URGENT!! Please help me with this as soon as possible!!
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Answer: Choice B.
The statement is valid because the measure of the vertices located in the center of the pentagon is the quotient of 360 and 5, and the sum of two base angles in the given isosceles triangle is 108.
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Explanation:
Check out the diagram below. I have added x and y such that
x = base angle
y = vertex angle (located adjacent to center of polygon)
The pentagon is sliced up like a cake into 5 equal portions. Each vertex angle is y = 360/5 = 72 degrees.
The two base angles x must add with the vertex angle y to get 180
(angle1)+(angle2)+(angle3) = 180
x+x+y = 180
2x+y = 180
2x+72 = 180
2x = 180-72
2x = 108
This is true for any one of the five triangles. Notice that for angle LMN, we can divide it into LMQ and QMN where Q is the center of the polygon. Both of these angles are x. Since we've shown 2x = 108, we can see that LMN must also be 108 as well.
Choice A is close, but we wouldn't use the exterior angle theorem. Choice B is the better answer.
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Answers
Answer:
see the picture for the answer
Answer:
Step-by-step explanation:
F(x) = 2x - 1; G(x) = 3x + 2;
F[G(x)] - F(x).= F(3x+2) -F(x) = 2(3x+2)-1 -(2x-1) = 6x +4-1-2x+1
F[G(x)] - F(x).= 4x+4
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Answer:
x^2+4x+4=x(x−2)
Step-by-step explanation:
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Answers
move y by itself
7x+5y=20
minus 7x from both sides
7x-7x+5y=20-7x
0+5y=20-7x
5y=20-7x
divide both sides by 5
5y/5=(20-7x)/5
y=(20-7x)/5
y=(20/5)+(-7x/5)
a. An equilateral triangle inscribed in a polygon
b. A square inscribed in a circle
c. A regular pentagon inscribed in a polygon
d. A regular hexagon inscribed in a polygon
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Answers
x³ + 3x² - 3x² + 9x - 9x - 27 = 0
x³ + 3x² + 9x - 3x² - 9x - 27 = 0
x(x²) + x(3x) + x(9) - 3(x²) - 3(3x) - 3(9) = 0
x(x² + 3x + 9) - 3(x² + 3x + 9) = 0
(x - 3)(x² + 3x + 9) = 0
x - 3 = 0 or x² + 3x + 9 = 0
+ 3 + 3 x = -(3) ± √((3)² - 4(1)(9))
x = 3 2(1)
x = -3 ± √(9 - 36)
2
x = -3 ± √(-27)
2
x = -3 ± 3i√(3)
2
x = -1.5 ± 1.5i√(3)
x = -1.5 + 1.5i√(3) or x = -1.5 - 1.5i√(3)
In each step I have to factor the equation to a trinomial so that it could be factored into the sums and differences of cubes. Then, I have to use the quadratic equation in order to find the two solutions of the second factor, which is equal to -1.5 + 1.5i√(3) and -1.5 - 1.5i√(3), while on the first factor I have to find the solution to the first factor, making the solution equal to 3. So the solutions to the polynomial equation are the numbers 3, -1.5 + 1.5i√(3), and -1.5 - 1.5i√(3).
Solution Set: {3, -1.5 ± 1.5i√(3)}