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tino4ka555 [31]
3 years ago
15

The four points (−2, 5), (−2, −1), (6, −1), and (3, 5) are the vertices of a polygon. What is the area, in square units, of this

polygon?

Mathematics
2 answers:
Norma-Jean [14]3 years ago
6 0

Answer:

39 square units

Step-by-step explanation:

crimeas [40]3 years ago
5 0
Look at the picture.

The polygon is a right-angled trapezoid.

The area is:
A=\frac{a+b}{2} \times h

The points (-2,-1) and (6,-1) lie on the same horizontal line, so the distance between them is 6-(-2)=6+2=8. The length of a is 8 units.
The points (-2,5) and (3,5) lie on the same horizontal line, so the distance between them is 3-(-2)=3+2=5. The length of b is 5 units.
The points (-2,5) and (-2,-1) lie on the same vertical line, so the distance between them is 5-(-1)=5+1=6. The length of h is 6 units.

A=\frac{8+5}{2} \times 6=\frac{13}{2} \times 6=13 \times 3=39

The area is 39 square units.

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Complementary angles have a sum of 90

32+2x+14=90

2x+46=90

2x=44

x=22 the second angle is 58

2(22)+14=58

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15 is what percent of 70​
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Answer:

21%

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Read 2 more answers
What is the antiderivative of 3x/((x-1)^2)
Maslowich

Answer:

\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C

Step-by-step explanation:

Given

\int \:\:3\cdot \frac{x}{\left(x-1\right)^2}dx

\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx

=3\cdot \int \frac{x}{\left(x-1\right)^2}dx

\mathrm{Apply\:u-substitution:}\:u=x-1

=3\cdot \int \frac{u+1}{u^2}du

\mathrm{Expand}\:\frac{u+1}{u^2}:\quad \frac{1}{u}+\frac{1}{u^2}

=3\cdot \int \frac{1}{u}+\frac{1}{u^2}du

\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx

=3\left(\int \frac{1}{u}du+\int \frac{1}{u^2}du\right)

as

\int \frac{1}{u}du=\ln \left|u\right|     ∵ \mathrm{Use\:the\:common\:integral}:\quad \int \frac{1}{u}du=\ln \left(\left|u\right|\right)

\int \frac{1}{u^2}du=-\frac{1}{u}        ∵     \mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1

so

=3\left(\ln \left|u\right|-\frac{1}{u}\right)

\mathrm{Substitute\:back}\:u=x-1

=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)

\mathrm{Add\:a\:constant\:to\:the\:solution}

=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C

Therefore,

\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C

4 0
3 years ago
What is the volume of the figure below, which is composed of two cubes with side lengths of 9 units
Rufina [12.5K]
9 cubed. times 2 =
1458
6 0
3 years ago
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