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Bumek [7]
3 years ago
5

What is the area of the given circle in terms of pi

Mathematics
1 answer:
ser-zykov [4K]3 years ago
6 0
3.14(3.3)(3.3)
3.13(10.89)

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Alina [70]

a) 2^13 / 2^18

b) 216

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Name the triangles that are classified by angles
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<span>2. scalene, isosceles, equilateral</span>
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On paper, make a rectangle that has points A and B as two of its vertices and has a perimeter of 40 units. Draw and label the tw
Westkost [7]

Answer:

A(x,y),B(x+10,y),C(x+10,y-10) \:and\: D(x,y-10)

Step-by-step explanation:

If a Rectangle has a perimeter of 40 units, then we can write that formula and find the a formula for the coordinates:

\left\{\begin{matrix}w+l=20 & \\ w*l=100 & \end{matrix}\right.\Rightarrow w=20-l\Rightarrow (20-l)l=100\Rightarrow -l^2+20l-100=0\Rightarrow l^2-20l+100\Rightarrow (l-10)(l-10)\Rightarrow l=10\therefore l=10\: and\: w=10

Therefore we can say:

A(x,y),B(x+10,y),C(x+10,y-10) \:and\: D(x,y-10)

Technically when w=lin this rectangle is a square, a degenerate square.

8 0
3 years ago
PLEASE LOOK AT PHOTO! WHOEVER ANSWERS CORRECTLY I WILL MARK BRAINIEST!!
marissa [1.9K]

Answer:

53°

Step-by-step explanation:

It is given that the total measurement of the two angles combined would equate to 116°.

It is also given that m∠WXY is 10° more then m∠ZXY.

Set the system of equation:

m∠1 + m∠2 = 116°

m∠1 = m∠2 + 10°

First, plug in "m∠2 + 10" for m∠1 in the first equation:

m∠1 + m∠2 = 116°

(m∠2 + 10) + m∠2 = 116°

Simplify. Combine like terms:

2(m∠2) + 10 = 116

Next, isolate the <em>variable</em>, m∠2. Note the equal sign, what you do to one side, you do to the other. Do the opposite of PEMDAS.

First, subtract 10 from both sides of the equation:

2(m∠2) + 10 (-10) = 116 (-10)

2(m∠2) = 116 - 10

2(m∠2) = 106

Next, divide 2 from both sides of the equation:

(2(m∠2))/2 = (106)/2

m∠2 = 106/2 = 53°

53° is your answer.

~

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3 years ago
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Define "retirement" as presented in the lesson.
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whenever u get old and feellike it is time to settle down and quit wrking for whatever company you were working for


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