It is known that a displacement S = (2lg5,1) is produced by a common point force F1 = (LG2, LG2) and F2 = (lg5, LG2) If it is known that a displacement S = (2lg5,1) is produced by a common point force F1 = (LG2, LG2), F2 = (lg5, LG2), then the work done by the common point force on the object is: a.lg2 b.lg5 c.1d. Please do it mathematically.)

Resultant force F = F1 vector + F2 vector = (LG2 + lg5, LG2 + LG2) = (1, 2lg2)
The point product of resultant force and displacement = f * s = (1, 2lg2) * (2lg5, 1) = 1 * 2lg5 + 2lg2 * 1 = 2lg5 + 2lg2 = 2
Choose D

Through observation, write the sequence 1,2 / / 3,1 / 2,2 / 5 A general term formula of

2/(n+1)

3 / 4 of a number is equal to the product of 20 and 4

20×4÷（3/4）=320/3

Write 546 in the form ()

3X2X7X13=546

A necessary and sufficient condition for function f (x) = {[radical (1-x ^ 2)] / (x-3)} - m to have zeros?

F (x) = √ (1-x ^ 2) / (x-3) - M has zero √ (1-x ^ 2) / (x-3) = m has real number solution √ (1-x ^ 2) = m (x-3) has real number solution y = √ (1-x ^ 2) and y = m (x-3) image has intersection function y = √ (1-x ^ 2) (- 1 ≤ x ≤ 1,0 ≤ y ≤ 1) x ^ 2 + y ^ 2 = 1 (- 1 ≤ x ≤ 1,0 ≤ y ≤ 1) image is the upper part of circle x ^ 2 + y ^ 2 = 1 (...)

What is the law between the groups of 0 58 17 24 37?

1 ^ 2-1 = 0 2 ^ 2 + 1 = 5 3 ^ 2-1 = 8 4 ^ 2 + 1 = 17 5 ^ 2-1 = 24 6 ^ 2 + 1 = 37 should be this rule

Given that the sum of two natural numbers is 50, the greatest common divisor is 5, find the least common multiple of two numbers

50÷5=10
10=1+9=3+7
So the two numbers are 5 and 45 or 15 and 35
So the least common multiple is 45 or 105

Can the sliding friction formula F = μ n be used for static friction

The maximum static friction is a fixed value, but the maximum static friction coefficient μ and sliding friction coefficient μ are generally different. Generally, when solving the problem, the topic will tell that the maximum static friction can be replaced by sliding friction, In this case, the formula of sliding friction can be applied directly

(): 4 = fraction = 0.125 = ()% = (): () = fraction

( 0.5 ):4=1/8=0.125=( 12.5)%=( 12.5 ):( 100 )=1/8
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The sum of the largest divisor and the second largest divisor of a natural number is 111______ ．

Because 111 is odd, and odd = odd + even, the maximum divisor and the second approximation of the number must be an odd even. The maximum divisor of a number is itself. If a number has an even divisor, the number must be even, and the second approximation of an even number should be 12 of the even number