There is many answers to this question but angle B and C have to add up to 140 so angle B and C could both be 70 that should work
2 1/3 / 1 3/5 7/3 / 8/5 7/3 * 5/8 7*5 = 35 3*8 = 24 35 / 24 1 and 11/24 1 and 11/24 1 and 11/24
Hope this helps!
<span><u>Answer </u>
A) 〖128.6〗^o
<u>Explanation </u>
The question requires us to find the interior angle of a regular heptagon.
To do this first calculate the exterior angle of that polygon.
The sum of exterior angles is 360o. A heptagon has 7 sides.
So, one exterior angle = 〖360〗^o/7=〖51.4〗^o
interior angle+exterior angle=〖180〗^o
exterior=180-51.4=〖128.6〗^o
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<h3>Answer: 1.15</h3>
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Work Shown:
QR = 73
QP = 55
PR = x
Pythagorean Theorem
a^2 + b^2 = c^2
(QP)^2 + (PR)^2 = (QR)^2
(55)^2 + (x)^2 = (73)^2
3025 + x^2 = 5329
x^2 = 5329-3025
x^2 = 2304
x = sqrt(2304)
x = 48
PR = 48
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tan(angle) = opposite/adjacent
tan(R) = QP/PR
tan(R) = 55/48
tan(R) = 1.1458333 approximately
tan(R) = 1.15
Problem
For a quadratic equation function that models the height above ground of a projectile, how do you determine the maximum height, y, and time, x , when the projectile reaches the ground
Solution
We know that the x coordinate of a quadratic function is given by:
Vx= -b/2a
And the y coordinate correspond to the maximum value of y.
Then the best options are C and D but the best option is:
D) The maximum height is a y coordinate of the vertex of the quadratic function, which occurs when x = -b/2a
The projectile reaches the ground when the height is zero. The time when this occurs is the x-intercept of the zero of the function that is farthest to the right.
