Prove that a regular polygon has 5 sides if its exterior angle is 2/3 of its interior angle

first of all assume that the values of the interior and exterior angles of the polygon are x.

Therefore interior angle is x and exterior angle is 2/3 x.

We know that the interior angle plus the exterior angle add up to 180 degrees hence

x+2/3 x=180

solve for the value of x

multiply everything by 3

3x+2x=540

add the like terms

5x=540

then simplify

x=108.

hence one interior angle is 108 degrees.

in order to find the value of one interior angle we use the formula:

(2n-4)90

n

hence:

(2n-4)90

=108.

n

open the brackets:

180n-360

=108

n

multiply all sides by the denominator to eliminate it from the equation:

180n-360=108n.

collect like terms together:

180n-108n=360.

subtract 108 from 180:

180n-108n=72n.

hence 72n=360.

simplify the equation to get the value of n:

72n=360

divide both sides by 72

72n 360

=

72 72

hence n=5

38.5

step-by-step explanation:

lets say, first number = r

second number = 3r+7

total = 49

(r)+(3r+7)= 49

4r+7 = 49

4r = 42

r = 42/4

r= 10.5

3r+7 = 3(10.5)+7 = 38.5

hope this

set each factor equal to 0

4x-3 = 0

2x-1 = 0

solve for x.

4x-3=0

4x=3

x= 3/4

solve for x. 

2x-1= 0

2x= 1

x= 1/2

eventually after a few more steps, your answer should be 1/2 is less than or equal to x and 3/4 is greater than or equal to x