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3 years ago

7

Working together, two pumps can drain a certain pool in 6 hours. If it takes the older pump 15 hours to drain the pool by itself

, how pump to drain the pool on its own?

1 answer:

stellarik [79]3 years ago

5 0

Answer:

10

Step-by-step explanation:

Rate of drainage for pump A and B = [1 pool drained]/[6 hours]

Rate of drainage for pump A (older pump) = [1 pool drained]/[15 hours]

Rate for pump B (newer pump) = [1 pool drained]/[t hours]

Thus, adding the rate of drainage of A and B,

1/t + 1/15 = 1/6

t=10

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Step-by-step explanation:

$\frac{1}{2} ^{x-4}$ - 3 = $4^{x-3}$ - 2

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$\frac{1}{2} ^{x-4}$ - 3 = $(2^{2})^{x-3}$ - 2

Next, convert $\frac{1}{2} ^{x-4}$ to base 2:

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Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

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$2^{-1*(x-4)}$ - 3 = $2^{2(x-3)}$ - 2

Apply exponent rule: $a^{b+c}$ = $a^{b}$ $a^{c}$ :

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$2^{-1 * x}$ * $2^{4}$ - 3 = $2^{2x}$ * $2^{-6}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$2^{-1x}$ = $(2^{x})^{-1}$ , $2^{2x}$ = $(2^{x})^{2}$

$(2^{x})^{-1}$ * $2^{4}$ - 3 = $(2^{x})^{2}$ * $2^{-6}$ - 2

Rewrite the equation with $2^{x} = u$ :

$(u)^{-1}$ * $2^{4}$ - 3 = $(u)^{2}$ * $2^{-6}$ - 2

Solve $u^{-1}$ * $2^{4}$ - 3 = $u^{2}$ * $2^{-6}$ - 2:

$u^{-1}$ * $2^{4}$ - 3 = $u^{2}$ * $2^{-6}$ - 2

Refine:

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$\frac{16}{u}$ - 3 + 3 = $\frac{1}{64}$ $u^{2}$ - 2 + 3

Simplify:

$\frac{16}{u}$ = $\frac{1}{64}$ $u^{2}$ + 1

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Simplify:

$\frac{16}{u}$ * 64u = $\frac{1}{64}$ $u^{2}$ + 1 * 64u

Simplify $\frac{16}{u}$ * 64u:

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Substitute:

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Solve for u:

u = 8

Substitute back u = $2^{x}$ :

8 = $2^{x}$

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