9+1999.9+199.9+19.9+1.9
=(20000+2000+200+20+2)-0.5
=22222-0.5
=22221.5
Not afraid to move forward, you can also have a brave heart, resist all difficulties! Come on!



A natural number greater than 10 is removed by 90 and 164, and the sum of the two remainder is equal to that of the natural number after 220
If 90 and 164 are removed from a natural number greater than 10, the sum of the two remainder is equal to the remainder after 220 is removed from the natural number. What is the natural number?


The number is 17
(90+164-220)/2=17
checking calculation
90/17=5…… five
164/17=9…… eleven
220/17=12…… sixteen
5+11=16
I think 17 should be the right answer



If the sum of the two residuals obtained after 90 and 164 are removed from a natural number greater than 10 is equal to the residuals obtained after 220 is removed from the natural number, then the number is?


The number is 17



If the quotient of natural number a is 77, what is a and what is the remainder?
Pull the process!


Let 2006 = 77A + R, 0 ≤ R



There is a natural number, which can be used to remove 63, 90130 and have remainder respectively. The sum of the three remainder is 25. What is the largest of the three remainder?


Let this natural number be m, and M remove 63, 90130, and the remainder is a, B, and C respectively, then 63-a, 90-b, and 130-c are all multiples of M. it can be obtained that: (63-a) + (90-b) + (130-c) = 283 - (a + B + C) = 283-25 = 258 is also a multiple of M. and 258 = 2 × 3 × 43. Then it may be 2 or 3 or 6 or 43; a + B + C = 25, so at least one of a, B, and C must be greater than 8 Must be greater than the remainder, can be determined = 43. Thus a = 20, B = 4, C = 1. Obviously, 20 is the largest of the three remainder. Answer: the largest of the three remainder is 20



There is a natural number, which can be used to remove 63, 90130 and have remainder respectively. The sum of the three remainder is 25. What is the largest of the three remainder?


Let this natural number be m, and the remainders obtained by removing 63, 90130 from m are a, B, and C respectively. Then 63-a, 90-b, and 130-c are all multiples of M. we can get that: (63-a) + (90-b) + (130-c) = 283 - (a + B + C) = 283-25 = 258 is also a multiple of M. and 258 = 2 × 3 × 43. Then it may be 2 or 3 or 6 or 43; a +



There is a natural number, which can be used to remove 63, 90130 and have remainder respectively. The sum of the three remainder is 25. What is the largest of the three remainder?


Let this natural number be m, and M remove 63, 90130, and the remainder is a, B, and C respectively, then 63-a, 90-b, and 130-c are all multiples of M. it can be obtained that: (63-a) + (90-b) + (130-c) = 283 - (a + B + C) = 283-25 = 258 is also a multiple of M. and 258 = 2 × 3 × 43. Then it may be 2 or 3 or 6 or 43; a + B + C = 25, so at least one of a, B, and C must be greater than 8 Must be greater than the remainder, can be determined = 43. Thus a = 20, B = 4, C = 1. Obviously, 20 is the largest of the three remainder. Answer: the largest of the three remainder is 20