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2 years ago

Help me plsssssssssssssssss

Mathematics

1 answer:

asambeis [7]2 years ago

8 0

Answer:

I would say 1.5in

Step-by-step explanation:

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A unicycle wheel has a diameter of 30 inches. How many inches will the unicycle travel in 5 revolutions? Use π = 3.14 and round

miv72 [106K]

Answer:

471.3inches

Step-by-step explanation:

Given data

Diameter= 30in

Radius= D/2= 30/2= 15in

The circumference is the same as the distance covered in one revolution

C= 2πr

Hence the distance covered in 5 revolutions is

C= 2*5*πr

Substitute

C=10*3.142*15

C= 471.3inches

3 0

2 years ago

What is the difference between -120 and 400?

Mice21 [21]

Answer:

-120 is a negative and 400 is a positive

Step-by-step explanation:

also there is a 520 difference

5 0

2 years ago

Please answer this correctly without making mistakes

marysya [2.9K]

Answer:

A digit that makes this sentence true is 4.

Step-by-step explanation:

Since the first digit in the number to the left is 3, you simply have to find a digit greater than 3. Here are the possibilities:

and

Out of any of these you can choose, I chose 4.

4 0

2 years ago

(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 :

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