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34kurt
3 years ago
15

Quadrilateral ABCD is rotated 270° clockwise about the origin. What is the orientation of the rotated figure, and what quadrant

does it lie in?

Mathematics
1 answer:
Basile [38]3 years ago
3 0

Answer:

the second quadrant

Step-by-step explanation:

the second quadrant because it is in the 4th. If you rotate it 180 degrees then 90 it ends up being in the second quadrant.

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I honestly don’t understand this ::))
GrogVix [38]

Answer:

Answer is 100 sq m

Step-by-step explanation:

A = a^{2}

A = 10^{2}

A = 100 sq m

HOPE IT HELPS

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7 0
3 years ago
Here are 10 test scores: 30, 74, 76, 77, 78, 79, 80, 80, 82, 84. The mean of these scores is 74.
creativ13 [48]
We add up all those numbers (excluding 30) and divide by the number of numbers (9). This results in 78.8 (round up to 79).

79-74 = 5. So the mean has increased by 5 points. (C)
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3 years ago
SHOW YOUR WORK, <br><br>PLEASE HELP!!!!!!!!!
Shkiper50 [21]
To find the inverse, we swap the variables y and x, then solve for the new y.

3a. y=\frac{3}{x-1}

Swapping the variables: x=\frac{3}{y-1}
Solving for y: x(y-1)=3 \\ y-1= \frac{3}{x} \\ y=1+\frac{3}{x}
The domain of this inverse is x ≠ 0.
3b. y=x^2-1

Swapping: x = y^2 - 1
Solving for y: y^2 = x + 1 \\ y = \sqrt{x+1}
The domain of this inverse is x ≥ -1.
3c. y=\sqrt[3]{\frac{x-7}{3}}
Swapping: x=\sqrt[3]{\frac{y-7}{3}}
Solving for y: x^3=\frac{y-7}{3} \\ y-7=3x^3 \\ y=3x^3+7
The domain of this inverse is all real numbers.
4a. y=\frac{3}{x-1}, y=1+\frac{3}{x}
y=\frac{3}{(1+\frac{3}{x})-1} \\ y=\frac{3}{(\frac{3}{x})} \\ y=x
y=1+\frac{3}{(\frac{3}{x-1})} \\ y = 1+(x-1) \\ y = x

4c. y=\sqrt[3]{\frac{x-7}{3}}, y=3x^3+7
y=\sqrt[3]{\frac{(3x^3+7)-7}{3}} \\ y=\sqrt[3]{\frac{3x^3}{3}} \\ y=\sqrt[3]{x^3} \\ y=x
y=3(\sqrt[3]{\frac{x-7}{3}})^3+7 \\ y = 3({\frac{x-7}{3}})+7 \\ y = (x-7)+7 \\ y=x



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3 years ago
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Answer:

a,b,c,e

Step-by-step explanation:

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