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2 years ago

I don't understand can someone help me PLZ

Mathematics

2 answers:

Tpy6a [65]2 years ago

6 0

The answer is 1/2. Count how many a out of the total.

Inessa05 [86]2 years ago

4 0

Answer:

1/2

Step-by-step explanation:

the circle has 8 pieces so that's how you get the denominator

you count how many A's there are and that is your numerator

so now 4/8 and reduce to get a half

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Pls help It’s due tomorrow question is in the picture

Slav-nsk [51]

Answer:

Step-by-step explanation:

Let's find the area of each of the two circles.

First Circle:

The first circle has a diameter of 10 inches. That means the radius is 5 inches. So, its area is:

$A=\pi r^2\\A=\pi (5)^2\\A=25\pi$

Second Circle:

The second circle has a diameter twice that of the first. So, the diameter is 20 inches. This means that the radius is 10 inches. Find the area:

$A=\pi r^2\\A=\pi (10)^2\\A=100\pi$

Now, find the ratio between them by dividing:

$\frac{\text{Smaller}}{\text{Larger}}=\frac{25\pi}{100\pi}$

Simplify:

$\frac{\text{Smaller}}{\text{Larger}}=\frac{1}{4}$

1/4 is the same as 1:4

So, our answer is C :)

5 0

3 years ago

The teacher said estimate this with out counting 1 by 1 then divide it by 20 HELP PLEASSEE

WARRIOR [948]

Well, first fin dyour best estimate mine is...600
Then after you got your best estimate divide that my 20
600/20 = 30
Hope this helped :D

7 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 :

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