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3 years ago

What is15 qt/h= gal min

Mathematics

2 answers:

Brilliant_brown [7]3 years ago

5 0

we have

$15 \frac{qt}{h}$

we know that

1 quarter is equal to 0.25 gallon

1 hour is equal to 60 minutes

convert $\frac{qt}{h}$ to $\frac{gal}{min}$

$15 \frac{qt}{h} =15* \frac{0.25}{60} = 0.0625\frac{gal}{min}$

therefore

the answer is

$0.0625\frac{gal}{min}$

Goryan [66]3 years ago

5 0

Answer:

15 quart per hour = 0.0625 gal/min

Step-by-step explanation:

Consider the provided information.

We need to convert 15 quart per hours in gallon per min.

240 quart per hour = 1 gal/min

Divide both sides by 240.

240/240 quart per hour = 1/240 gal/min

1 quart per hour = 1/240 gal/min

Multiply both the side by 15 as shown.

1×15 quart per hour = 1/240 × 15 gal/min

15 quart per hour = 1/16 gal/min

15 quart per hour = 0.0625 gal/min

Hence, 15 quart per hour = 0.0625 gal/min

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3 years ago

Read 2 more answers

A cable runs along the wall from C to P at a cost of $4 per meter, and straight from P to M at a cost of $5 per meter. If M i

a_sh-v [17]

Answer:

The minimum cost of installing the cable is $156.

Step-by-step explanation:

We have an optimization problem.

We have to minimize the cost of the cable.

We will use the variable x to express the the length of cable CP and PM, accordingly to the attache picture.

The length of the cable that goes from C to P (let's call it CP) is x.

$\bar{CP}=x$

Then, the length of the cable that goes from P to M (PM) can be calcualted usign the Pithagorean theorem:

$\bar{PM}=\sqrt{(33-x)^2+9^2}$

The cost function Y is:

$Y=4*\bar{CP}+5*\bar{PM}=4x+5\sqrt{(33-x)^2+9^2}$

To optimize this cost funtion we have to derive and equal to 0:

$\dfrac{dY}{dx}=0\\\\\\\dfrac{dY}{dx}=4+5(\dfrac{1}{2})((33-x)^2+9^2)^{-1/2} *(-2)(33-x)\\\\\\\dfrac{dY}{dx}=4+5\dfrac{x-33}{\sqrt{(33-x)^2+81}}=0\\\\\\\dfrac{x-33}{\sqrt{(33-x)^2+81}}=-\dfrac{4}{5}\\\\\\(x-33)=-\dfrac{4}{5}\sqrt{(33-x)^2+81}\\\\\\(x-33)^2=(-\dfrac{4}{5})^2[(x-33)^2+81]\\\\\\(x-33)^2=\dfrac{16}{25}(x-33)^2+\dfrac{1296}{25}\\\\\\\dfrac{25-16}{25} (x-33)^2=\dfrac{1296}{25}\\\\\\9(x-33)^2=1296\\\\\\x-33=\sqrt{\dfrac{1296}{9}}=\sqrt{144}=\pm12\\\\\\x=33\pm12\\\\\\x_1=33-12=21\\\\x_2=33+12=45$

The valid solution is x=21, as x can not phisically larger than 33.

The cost then becomes:

$Y=4*\bar{CP}+5*\bar{PM}=4x+5\sqrt{(33-x)^2+9^2}\\\\\\Y=4*21+5\sqrt{(33-21)^2+81}\\\\Y=81+5\sqrt{144+81}\\\\Y=81+5\sqrt{225}\\\\Y=81+5*15\\\\Y=81+75\\\\Y=156$