150 cubic centimeters

A hose is leaking water at a rate of 2 oz/h. How many cups of water will leak out after 2 weeks?

Over [174]

Answer:

84 cups of water will leak out after 2 weeks

Step-by-step explanation:

* Lets how many how to change oz to cup

- One cup contains 8 oz

- We can use the ratio to solve this problem

- A hose is leaking water at a rate of 2 oz/h

- We need to find how many cups of water will leak out after 2 weeks

∵ The rate of the water leak out is 2 oz/hr

∵ 1 cup contains 8 oz

∴ 1 oz = $\frac{1}{8}$ cup

∴ 2 oz = 2 × $\frac{1}{8}$ = $\frac{2}{8}$ = $\frac{1}{4}$ cup

∴ The rate of the water leak out = $\frac{1}{4}$ cup/h

∵ 1 day = 24 hours

∵ 1 week = 7 days

∴ 1 week = 7 × 24 = 168 hours

∵ The number of hours in 1 week is 168

∴ The number of hours in 2 weeks = 168 × 2 = 336 hours

⇒ cups : hours

⇒ $\frac{1}{4}$ : 1

⇒ x : 336

- By using cross multiplication

∴ x = 336 × $\frac{1}{4}$

∴ x = 84

- x represents the number of cups of water leak out

<em>84 cups of water will leak out after 2 weeks</em>

5 0

3 years ago

Can someone help me please

Trava [24]

Answer:

5n units

Step-by-step explanation:

C=2πr =2·π·2.5

5 0

2 years ago

2x-32=12<br> means they spent how much

Hoochie [10]

Answer:

22

Step-by-step explanation:

6 0

2 years ago

In this problem we consider an equation in differential form Mdx+Ndy=0. (4x+2y)dx+(2x+8y)dy=0 Find My= 2 Nx= 2 If the problem is

zheka24 [161]

Answer:

$f(x,y)=2x^2+4y^2+2xy=C_1\\\\Where\\\\y(x)=\frac{1}{4} (-x\pm \sqrt{-7x^2+C_1} )$

Step-by-step explanation:

Let:

$M(x,y)=4x+2y\\\\and\\\\N(x,y)=2x+8y$

This is and exact equation, because:

$\frac{\partial M(x,y)}{\partial y} =2=\frac{\partial N}{\partial x}$

So, define f(x,y) such that:

$\frac{\partial f(x,y)}{\partial x} =M(x,y)\\\\and\\\\\frac{\partial f(x,y)}{\partial y} =N(x,y)$

The solution will be given by:

$f(x,y)=C_1$

Where C1 is an arbitrary constant

Integrate $\frac{\partial f(x,y)}{\partial x}$ with respect to x in order to find f(x,y):

$f(x,y)=\int\ {4x+2y} \, dx =2x^2+2xy+g(y)$

Where g(y) is an arbitrary function of y.

Differentiate f(x,y) with respect to y in order to find g(y):

$\frac{\partial f(x,y)}{\partial y} =2x+\frac{d g(y)}{dy}$

Substitute into $\frac{\partial f(x,y)}{\partial y} =N(x,y)$

$2x+\frac{dg(y)}{dy} =2x+8y\\\\Solve\hspace{3}for\hspace{3}\frac{dg(y)}{dy}\\\\\frac{dg(y)}{dy}=8y$

Integrate $\frac{dg(y)}{dy}$ with respect to y:

$g(y)=\int\ {8y} \, dy =4y^2$

Substitute g(y) into f(x,y):

$f(x,y)=2x^2+4y^2+2xy$

The solution is f(x,y)=C1

$f(x,y)=2x^2+4y^2+2xy=C_1$

Solving y using quadratic formula:

$y(x)=\frac{1}{4} (-x\pm \sqrt{-7x^2+C_1} )$

4 0

3 years ago

Which statement is true about the equations –3x + 4y = 12 and 1/4x –1/3 y = 1?The system of the equations has exactly one soluti

iren [92.7K]

Answer:

The system of the equations has no solution; the two lines are parallel.

Step-by-step explanation:

The equations have the same slope. In standard form, this can be seen as the same coefficients for x and y. We will multiply the second equation by -12 to reveal this.

$\frac{1}{4} x-\frac{1}{3}y=1 \\-12(\frac{1}{4} x-\frac{1}{3}y=1)\\\frac{-12}{4} x-\frac{-12}{3}y=-12\\-3x+4y=-12$

This means the equations are parallel and will never cross. There is no solution.

4 0

2 years ago