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3 years ago

Find the sum. 12x2 + 9x2 = 21x2 21x4 22x2 22x4

1 answer:

8 0

First, we need to find what 12*2 is.
12*2=24
Next,we need to find out what 9*2 is.
9*2=18
24=18=42.
Finally, we need to find out which of the following equals 42.
21*2=42, 21*4=84, 22*2=44, and 22*4=88
21*2=42,so that is your answer.
Hope this helped

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Carpet, padding, and installation costs $23.23 per sq. yd. how much would it cost to carpet both bedrooms.

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$46.46 for both rooms to have carpet

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4 years ago

Solve for x 5x - 10 = -10

Zigmanuir [339]

Answer:

the answer is x = 0

Step-by-step explanation:

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3 years ago

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4 years ago

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

The distance between points (5,9) and (-7,-7)

ankoles [38]

You need to find the change in x and y

change in x = 12
change in y = 16

to find the distance you use a^2+ b^2 = c^2
or in this case x^2 + y^2 = d^2

144+256 = 400

$\sqrt{400}$

=20
so distance equals 20

I hope this helps!

4 0

3 years ago