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stira [4]
4 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]4 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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Find the difference between 7 1/2 and 2 3/4
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Answer:

4\frac{3}{4} or 4.75

Step-by-step explanation:

first we're going to convert the number into improper fractions.

7\frac{1}{2}= \frac{15}{2} and 2\frac{3}{4}= \frac{11}{4}

For subtracting it is suitable to adjust both fractions to an equal denominator, so we're going to multiply \frac{15}{2}x2, it'll equal the same number.

So now we have \frac{30}{4}-\frac{11}{4}, 30-11= 19.

\frac{19}4} simplified is, 4\frac{3}{4}.

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I hope this helps you!

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Step-by-step explanation:

6 0
3 years ago
Find an equation of the plane that contains the points p(5,−1,1),q(9,1,5),and r(8,−6,0)p(5,−1,1),q(9,1,5),and r(8,−6,0).
topjm [15]
Given plane passes through:
p(5,-1,1), q(9,1,5), r(8,-6,0)

We need to find a plane that is parallel to the plane through all three points, we form the vectors of any two sides of the triangle pqr:
pq=p-q=<5-9,-1-1,1-5>=<-4,-2,-4>
pr=p-r=<5-8,-1-6,1-0>=<-3,5,1>

The vector product pq x pr gives a vector perpendicular to both pq and pr.  This vector is the normal vector of a plane passing through all three points
pq x pr
=
  i   j   k
-4 -2 -4
-3  5  1
=<-2+20,12+4,-20-6>
=<18,16,-26>

Since the length of the normal vector does not change the direction, we simplify the normal vector as
N = <9,8,-13>

The required plane must pass through all three points.
We know that the normal vector is perpendicular to the plane through the three points, so we just need to make sure the plane passes through one of the three points, say q(9,1,5).

The equation of the required plane is therefore
Π :  9(x-9)+8(y-1)-13(z-5)=0
expand and simplify, we get the equation
Π  :  9x+8y-13z=24

Check to see that the plane passes through all three points:
at p: 9(5)+8(-1)-13(1)=45-8-13=24
at q: 9(9)+8(1)-13(5)=81+9-65=24
at r: 9(8)+8(-6)-13(0)=72-48-0=24
So plane passes through all three points, as required.

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3 years ago
When we divide 2√3 by 2√3 it gives 3 how ?(2√3/2√3)
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Only if the individual doing the calculation makes a substantial mistake.

Any fraction with the same quantity in the numerator and denominator
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3 years ago
Read 2 more answers
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