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3 years ago

The picture is my problem.

Mathematics

1 answer:

Sidana [21]3 years ago

3 0

Answer:

Step-by-step explanation:

Your line has the wrong slope. Its slope is ½, but the equation tells you that the slope is ¾.

The point (0,2) is correct, but (-4,0) is incorrect. If x=-4 then y=-1, not 0.

Plot (-4,-1), then draw the line passing through (-4,-1) and (0,2).

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3 years ago

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3 years ago

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3 years ago

“Peaches at the Farmer’s Market cost $2.70 per pound.”

koban [17]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

∫<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bxdx%7D%7B%28x%5E%7B2%7D%2B4%29%5E%7B3%7D%20%7D" id="TexFormula1" title="\frac{xdx}{

photoshop1234 [79]

Substitute $u=x^2+4$ and $du=2x\,dx$ . Then the integral transforms to

$\displaystyle \int \frac{x\,dx}{(x^2+4)^3} = \frac12 \int \frac{du}{u^3}$

Apply the power rule.

$\displaystyle \int \frac{du}{u^3} = -\dfrac1{2u^2} + C$

Now put the result back in terms of $x$ .

$\displaystyle \int \frac{x\,dx}{(x^2+4)^3} = \frac12 \left(-\dfrac1{2u^2} + C\right) = -\dfrac1{4u^2} + C = \boxed{-\dfrac1{4(x^2+4)^2} + C}$

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2 years ago

The picture is my problem.​

The picture is my problem.