Answer:
Step-by-step explanation:
h(5) = 5 - 7 = -2
g(-2) = (-2)^2 = 4
12,,389+15
I’m not for sure but used photomath
Answer:
a. [-3, 4]
b. (-inf, -3]
c. [4, inf)
Step-by-step explanation:
Our intervals will represent the x-values
We know that since there's an arrow pointing to the left of the line that it goes on infinitely
Same thing when the arrow is going to the right
Then we can just looking at the x-values on the graph for the intervals where it starts and stops
Hope this helps
Best of luck
If tan theta is -1, we know immediately that theta is in either Quadrant II or Q IV. We need to focus on Q IV due to the restrictions on theta.
Because tan theta is -1, the ray representing theta makes a 45 degree angle with the horiz axis, and a 45 degree angle with the negative vert. axis. Thus the hypotenuse, by the Pythagorean Theorem, tells us that the hyp is sqrt(2).
Thus, the cosine of theta is adj / hyp, or +1 / sqrt(2), or [sqrt(2)]/2
The secant of theta is the reciprocal of that, and thus is
2 sqrt(2)
---------- * ------------ = sqrt(2) (answer)
sqrt(2) sqrt(2)
