I really need the answer to number 4
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Answer : option D
The rate of change is not constant and increases then decreases over time. The height of the ball above ground gets larger until 1.25 seconds and then gets smaller after that time.
Rate of change is nothing but the slope
To find slope we use formula
Time (seconds) Height (feet) Rate of change
0 -------------------------0
0.5 -----------------------16 --------------------
1 ----------------------24 --------------------
1.25 -----------------------25--------------------
1.5 -------------------------24 --------------------
2 ----------------------------16 --------------------
2.5 -------------------------0 --------------------
Rate of change is positive till 1.25 seconds and then it becomes negative. From this it is clear that, the rate of change is not constant and increases then decreases over time. The height of the ball above ground gets larger until 1.25 seconds and then gets smaller after that time.
We have that
csc ∅=13/12
sec ∅=-13/5
cot ∅=-5/12
we know that
csc ∅=1/sin ∅
sin ∅=1/ csc ∅------> sin ∅=12/13
sec ∅=-13/5
sec ∅=1/cos ∅
cos ∅=1/sec ∅------> cos ∅=-5/13
sin ∅ is positive and cos ∅ is negative
so
∅ belong to the II quadrant
therefore
<span>the coordinates of point (x,y) on the terminal ray of angle theta are
</span>x=-5
y=12
the answer is
point (-5,12)
see the attached figure
csc ∅=13/12
sec ∅=-13/5
cot ∅=-5/12
we know that
csc ∅=1/sin ∅
sin ∅=1/ csc ∅------> sin ∅=12/13
sec ∅=-13/5
sec ∅=1/cos ∅
cos ∅=1/sec ∅------> cos ∅=-5/13
sin ∅ is positive and cos ∅ is negative
so
∅ belong to the II quadrant
therefore
<span>the coordinates of point (x,y) on the terminal ray of angle theta are
</span>x=-5
y=12
the answer is
point (-5,12)
see the attached figure