Well, +10 and -10 can do that job.
D(t)=-16t²+32t+88
a] Maximum height
At maximum height d'(t)=0
from d(t);
d'(t)=-32t+32=0
thus
t=1
hence the maximum height will be:
d(t)=-16t²+32t+88
d(1)=-16(1)^2+32(1)+88
d(1)=104 ft
2] <span> how many seconds does it take to reach the ground after first being thrown upward?
Here we solve for t intercept
</span>when d(t)=0
<span>thus</span>
0=-16t²+32t+88
<span>solving for t, we get:
t=1-</span>√13/2
or
<span>t=1+</span>√13/2
<span>
t=-1.55 or t=3.55 sec
Thus the answer is t=3.55 sec
</span>
Answer:
It'd help to show the graph but, count out how far it goes right, than count up how far it goes up, the first number is x, the second number is y
Step-by-step explanation:
9514 1404 393
Answer:
(i) ∠CDF = 86°
(ii) ∠BAD = 141°
Step-by-step explanation:
(i) Angles CDF and GFD are "alternate interior" angles, so are congruent.
∠CDF = 86°
__
(ii) Angles BAD and CDA are "alternate interior angles, so are congruent. Angle CDA is the sum of angles CDH and HDA. Angle CDH is supplementary to angle CDF, so is ...
∠CDH = 180° -∠CDF = 180° -86° = 94°
Then ...
∠CDA = ∠CDH +∠HDA = 94° +47°
∠CDA = 141°