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2 years ago

Will mark brainliest please help

Mathematics

1 answer:

Brut [27]2 years ago

5 0

Answer:

6*10^-9

Step-by-step explanation:

Move the decimal so there is one non-zero digit to the left of the decimal point. The number of decimal places you move will be the exponent on the

. If the decimal is being moved to the right, the exponent will be negative. If the decimal is being moved to the left, the exponent will be positive.

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HELP ME WITH THIS EQUATION! 11 = y – 6 *

Blizzard [7]

Answer:

Step-by-step explanation:

here you go mate'

step 1

11 = y-6 equation

step 2

11+6=y-6+6 add 6 to the sides

answer

y=17

can i get brainliest if you dont mind?

8 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

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

5 0

3 years ago

What is the area of this figure

konstantin123 [22]

Answer:

108

Step-by-step explanation:

add all of the numbers together and thats what you get :)

4 0

2 years ago

Jimmy loaded three bags of sand into his wheelbarrow. The bags weighed 24

Rudik [331]

The whole numbers add up to 100. So now we are left with 1/2 or 0.5, 3/5 or .60, and 1/8 or .125. Those add up to 1.225 so the amount of sand in pounds in the wheelbarrow is 101.225. Hope I helped! If you have any questions just leave a comment.

3 0

2 years ago

Plz answer these i dont know them

Black_prince [1.1K]

Answer:

117, 55, 54, 77, 130, 7h + 5x, 14

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers