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3 years ago

A hose fills a tank at a rate of 20 gallons per minute.

Mathematics

2 answers:

AfilCa [17]3 years ago

7 0

Answer:

$\frac{4548}{1}$ liters per hour.

Step-by-step explanation:

Given : A hose fills a tank at a rate of 20 gallons per minute.

To find : What is the rate in liters per hour?

Solution : We have given

A tank can fill hose in 1 minute = 20 gallon .

1 gallon = 3.79 liters.

20 gallon = 20 * 3.79 = 75.8 liters.

20 gallon = 75 .8 liters.

So,

A tank can fill hose in 1 minute = 75.8 liters.

A tank can fill hose in 60 minute = 75.8 * 60 liters = 4548 liters.

A tank can fill hose in 1 hour = 4548 liters.

Rate = $\frac{4548}{1}$ liters per hour.

Therefore, $\frac{4548}{1}$ liters per hour.

bija089 [108]3 years ago

6 0

First, you need to know how many gallons will be in the tank after 1 hour. So:

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Read 2 more answers

1+-w2+9w and I need help cuz I’m on 76 and I’m sooo close help

Gnesinka [82]

$\huge \boxed{\mathfrak{Question} \downarrow}$

Simplify :- 1 + - w² + 9w.

$\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}$

$\large \sf1 + - w ^ { 2 } + 9 w$

Quadratic polynomial can be factored using the transformation $\sf \: ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$ , where $\sf x_{1} and x_{2}$ are the solutions of the quadratic equation $\sf \: ax^{2}+bx+c=0$ .

$\large \sf-w^{2}+9w+1=0$

All equations of the form $\sf\:ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\sf\frac{-b±\sqrt{b^{2}-4ac}}{2a}$ . The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$\large \sf \: w=\frac{-9±\sqrt{9^{2}-4\left(-1\right)}}{2\left(-1\right)} \\$

Square 9.

$\large \sf \: w=\frac{-9±\sqrt{81-4\left(-1\right)}}{2\left(-1\right)} \\$

Multiply -4 times -1.

$\large \sf \: w=\frac{-9±\sqrt{81+4}}{2\left(-1\right)} \\$

Add 81 to 4.

$\large \sf \: w=\frac{-9±\sqrt{85}}{2\left(-1\right)} \\$

Multiply 2 times -1.

$\large \sf \: w=\frac{-9±\sqrt{85}}{-2} \\$

Now solve the equation $\sf\:w=\frac{-9±\sqrt{85}}{-2}$ when ± is plus. Add -9 to $\sf\sqrt{85}$ .

$\large \sf \: w=\frac{\sqrt{85}-9}{-2} \\$

Divide -9+ $\sf\sqrt{85}$ by -2.

$\large \boxed{ \sf \: w=\frac{9-\sqrt{85}}{2}} \\$

Now solve the equation $\sf\:w=\frac{-9±\sqrt{85}}{-2}$ when ± is minus. Subtract $\sf\sqrt{85}$ from -9.

$\large \sf \: w=\frac{-\sqrt{85}-9}{-2} \\$

Divide $\sf-9-\sqrt{85}$ by -2.

$\large \boxed{ \sf \: w=\frac{\sqrt{85}+9}{2}} \\$

Factor the original expression using $\sf\:ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$ . Substitute $\sf\frac{9-\sqrt{85}}{2}$ for $\sf\:x_{1}$ and $\sf\frac{9+\sqrt{85}}{2}$ for $\sf\:x_{2}$ .

$\large \boxed{ \boxed {\mathfrak{-w^{2}+9w+1=-\left(w-\frac{9-\sqrt{85}}{2}\right)\left(w-\frac{\sqrt{85}+9}{2}\right) }}}$

Well, in the picture you inserted it says that it's 8th grade mathematics. So, I'm not sure if you have learned simplification with the help of biquadratic formula. So, if you want the answer simplified only according to like terms then your answer will be ⇨

$\large \sf \: 1 + - w {}^{2} + 9w \\ =\large \boxed{\bf \: 1 - {w}^{2} + 9w}$

This cannot be further simplified as there are no more like terms (you can use the biquadratic formula if you've learned it.)

4 0

2 years ago

Myra is a scientist and needs 45 gallons of a 16% acid solution. The lab is currently stocked only with a 20% acid solution and

asambeis [7]

Answer:

The volume of the solution with 20% acid is 27 gallons and the one with 10% acid is 18 gallons

Step-by-step explanation:

Myra needs to mix "x" gallons of the solution with 20% and "y" gallons of the solution with 10%. The volume of the final solution must be 45 gallons, therefore:

x + y = 45

The concentration of acid of the final solution is:

0.2*x + 0.1*y = 45*0.16

0.2*x + 0.1*y = 7.2

Therefore we have a system of equation:

x + y = 45

0.2*x + 0.1*y = 7.2

We need to multiply the first equation by -0.1:

-0.1*x -0.1*y = -4.5

0.2*x + 0.1*y = 7.2

We now sum both equation:

0.1*x = 2.7

x = 2.7/0.1 = 27 gallons

y = 45 - 27 = 18 gallons

6 0

3 years ago