A ) h = -16 t² + 135 t + 76
Let : h = 0
0 = - 16 t² + 135 t + 76
B ) t 1/2 = (-b+/- √ ( b² - 4 ac ) / ( 2 a )
t 1/2 = (-135 - √(18,225 + 4,864))/ (-32) = ( - 135 - 151.95) / (- 32)=
= (-286.95) / (- 32) = 9.967 ≈ 9.0 s ( other solution is negative )
Answer:
2) 0 = -16 t² + 135 t + 76; 9 s
The correct answer to this question is 240
From the image i've seen, the RDC has an angle of 120. So the major arc DC (that passes A and B) has an angle of 240.
So this makes that the measure of the angle of DAC is 240 because of the angle formed by RDC which is 120.
Problem 1
x = measure of angle N
2x = measure of angle M, twice as large as N
3(2x) = 6x = measure of angle O, three times as large as M
The three angles add to 180 which is true of any triangle.
M+N+O = 180
x+2x+6x = 180
9x = 180
x = 180/9
x = 20 is the measure of angle N
Use this x value to find that 2x = 2*20 = 40 and 6x = 6*20 = 120 to represent the measures of angles M and O in that order.
<h3>Answers:</h3>
- Angle M = 40 degrees
- Angle N = 20 degrees
- Angle O = 120 degrees
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Problem 2
n = number of sides
S = sum of the interior angles of a polygon with n sides
S = 180(n-2)
2700 = 180(n-2)
n-2 = 2700/180
n-2 = 15
n = 15+2
n = 17
<h3>Answer: 17 sides</h3>
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Problem 3
x = smaller acute angle
3x = larger acute angle, three times as large
For any right triangle, the two acute angles always add to 90.
x+3x = 90
4x = 90
x = 90/4
x = 22.5
This leads to 3x = 3*22.5 = 67.5
<h3>Answers:</h3>
- Smaller acute angle = 22.5 degrees
- Larger acute angle = 67.5 degrees
4 and 2 are the porportions
F(1) = (1)^3 + 2(1)^2 - 5(1) - 6 = 1 + 2 - 5 - 6 = -8; x = 1 is not a solution to the polynomial.
f(-1) = (-1)^3 + 2(-1)^2 - 5(-1) - 6 = -1 + 2 + 5 - 6 = 0; x = -1 is a solution to polynomial.
From the options, a or c has x = -1, both has x = , so lets check for x = 3
f(3) = (3)^3 + 2(3)^2 - 5(3) - 6 = 27 + 18 - 15 - 6 = 24; x = 3 is not a solution to the polynomial.
Therefore the solutions to the polynomial are x = -3, x = -1, x = 2
