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1 year ago

the radius of a cylindrical construction pipe is 3 ft . If the pipe is 25ft long , what is it’s volume ?

Mathematics

1 answer:

ki77a [65]1 year ago

7 0

Given:

radius of a cylindrical construction pipe is 3 ft

lenghtis 25ft long

Required:

what is it’s volume

Explanation:

volume of cylinder:

$\begin{gathered} V=\pi r^2h \\ \\ here\text{ h is the length of the pipe} \\ \\ V=3.14\times(3)^2\times25 \\ \\ V=3.14\times9\times25 \\ \\ V=707\text{ ft}^3 \end{gathered}$

Required answer:

$V=707\text{ ft}^3$

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. Orlando needs to fix windows in his house. He must buy 3 feets of wood,

ira [324]

Answer:(7x3)+(4x24)=117

Step-by-step explanation:

Since he needs 3 feet of wood and it is 7 dollar per foot, you need to multiply 7 by 3. Since he needs 4 pieces of glass and its 24 dollars per glass multiply 4 by 24. Add the products. Ur done!

4 0

2 years ago

Read 2 more answers

(x^2y+e^x)dx-x^2dy=0

klio [65]

It looks like the differential equation is

$\left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0$

Check for exactness:

$\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x$

As is, the DE is not exact, so let's try to find an integrating factor µ(x, y) such that

$\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0$

*is* exact. If this modified DE is exact, then

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}$

We have

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu$

Notice that if we let µ(x, y) = µ(x) be independent of y, then ∂µ/∂y = 0 and we can solve for µ :

$x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}$

The modified DE,

$\left(e^{-x}y + \dfrac1{x^2}\right) \,\mathrm dx - e^{-x}\,\mathrm dy = 0$

is now exact:

$\dfrac{\partial\left(e^{-x}y+\frac1{x^2}\right)}{\partial y} = e^{-x} \\\\ \dfrac{\partial\left(-e^{-x}\right)}{\partial x} = e^{-x}$

So we look for a solution of the form F(x, y) = C. This solution is such that

$\dfrac{\partial F}{\partial x} = e^{-x}y + \dfrac1{x^2} \\\\ \dfrac{\partial F}{\partial y} = e^{-x}$

Integrate both sides of the first condition with respect to x :

$F(x,y) = -e^{-x}y - \dfrac1x + g(y)$

Differentiate both sides of this with respect to y :

$\dfrac{\partial F}{\partial y} = -e^{-x}+\dfrac{\mathrm dg}{\mathrm dy} = e^{-x} \\\\ \implies \dfrac{\mathrm dg}{\mathrm dy} = 0 \implies g(y) = C$

Then the general solution to the DE is

$F(x,y) = \boxed{-e^{-x}y-\dfrac1x = C}$

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3 years ago

List the factors of 48 and 54

sergeinik [125]

48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48

54: 1, 2, 3, 6, 9, 18, 27, 54

gcf: 6

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3 years ago

Y=3/8x−4 write in standard form

Naddik [55]

Answer:

y=3/8x-4
Move the expression to the left
y-3/8x=-4
Multiply both sides by 8
8y-3x=-32
Reorder the terms
-3x+8y=-32
Change the signs and standard form is...
3x-8y=32

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3 years ago

Read 2 more answers

Please help with this

GalinKa [24]

The answer is 20

Reason: so the W is 4 and the L is 6 so you do 4+4=8 then you do 6+6=12 then you add the two together and get 20

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3 years ago