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3 years ago

Simplify 15(5b+3) Please help me I don't understand at all

1 answer:

3 0

$15(5b+3) \\ \\ 15 \times 5b + 15 \times 3 \ / \ distribute \\ \\ 75b + 15 \times 3 \ / \ simplify \\ \\ 75b + 45 \ / \ simplify \\ \\$

The final result is: 75b + 45.

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ABCD is a parallelogram. If side AB = 12x - 7, side BC = 5x + 7, and side CD = 7x + 18, find the value of x.

fredd [130]

Answer:

Step-by-step explanation:

AB = CD. so 12x - 7 = 7x +18. so x=5

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3 years ago

2\times 3x\times y\times 4z\times 2x\times y\times z

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3 years ago

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Find three consecutive even integers such that the product of the second and

Rina8888 [55]

Step-by-step explanation:

first number=x

second number=x+2

third number=x+4

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using that, you can find the first number,

then use the data to find the other two.

7 0

3 years ago

If a – 2 = -12, then a = ______.

nata0808 [166]

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3 years ago

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(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 :

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