All categories

3 years ago

when full the gas tank of a car holds 15 gallons. it now contains 12 gallons. what percent represents how full the tank is?

Mathematics

1 answer:

Masteriza [31]3 years ago

5 0

Hello!

to find the percent of a whole, use this equation here:

(remaining) / (total) = X

X * 100 = %

in this case:

12 / 15 = 0.8

0.8 * 100 = 80

your answer is 80%

I hope this helps, and have a nice day!

You might be interested in

Plot the points A(−2,−3) , B(−3,0) , C(3,2) , and D(4,−1) . What type of quadrilateral is ABCD ? Justify your answer using mid-p

Alika [10]

Obtuse quadrilaateral. midpoint(-1,3) with slooe 1/2
distance 2

6 0

3 years ago

Find the measure of z.

liraira [26]

I think it would be 120

8 0

3 years ago

Read 2 more answers

Is (-4x+9) squared a perfect square

kaheart [24]

Answer:

+ 6x + 9 ...so x2 + 6x + 9 is a perfect square trinomial. ... The first term, 16x2, is the square of 4x, and the last term, 36, is the square of 6. (4x)2 – 48x + 62.

Step-by-step explanation:

5 0

3 years ago

A 1000-liter (L) tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flo

djverab [1.8K]

Answer:

a) $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) $31690.7 g/L$

Step-by-step explanation:

By definition, we have that the change rate of salt in the tank is $\frac{dy}{dt}=R_{i}-R_{o}$ , where $R_{i}$ is the rate of salt entering and $R_{o}$ is the rate of salt going outside.

Then we have, $R_{i}=80\frac{L}{min}*50\frac{g}{L}=4000\frac{g}{min}$ , and

$R_{o}=40\frac{L}{min}*\frac{y}{500} \frac{g}{L}=\frac{2y}{25}\frac{g}{min}$

So we obtain. $\frac{dy}{dt}=4000-\frac{2y}{25}$ , then

$\frac{dy}{dt}+\frac{2y}{25}=4000$ , and using the integrating factor $e^{\int {\frac{2}{25}} \, dt=e^{\frac{2t}{25}$ , therefore $(\frac{dy }{dt}+\frac{2y}{25}}=4000)e^{\frac{2t}{25}$ , we get $\frac{d}{dt}(y*e^{\frac{2t}{25}})= 4000 e^{\frac{2t}{25}$ , after integrating both sides $y*e^{\frac{2t}{25}}= 50000 e^{\frac{2t}{25}}+C$ , therefore $y(t)= 50000 +Ce^{\frac{-2t}{25}}$ , to find $C$ we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions $y(0)=10$ , so $10= 50000+Ce^{\frac{-0*2}{25}}$

$10=50000+C\\C=10-50000=-49990$

Finally we can write an expression for the amount of salt in the tank at any time t, it is $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) The tank will overflow due Rin>Rout, at a rate of $80 L/min-40L/min=40L/min$ , due we have 500 L to overflow $\frac{500L}{40L/min} =\frac{25}{2} min=t$ , so we can evualuate the expression of a) $y(25/2)=50000-49990e^{\frac{-2}{25}\frac{25}{2}}=50000-49990e^{-1}=31690.7$ , is the salt concentration when the tank overflows

4 0

3 years ago

Please help me with this!

morpeh [17]

Answer:

Step-by-step explan

6 0

3 years ago