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Ivanshal [37]
2 years ago
13

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

Mathematics
1 answer:
Jobisdone [24]2 years ago
6 0

Using the distance formula, the area of the rectangle is: 16 units².

<h3>What is the Area of a Rectangle?</h3>

Area of a rectangle = (length)(width).

Given the vertices:

  • A(−3, −1)
  • B(1, 3)
  • C(3, 1)
  • D(−1, −3)

Area = (AB)(BC)

Apply the distance formula, d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}, to find AB and BC:

AB = √[(1−(−3))² + (3−(−1))²]

AB = √[(4)² + (4)²]

AB = √32

BC = √[(1−3)² + (3−1)²]

BC = √[(−2)² + (2)²]

BC = √8

Area = (√32)(√8) = √256

Area = 16 units²

Learn more about area of rectangle on:

brainly.com/question/25292087

#SPJ1

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Kitty [74]
1) 3n+1=16-2n - add 2n to both sides - you should now be left with 5n+1=16 (remember the 2n gets canceled out) - subtract 1 from both sides - you should now be left with 5n=15 - now you divide by 5 to get n by itself - 15 divided by 5 is 3 - n=3 2) substitute n with 3 in to the original equation - so 3n+1=16-2n now becomes 3(5)+1=16-2(3) - you should get 10 for both so 10=10 3) according to the segment addition postualte , XR + YR = XY. - add 10+10 which stands for the lengths of XR and YR Answer - XY = 20
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6 0
3 years ago
Can plz help answer number 4
enot [183]

Answer:

It was 6 7/8 pounds of fruit in the fruit salad.

Step-by-step explanation:

Regina bought in total (in pounds) of fruit: B

B=3\frac{1}{2}+4\frac{3}{4}\\ B=\frac{3(2)+1}{2}+\frac{4(4)+3}{4}\\ B=\frac{6+1}{2}+\frac{16+3}{4}\\ B=\frac{7}{2}+\frac{19}{4}\\ B=\frac{7}{2}.1+\frac{19}{4}\\ B=\frac{7}{2}.\frac{2}{2}+\frac{19}{4}\\ B=\frac{7(2)}{2(2)}+\frac{19}{4}\\ B=\frac{14}{4}+\frac{19}{4}\\ B=\frac{14+19}{4}\\ B=\frac{33}{4}

Regina bought in total 33/4 pounds of fruit

She used all (33/4 pounds) but 1 3/8 pounds of the fruit in a fruit salad, then she used in the fruit salad (in pounds): S

S=B-1\frac{3}{8}\\ S=\frac{33}{4}-\frac{1(8)+3}{8}\\ S=\frac{33}{4}-\frac{8+3}{8}\\ S=\frac{33}{4}-\frac{11}{8}\\ S=\frac{33}{4}.1-\frac{11}{8}\\ S=\frac{33}{4}.\frac{2}{2}-\frac{11}{8}\\ S=\frac{33(2)}{4(2)}-\frac{11}{8}\\ S=\frac{66}{8}-\frac{11}{8}\\ S=\frac{66-11}{8}\\ S=\frac{55}{8}\\ S=\frac{48+7}{8}\\ S=\frac{48}{8}+\frac{7}{8}\\ S=6+\frac{7}{8}\\ S=6\frac{7}{8}

It was 6 7/8 pounds of fruit in the fruit salad.

7 0
3 years ago
The average (arithmetic mean) of 0, a, and b is 2a. What is the value of b in terms of a?
kramer
0+a+b
a=2
b=10
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5a is the answer.


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3 years ago
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