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GrogVix [38]
3 years ago
11

Help me pls i keep getting 3/2 but thats not the answer

Mathematics
2 answers:
Maksim231197 [3]3 years ago
4 0

\huge =  >  \frac{ {6x}^{2} }{ {y {}^{} }^{3} }

\huge  =  >  \frac{6 \times (4) {}^{2} }{ {(4)}^{3} }

\huge =  >  \frac{6 \times 16}{64}

\huge =  >  \frac{96}{64}

Gemiola [76]3 years ago
3 0

\huge\text{Hey there!}

\large\boxed{\mathsf{\dfrac{6x^2}{y^3}}}\\\\\large\boxed{= \mathsf{\dfrac{6(4^2)}{4^3}}}\\\\\large\boxed{= \mathsf{\dfrac{6(4\times4)}{4\times4\times4}}}\\\\\large\boxed{= \mathsf{\dfrac{96}{16\times4}}}}\\\\\large\boxed{= \mathsf{\dfrac{96}{64}}}\\\large\boxed{\approx \mathsf{\dfrac{96\div32}{64\div32}}}\\\large\boxed{\approx \mathsf{\dfrac{3}{2} \rightarrow \bf \dfrac{96}{64}}}

\huge\text{Thus, your answer is: \boxed{\mathsf{Option\ A. } \  \mathsf{\dfrac{96}{64}}}}\huge\checkmark

\huge\text{Good luck on your assignment \& enjoy}\\\huge\text{your day!}

~\frak{Amphitrite1040:)}

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Answer:

It will take the smaller pipe 12 hours working alone

Step-by-step explanation:

In this question, we are asked to calculate the time it will take for a smaller pipe when working alone to fill a tank if the rate at which a larger one fills the tank is given and the rate at which they both fill when working together is given.

First, let’s represent the the volume to fill with say x cubic feet

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Now, the larger pipe can fill the tank in 6 hours less time. Let’s say the time taken for the smaller y, for the bigger, it definitely would be y-6

Hence rate of bigger will be x/(y-6)

For the smaller, the rate will be x/y

Now when we add both rates together, we have x/4 total rate for both

Let’s do tihis;

x/y + x/(y-6) = x/4

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Cross multiply;

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8y-24 = y^2 -6y

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(y-12)(y-2) = 0

y = 12 or 2

y cannot be 2 because since it take the bigger 6 hours less, 2-6 = -4 is not possible as hours cannot be negative

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