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4 years ago

HELPPP!!!! 10 PTS!! Find the missing information x = _ m angle A =

Mathematics

2 answers:

7nadin3 [17]4 years ago

5 0

Answer:

170/8

Step-by-step explanation:

the steps are in the pic

Rudiy274 years ago

5 0

Good morning,

Answer:

x = (180 - 2× m∠A)/2 = 90 - m∠A = 90 - (3x+5)

then x = 85-3x

then 4x = 85

then x = 85/4.

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A parallelogram is shown below:

joja [24]

Answer:

part A. area = L X H

9 x 8/9 = 72/9 = 8 square feet

Part B. Draw a line diagonally either between Point A & C or B & D, this would make 2 triangles.

the area of the triangle would be half of the full area so 8 / 2 = 4 feet square

Step-by-step explanation:

3 0

4 years ago

Why are exponents used in mathematics? Please help!! 10 points!!

Oksi-84 [34.3K]

Hi There!

Exponents are used in math so its easier to multiply. For example if you have to do 8 * 8 * 8 * 8 * 8 * 8 * 8 * 8, then to make it easier to multiply you could change into the exponent form $8^{8}$ that is what exponents are used for in math

Hope This Helps :)

4 0

4 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 :

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