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stira [4]
3 years ago
12

A cricular arena is lit by 5 lights equally spaced around the perimeter of the arena. What is the measure of each angle formed b

y the lights on the perimeter?

Mathematics
1 answer:
Kitty [74]3 years ago
7 0

Answer:

108°

Step-by-step explanation:

Suppose that circle with center A is a circular arena. Points B, C, D, E and F are 5 lights. These 5 points form regular pentagon (because these 5 lights are equally spaced around the perimeter of the arena).

The sum of all interior angles of pentagon can be calculated using following formula

(n-2)\cdot 180^{\circ},\\ \\(5-2)\cdot 180^{\circ}=3\cdot 180^{\circ}=540^{\circ}

All interior angles in regular pentagon are of equal measure, so

\dfrac{540^{\circ}}{5}=108^{\circ}

Thus, the measure of each angle formed by the lights on the perimeter is 108°.

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The Chinese Buffet costs $9.95 a person. Johnny, his mother, and his father have dinner there, They receive a 25% discount. What
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$22.39

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hope it helps :)

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2 years ago
g During winter, red foxes hunt small rodents by jumping into thick snow cover, without any visual clue. Researchers examined th
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Events can be classed as being dependent or independent depending on whether the outcome of a certain event affects the outcome of another. In the scenario above, the success rate of jumps in the Northeast direction is higher than in the southwest direction and even much than in other directions. With these uneven rate of success, the probability that a fox will have a successful jump will depend on the direction in which the fox is jumping. With an high expected success probability in the Northeast and southwest than in other directions

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2 years ago
2 friends share 3 fruit bars equally. How much does each friend get?
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5 0
3 years ago
Read 2 more answers
In a right triangle ABC, CD is an altitude, such that AD=BC. Find AC, if AB=3 cm, and CD= 2 cm.
konstantin123 [22]

Consider right triangle ΔABC with legs AC and BC and hypotenuse AB. Draw the altitude CD.

1. Theorem: The length of each leg of a right triangle is the geometric mean of the length of the hypotenuse and the length of the segment of the hypotenuse adjacent to that leg.

According to this theorem,

BC^2=BD\cdot AB.

Let BC=x cm, then AD=BC=x cm and BD=AB-AD=3-x cm. Then

x^2=(3-x)\cdot 3,\\ \\x^2=9-3x,\\ \\x^2+3x-9=0,\\ \\D=3^2-4\cdot (-9)=9+36=45,\\ \\\sqrt{D}=\sqrt{45}=3\sqrt{5},\\ \\x_1=\dfrac{-3-3\sqrt{5} }{2}0.

Take positive value x. You get

AD=BC=\dfrac{-3+3\sqrt{5} }{2}\ cm.

2. According to the previous theorem,

AC^2=AD\cdot AB.

Then

AC^2=\dfrac{-3+3\sqrt{5} }{2}\cdot 3=\dfrac{-9+9\sqrt{5} }{2},\\ \\AC=\sqrt{\dfrac{-9+9\sqrt{5} }{2}}\ cm.

Answer: AC=\sqrt{\dfrac{-9+9\sqrt{5} }{2}}\ cm.

This solution doesn't need CD=2 cm. Note that if AB=3cm and CD=2cm, then

CD^2=AD\cdot DB,\\ \\2^2=AD\cdot (3-AD),\\ \\AD^2-3AD+4=0,\\ \\D

This means that you cannot find solutions of this equation. Then CD≠2 cm.

8 0
3 years ago
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