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bija089 [108]
3 years ago
15

What is the area of a parallelogram whose vertices are A(−12, 2) , B(6, 2) , C(−2, −3) , and D(−20, −3) ? Enter your answer in t

he box.
Mathematics
1 answer:
agasfer [191]3 years ago
4 0

Answer:

  \boxed{90}\,\text{units$^2$}

Step-by-step explanation:

The sides AB and CD are parallel to the x-axis and 5 units apart. The length of side AB is 6-(-12) = 18 units. The area is the product of these dimensions:

  Area = (Base)(Height) = (18 units)(5 units) = 90 units²

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Analyze the diagram below and complete the instructions that follow. Find the area of QRS.
11Alexandr11 [23.1K]

<u>Answer-</u>

<em>The area of QRS is </em><em>238.7</em><em> sq units.</em>

<u>Solution-</u>

According to properties of supplementary angles,

m\angle R+124=180\\\\m\angle Q+132=180

So,

\Rightarrow m\angle R=180-124=56,\ m\angle Q=180-132=48

As the sum of measurement of all the angles in a triangle is 180, so

\Rightarrow m\angle S+m\angle R+m\angle Q=180

\Rightarrow m\angle S=180-m\angle R-m\angle Q

\Rightarrow m\angle S=180-56-48=76

Then the area of the triangle will be,

\text{Area}=\dfrac{1}{2}qr\sin S

=\dfrac{1}{2}\times 21\times 23.43\times \sin 76

=238.7\text{sq.units}

8 0
3 years ago
What would my grade be if i had a 76% in the class, then got a 0% on a test worth 70% of my grade?
ExtremeBDS [4]

Answer:

I believe it would be 6

7 0
3 years ago
Please I really need help on this.
Molodets [167]

Answer:

C

Step-by-step explanation:

\sqrt{23}=4.79

3 0
3 years ago
Read 2 more answers
What is the slope of the line that passes through the points (-3, 6) and (1,2)
Nadya [2.5K]

Answer:

the slope is -1

m = -1

Step-by-step explanation:

y2 - y1     y

--------- = ----

x2 - x1     x

= y/x = m

2 - 6

-------

1 - - 3

= -4/4

m = -1

8 0
3 years ago
Evaluate the integral by changing to polar coordinatesye^x dA, where R is in the first quadrant enclosed by the circle x^2+y^2=2
34kurt
\displaystyle\iint_Rye^x\,\mathrm dA=\int_{\theta=0}^{\theta=\pi/2}\int_{r=0}^{r=5}r^2\sin\theta e^{r\cos\theta}\,\mathrm dr\,\mathrm d\theta

which follows from the usual change of coordinates via

\begin{cases}\mathbf x(r,\theta)=r\cos\theta\\\mathbf y(r,\theta)=r\sin\theta\end{cases}

and Jacobian determinant

|\det J|=\left|\begin{vmatrix}\mathbf x_r&\mathbf x_\theta\\\mathbf y_r&\mathbf y_\theta\end{vmatrix}\right|=|r|

Swap the order, so that the integral is

\displaystyle\int_{r=0}^{r=5}\int_{\theta=0}^{\theta=\pi/2}r^2\sin\theta e^{r\cos\theta}\,\mathrm d\theta\,\mathrm dr

and now let \sigma=r\cos\theta, so that \mathrm d\sigma=-r\sin\theta. Now, you have

\displaystyle\int_{r=0}^{r=5}\int_{\sigma=0}^{\sigma=r}re^\sigma\,\mathrm d\sigma\,\mathrm dr=\int_{r=0}^{r=5}r(e^r-1)\,\mathrm dr=4e^5-\dfrac{23}2
7 0
3 years ago
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