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artcher [175]
2 years ago
5

Please help!! Giving brainliest and extra points!

Mathematics
1 answer:
alexgriva [62]2 years ago
6 0

Answer:

radius of circle is 11 cm

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<u><em>The answer is 129  </em></u><em> i hope this helps </em>

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3 years ago
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WILL GIVE BRAINLIEST!! (Look at photo if confused)
andrey2020 [161]

Answer:

1. B. 9

2. A. 4

3. D. 9 × 9 × 9 × 9

4. C. 6,561

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Find the measure of each exterior angle of the polygon.
Karo-lina-s [1.5K]
Since all the inside angles are 60, in order to add up to 180, the outside angles need to be 120.
7 0
2 years ago
The length of a rectangle is 2 times it’s width if the perimeter of the rectangle is 36 inches find it’s length and width
Tamiku [17]

P = Perimeter            

L = Length          

W = Width

Perimeter of rectangle = L + L + W + W        

or P = 2L + 2W

You know:

P = 36 inches

L = 2W          [length is(=) 2 times it's width]

W = ?

P = 2L + 2W    

Substitute/plug in what you know, plug in 2W for L since L = 2W

36 = 2(2W) + 2W         Simplify

36 = 4W + 2W

36 = 6W        Divide 6 on both sides

6 = W              Now that you know the width, you can find the length:

L = 2W

L = 2(6)

L = 12

L = 12 in

W = 6 in

PROOF

P = 2(12) + 2(6)

P = 24 + 12

P = 36

6 0
3 years ago
How do you do this question?
alina1380 [7]

Answer:

(8√2) / 15

Step-by-step explanation:

A curve bounded by the y-axis is represented by in terms of dy;

\int \:x\:dt

When the curve crosses the y-axis, x will be 0. In this case x is the function of t, so we have to solve for x(t) = 0;

0 = t^2 + 2t --- (1)

Solution(s) => t = 0, t = 2

dy = (1/2 * 1/√t)dt --- (2)

Our solutions (0, 2) are our limits. The area of the curve is in the form A\:=\:\int _b^a\:f\left(t\right)g'\left(t\right)dt , so now let's introduce the limits of integration, x(t) and dy/dt. Remember, dy/dt = (1/2 * 1/√t) (second equation). 1/2 * 1/√t can be rewritten as 1/2 * t^(-1/2)....

A\:=\:\int _2^0\:\left(t^2-2t\right)\left(\frac{1}{2}t^{-\frac{1}{2}}\right)dt\\\\= \int _2^0\:\left(\frac{1}{2}t^{\frac{3}{2}}-t^{\frac{1}{2}}\right)dt\\\\= \left[\frac{t^{\frac{5}{2}}}{5}-\frac{2t^{\frac{3}{2}}}{3}\right]_2^0\\\\= 0\:-\:\left(\frac{4\sqrt{2}}{5}-\frac{4\sqrt{2}}{3}\right)\\\\= \frac{8\sqrt{2}}{15}

Your solution is 8√2 / 15

7 0
2 years ago
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