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3 years ago

8

Solve the equation P=4s for s

2 answers:

Bess [88]3 years ago

6 0

Solve for s.

p=4s

Flip the equation.

4s=p

Divide both sides by 4.

4s/4=p/4

s=1/4p

Andrej [43]3 years ago

3 0

Answer:

solve for s.

p=4s

Flip the equation.

4s=p

Divide both sides by 4.

4s/4=p/4

s=1/4p

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One container is filled with mixture that is 25% acid. A second container is filled with a mixture that is 55% acid. The second

algol13

Answer:

Therefore the third container contains $\frac{735}{17}\%$ = 43.23 % acid.

Step-by-step explanation:

Given that, One container is filled with the mixture that 25% acid. 55% acid is contain by second container.

Let the volume of first container be x cubic unit.

Since the volume of second container is 55% larger than the first.

Then the volume of the second container is

$= (V\times \frac{100+55}{100})$ cubic unit.

$=\frac{155}{100}V$ cubic unit.

The amount of acid in first container is

$=V\times \frac{25}{100}$ cubic unit.

$=\frac{25}{100}V$ cubic unit.

The amount of acid in second container is

$=(\frac{155}{100}V \times \frac{55}{100})$ cubic unit.

$=\frac{8525}{10000}V$ cubic unit.

Total amount of acid $=(\frac{25}{100}V+\frac{8525}{10000}V)$ cubic unit.

$=(\frac{2500+8525}{10000}V)$ cubic unit.

$=(\frac{11025}{10000}V)$ cubic unit.

Total volume of mixture $=(V+\frac{155}{100}V)$ cubic unit.

$=\frac{100+155}{100}V$ cubic unit.

$=\frac{255}{100}V$ cubic unit.

The amount of acid in the mixture is

$=\frac{\textrm{The volume of acid}}{\textrm{The amount of mixture}}\times 100 \%$

$=\frac {\frac{11025}{10000}V}{\frac{255}{100}V}\times 100\%$

$=\frac{735}{17}\%$

Therefore the third container contains $\frac{735}{17}\%$ acid.

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3 years ago