Given:
To find:
The quadrant of the terminal side of 
and find the value of 
.
Solution:
We know that,
In Quadrant I, all trigonometric ratios are positive.
In Quadrant II: Only sin and cosec are positive.
In Quadrant III: Only tan and cot are positive.
In Quadrant IV: Only cos and sec are positive.
It is given that,
Here cos is positive and sine is negative. So, 
must be lies in Quadrant IV.
We know that,
It is only negative because lies in Quadrant IV. So,
After substituting , we get
Therefore, the correct option is B.
Answer:
4m3+18m2−34m+12
Step-by-step explanation:
=(4m+−2)(m2+5m+−6)
=(4m)(m2)+(4m)(5m)+(4m)(−6)+(−2)(m2)+(−2)(5m)+(−2)(−6)
=4m3+20m2−24m−2m2−10m+12
(0,-3)
(2,0)
(4,3)
These are your answers
Answer:
h(d) = (17/3249)(-d² +114d)
Step-by-step explanation:
For this purpose, it is convenient to translate and scale a quadratic parent function so it has the desired characteristics. We can start with the function ...
f(x) = 1 -x² . . . . . . . has zeros at x = ±1 and a vertex at (0, 1)
We want to horizontally expand this function by a factor of 57, so we can replace x by x/57. We want to vertically scale it by a factor of 17, so the vertex is at (0, 17). Finally, we want to translate the function 57 m to the right, which requires replacing x with x-57. After these transformations, we have ...
f(x) = 17(1 -((x-57)/57)²) = (17/3249)(-x²+114x)
Using the appropriate function name and variable, we have ...
h(d) = (17/3249)(-d² +114d)
