All Answers

#1

+1090

+5

Given the description, it appears to be an GCF problem.

So that means we have to find the greatest common factor common to all 3 volumes:

I'll start by listing factors of the largest number, 1000:

1000

1 * 1000

2 * 500 -> That matches 500, so it's a factor of 500, but it's too large to be a factor of 250

4 * 250 -> Matches 250, so it's a factor of 250.

Having found a factor for both 1000 and 250, we can now check if 500 also has that factor:

500

1 * 500

2 * 250

If I interpreted the problem correctly, 250mL should be the answer.

MathematicianSep 27, 2016

#1

0

First, I think you should have a SOLID understanding of what "rational" and "irrational" words mean. I recommend that you study this carefully at some Math site such as this link:

http://www.mathwarehouse.com/arithmetic/numbers/rational-and-irrational-numbers-with-examples.php

GuestSep 27, 2016

#3

+1904

0

\(700\) as a prime factor

700

/ \

2 350

/ \

2 175

/ \

5 35

/ \

5 7

\({2}^{2}\times{5}^{2}\times7=700\)

.

gibsonj338Sep 27, 2016

#1

0

Possible derivation:
d/dr(1/(log(1+r/100)))
Using the chain rule, d/dr(1/(log(r/100+1))) = d/( du)1/u 0, where u = log(r/100+1) and ( d)/( du)(1/u) = -1/u^2:
= -(d/dr(log(1+r/100)))/(log^2(1+r/100))
Using the chain rule, d/dr(log(r/100+1)) = ( dlog(u))/( du) 0, where u = r/100+1 and ( d)/( du)(log(u)) = 1/u:
= -1/(log^2(1+r/100)) (d/dr(1+r/100))/(r/100+1)
Differentiate the sum term by term and factor out constants:
= -1/((1+r/100) log^2(1+r/100)) d/dr(1)+(d/dr(r))/100
The derivative of 1 is zero:
= -(1/100 (d/dr(r))+0)/((1+r/100) log^2(1+r/100))
Simplify the expression:
= -(d/dr(r))/(100 (1+r/100) log^2(1+r/100))
The derivative of r is 1:
Answer: |= -1/(100 (1+r/100) log^2(1+r/100)) 1