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Vitek1552 [10]
3 years ago
11

A . expression A

Mathematics
2 answers:
yan [13]3 years ago
4 0

Answer:

answer choice B.

Step-by-step explanation:

Likurg_2 [28]3 years ago
3 0
B hope this helps. :)
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Squirrels forget where they hide about half of their nuts
Anna35 [415]

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really?

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2 years ago
_____ compare one quantity to 100.
Alexxx [7]

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c

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6 0
3 years ago
Use the Pythagorean theorem to find the unknown side of the right triangle.
Nesterboy [21]

The length of hypotenuse is 39.

Step-by-step explanation:

Given,

Vertical side = a = 15

Horizontal side = b = 36

Hypotenuse = c

Using pythagorean theorem

a^2+b^2=c^2\\(15)^2+(36)^2=c^2\\225+1296=c^2\\1521=c^2\\c^2=1521

Taking square root on both sides

\sqrt{c^2}=\sqrt{1521}\\c=39

The length of hypotenuse is 39.

Keywords: square root, addition

Learn more about square roots at:

  • brainly.com/question/9196410
  • brainly.com/question/9178881

#LearnwithBrainly

8 0
3 years ago
Simplify the expression 14y +6+3y
umka2103 [35]

Answer:

The simplified version is 17y + 6.

Step-by-step explanation:

14y + 6 + 3y

17y + 6

4 0
3 years ago
Read 2 more answers
Solve dis attachment and show all work ( I got it all wrong and I want to know how to solve it )
DedPeter [7]
(a) First find the intersections of y=e^{2x-x^2} and y=2:

2=e^{2x-x^2}\implies \ln2=2x-x^2\implies x=1\pm\sqrt{1-\ln2}

So the area of R is given by

\displaystyle\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\left(e^{2x-x^2}-2\right)\,\mathrm dx

If you're not familiar with the error function \mathrm{erf}(x), then you will not be able to find an exact answer. Fortunately, I see this is a question on a calculator based exam, so you can use whatever built-in function you have on your calculator to evaluate the integral. You should get something around 0.5141.

(b) Find the intersections of the line y=1 with y=e^{2x-x^2}.

1=e^{2x-x^2}\implies 0=2x-x^2\implies x=0,x=2

So the area of S is given by

\displaystyle\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}(2-1)\,\mathrm dx+\int_{1+\sqrt{1-\ln2}}^2\left(e^{2x-x^2}-1\right)\,\mathrm dx
\displaystyle=2\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\mathrm dx

which is approximately 1.546.

(c) The easiest method for finding the volume of the solid of revolution is via the disk method. Each cross-section of the solid is a circle with radius perpendicular to the x-axis, determined by the vertical distance from the curve y=e^{2x-x^2} and the line y=1, or e^{2x-x^2}-1. The area of any such circle is \pi times the square of its radius. Since the curve intersects the axis of revolution at x=0 and x=2, the volume would be given by

\displaystyle\pi\int_0^2\left(e^{2x-x^2}-1\right)^2\,\mathrm dx
5 0
3 years ago
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