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Andrew [12]
3 years ago
9

In a circle, an angle measuring 2π radians intercepts an arc of length 16π. Find the radius of the circle in simplest form.

Mathematics
1 answer:
sdas [7]3 years ago
7 0
I believe if i am understanding the question right the radius=4
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1 hour and 35 minutes!! :)

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Please tell me the answer! I'm crying in confusion! :(
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Well, if 14 ounces of the drink has 10 and a half teaspoons of sugar, and the question asks how much sugar is in 1 ounce, the equation would be, 14 / 10.5 = 1. (repeating) 3 so...  1 and 1/3 teaspoons of sugar. I hope that helps! Please mark as brainliest! :)
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consider a wire 2 ft long cut into two pieces. one piece forms a circle with radius r and the other forms a square with side len
makkiz [27]

The formula for the radius r in terms of x . and for the maximum areas is x=2/\pi+4

Given that,

y forms a circle of radius r

y=2\pir

r=y/2\pi

(2-y)- forms Square Side x

(2-y) = 4x

x=(2-y)/4

Now Sum of Area's=Area of Square +Area of Circle

Sum = \pir² + x²

Substitute the r and x values in above equation,

A(y)= y²/4\pi+(y-2)²/ 16

To maximize Area A(y)

A'(y)= 0

2y/4\pi + 2(y-2)/16 =0

y/2\pi + (y-2)/8 =0

y = 2\pi/\pi+4

Y max will be max, x to be maximum.

for maximum sum of areas,

x=2/\pi+4

Hence,The formula for the radius r in terms of x . and for the maximum areas is x=2/\pi+4

Learn more about Area :

brainly.com/question/28642423

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4 0
2 years ago
PLZ HELP It's homework: Compound Probability (Without Replacement) 1.) A box of chocolates contains six milk chocolates and six
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Answer:

c 1/4

Step-by-step explanation:

5 0
3 years ago
PLEASE HELP! 20 POINTS 1) A ball is thrown starting at a time of 0 and a height of 2 meters. The height of the ball follows the
drek231 [11]

Answer:

1) The height of the ball from 0 to 5  seconds are;

At t = 0 second, height = 2

At t = 1 second, height h = 22.1

At t = 2 seconds, height h = 32.4

At t = 3 seconds, height h = 32.9

At t = 4 seconds, height h = 23.6

At t = 5 seconds, height h = 4.5

2)  The correct option is;

D. -16·t² + 25·t + 1

Step-by-step explanation:

1) The equation of motion of the ball is given as follows;

H(t) = -4.9·t² + 25·t + 2

The height of the ball from 0 to 5 seconds are;

H(0) = -4.9×(0)² + 25×(0) + 2 = 2

H(1) = -4.9×(1)² + 25×(1) + 2 = 22.1

H(2) = -4.9×(2)² + 25×(2) + 2 = 32.4

H(3) = -4.9×(3)² + 25×(3) + 2 = 32.9

H(4) = -4.9×(4)² + 25×(4) + 2 = 23.6

H(5) = -4.9×(5)² + 25×(5) + 2 = 4.5

Therefore, we have;

The height of the ball are

At t = 0 second, height = 2

At t = 1 second, height h = 22.1

At t = 2 seconds, height h = 32.4

At t = 3 seconds, height h = 32.9

At t = 4 seconds, height h = 23.6

At t = 5 seconds, height h = 4.5

2) Given that the equation of the ball is that of a projectile motion, such as follows;

h = h₀ + v₀·sin(θ₀)·t - 1/2·g·t² which is equivalent to h = -1/2·g·t²+ h₀+v₀·sin(θ₀)·t

it is best represented by the quadratic equation of an upside down parabola which is option D. -16·t² + 25·t + 1

6 0
3 years ago
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