No, it never can be a right triangle.
First, all the angles of an equilateral triangle is 60 degrees because all angles are the same and all the angles add up to 180:
180/3 = 60.
So if it is a right angle it needs one angle 90 degrees which an equilateral triangle can't have,
Answer:
which statements is there meant to be a diagram? ?
ANSWER
D.

EXPLANATION
The given function is

Comparing to the general form of a trigonometric sine function,


The amplitude is absolute value of A


The period is given by



is a shift to the right of 4 units.
and

is an upward shift of 9 units
Therefore the phase shift is (4,9).
The correct answer is D
Answer:
m = -8
Step-by-step explanation:
have a nice day! :)
a) For this question we set h=59 and solve for t, in order to do so we use the general formula for second-degree equations:
![\begin{gathered} t=\frac{-124\pm\sqrt[]{124^2-4(-16)(-40)}}{2(-16)} \\ t=\frac{-124\pm113.21}{-32} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20t%3D%5Cfrac%7B-124%5Cpm%5Csqrt%5B%5D%7B124%5E2-4%28-16%29%28-40%29%7D%7D%7B2%28-16%29%7D%20%5C%5C%20t%3D%5Cfrac%7B-124%5Cpm113.21%7D%7B-32%7D%20%5Cend%7Bgathered%7D)
The height of the object will be 59 feet at t=7.41 seconds and t=0.34 seconds.
b) When the object reaches the ground, h=0 therefore:

Solving for t we get:
![\begin{gathered} t=\frac{-124\pm\sqrt[]{124^2-4(-16)(19)}}{2(-16)} \\ t=\frac{-124\pm\sqrt[]{16592}}{-32}=\frac{-124\pm128.81}{-32} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20t%3D%5Cfrac%7B-124%5Cpm%5Csqrt%5B%5D%7B124%5E2-4%28-16%29%2819%29%7D%7D%7B2%28-16%29%7D%20%5C%5C%20t%3D%5Cfrac%7B-124%5Cpm%5Csqrt%5B%5D%7B16592%7D%7D%7B-32%7D%3D%5Cfrac%7B-124%5Cpm128.81%7D%7B-32%7D%20%5Cend%7Bgathered%7D)
Therefore, since t cannot be negative the solution is t=7.9 seconds.