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goldenfox [79]
3 years ago
11

Help pls me !!!!!!!!!!

Mathematics
2 answers:
Fittoniya [83]3 years ago
8 0
You have to multiply 12 by 18. Then you get 216.
bearhunter [10]3 years ago
3 0
The students = 18/1 x 12
the students = 18 x 12
the students = 216

There are 216 students attending the school
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What’s the answer to this?
sattari [20]

Answer:

14.42

Step-by-step explanation:

6 0
3 years ago
A triangle has ∠A, ∠B, and ∠C.<br> ∠B is 20° more than ∠A.<br> ∠C is double ∠B.<br> How big is ∠A?
san4es73 [151]

Answer:

30°

Step-by-step explanation:

∠A + ∠B + ∠C = 180

∠B = ∠A + 20

∠C = 2 * ∠B

<A + (∠A + 20) + (2 * ∠B) = 180

<A + (∠A + 20) + (2 * (∠A + 20)) = 180

<A + (∠A + 20) + 2∠A + 40 = 180

4∠A + 60 = 180

4∠A = 120

∠A = 30

5 0
3 years ago
Read 2 more answers
Factor completely 5c5 + 60c4 + 180c3.
Mrac [35]
5c^5 + 60c^4 + 180c^3

find the GCF, 5c³

5c³(5c^5 + 60c^4 + 180c^3/ 5c^3)

5c³(c² + 12c + 36)

5c³(c² + 2(c)(6) + 6²)

5c³(c + 6)² <<< the answer.

hope this helps, God bless!
8 0
3 years ago
How many times farther is Mars than Venus from the sun?
Alexandra [31]

Answer:

75,000,000

Step-by-step explanation:

To find how much father Mars is to Venus to the sun you just subtract the two numbers.

7 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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