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Alex
3 years ago
5

A pump started filling an empty pool with water and continued at a constant rate until the pool was full. At noon the pool was 1

/3 full, and 114114 hours later it was 3/4 full. What was the total number of hours that it took the pump to fill the pool?
A. 213213
B. 223223
C. 3
D. 312312
E. 323
Mathematics
1 answer:
Gemiola [76]3 years ago
5 0

Answer:

3 hours

Step-by-step explanation:

The pool was filled at noon = \frac{1}{3}

After 1\frac{1}{4}=\frac{5}{4} hours the fool was filled = \frac{3}{4}

So,water filled in pool in \frac{5}{4} hours = \frac{3}{4}-\frac{1}{3}

So,water filled in pool in \frac{5}{4} hours = \frac{5}{12}

\frac{5}{12} of pool is filled in hours =  \frac{5}{4}

Whole pool is filled in hours =  \frac{\frac{5}{4}}{\frac{5}{12}}

Whole pool is filled in hours =  3

Hence it took 3 hours the pump to fill the pool .

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Answer:

The option "StartFraction negative 2 (x + 6) Over (x + 4) (x minus 4) EndFraction" is correct

That is \frac{-2(x+6)}{(x+4)(x-4)}

Therefore \frac{x}{x^2-16}-\frac{3}{x-4}=\frac{-2(x+6)}{(x+4)(x-4)}

Step-by-step explanation:

Given problem is StartFraction x Over x squared minus 16 EndFraction minus StartFraction 3 Over x minus 4 EndFraction

It can be written as below :

\frac{x}{x^2-16}-\frac{3}{x-4}

To solve the given expression

\frac{x}{x^2-16}-\frac{3}{x-4}

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

=\frac{x}{(x+4)(x-4)}-\frac{3}{x-4}  ( using the property a^2-b^2=(a+b)(a-b) )

=\frac{x-3(x+4)}{(x+4)(x-4)}

=\frac{x-3x-12}{(x+4)(x-4)} ( by using distributive property )

=\frac{-2x-12}{(x+4)(x-4)}

=\frac{-2(x+6)}{(x+4)(x-4)}

\frac{x}{x^2-16}-\frac{3}{x-4}=\frac{-2(x+6)}{(x+4)(x-4)}

Therefore \frac{x}{x^2-16}-\frac{3}{x-4}=\frac{-2(x+6)}{(x+4)(x-4)}

Therefore the option "StartFraction negative 2 (x + 6) Over (x + 4) (x minus 4) EndFraction" is correct

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Step-by-step explanation:

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x^2 + 8 = 3x + 36

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