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charle [14.2K]
3 years ago
8

What is the area of a rectangle with vertices at (−3, −1) , (1, 3) , (3, 1) , and (−1, −3) ?

Mathematics
2 answers:
kondaur [170]3 years ago
7 0
The length of the side joining points (-3, -1) and (1, 3) is given by

l= \sqrt{(1+3)^2+(3+1)^2}  \\  \\ = \sqrt{4^2+4^2} = \sqrt{16+16}  \\  \\ = \sqrt{32}

The length of the side joining points (1, 3) and (3, 1) is given by

w= \sqrt{(3-1)^2+(1-3)^2} \\ \\ = \sqrt{2^2+(-2)^2} = \sqrt{4+4} \\ \\ = \sqrt{8}

The area of a rectangle is given by length times width.

Thus, the area of the given rectangle is given by

Area=l\times w \\  \\ = \sqrt{32} \times \sqrt{8}  \\  \\ = \sqrt{32\times8} = \sqrt{256}  \\  \\ =\bold{16} \ units^2
NNADVOKAT [17]3 years ago
7 0
Hiya,

So I did this quiz just now and I have the answers. c:


1.) What is the area of a rectangle with vertices at 

<span><span>(−3, −1)</span><span>(−3, −1)</span></span> , <span><span>(1, 3)</span><span>(1, 3)</span></span> , <span><span>(3, 1)</span><span>(3, 1)</span></span> , and <span><span>(−1, −3)</span><span>(−1, −3)</span></span> ?

Enter your answer in the box. Do not round any side lengths.

Answer: 16 Units


2.) What is the perimeter of the rectangle shown on the coordinate plane, to the nearest tenth of a unit?

Answer: 26.8 Units


3.) What is the area of a triangle with vertices at 

<span>(0, −2)</span> ,  ​ <span>(8, −2)</span> ​ , and ​ <span>(9, 1)</span> ​ ?

Enter your answer in the box.

Answer: 12 Units

4.) <span>What is the perimeter of the triangle shown on the coordinate plane, to the nearest tenth of a unit?
Answer: 21.6 Units

5.) 
</span>What is the perimeter of a polygon with vertices at 

<span>(−2, 1)</span> , ​ <span>(−2, 4)</span> <span>​,  <span>(2, 7)</span> , ​ <span>(6, 4)</span> ​, and <span>(6, 1)</span> ​?</span>

<span>Enter your answer in the box. Do not round any side lengths.​
</span>

<span>Answer: 24 Units</span>

<span>
</span>

<span>I really hope this helps! Let me know if it did... or anyone else that used this info!</span>

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I need answer on this graphing va substitution.Problem 7 I tried to solve it but not sure that’s correct
Bond [772]

Given the system of equations:

\begin{gathered} 2x+9y=27 \\ x-3y=-24 \end{gathered}

To solve it by substitution, follow the steps below.

Step 1: Solve one linear equation for x in terms of y.

Let's choose the second equation. To solve it for x, add 3y to each side of the equations.

\begin{gathered} x-3y=-24 \\ x-3y+3y=-24+3y \\ x=-24+3y \end{gathered}

Step 2: Substitute the expression found for x in the first equation.

\begin{gathered} 2x+9y=27 \\ 2\cdot(-24+3y)+9y=27 \\ -48+6y+9y=27 \\ -48+15y=27 \end{gathered}

Step 3: Isolate y in the equation found in step 2.

To do it, first, add 48 to both sides.

\begin{gathered} -48+15y=27 \\ -48+15y+48=27+48 \\ 15y=75 \end{gathered}

Then, divide both sides by 15.

\begin{gathered} \frac{15y}{15}=\frac{75}{15} \\ y=5 \end{gathered}

Step 4: Substitute y by 5 in the relation found in step 1 to find x.

\begin{gathered} x=-24+3y \\ x=-24+3\cdot5 \\ x=-24+15 \\ x=-9 \end{gathered}

Answer:

x = -9

y = 5

or (-9, 5)

Also, you can graph the lines by choosing two points from each equation, according to the picture below.

7 0
1 year ago
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