The cheetah runs 2 meters/sec faster than the gazelle. Eventually the cheetah will overcome the gazelle. Since distance = rate times time, time = distance/rate.
Thus,
300 meters / (2 meters/sec) = 150 sec = 2 min 30 sec
1.) 16/100 and 0.16
2.) 7/10 and 0.7
3.) 6/10
4.) 73/100
5.) 6 9/10
6.)8 57/100
7.)0.70
8.)0.33
9.)7.20
10.)3.09
11.) 0.80
12.)0.48
13.)0.02
14.)0.55
15.)D
I hope all of these are correct and help
Answer:
1)The rocket hit the ground at 
2)The maximum height of the rocket = 12.468 feet
Step-by-step explanation:
<u><em>Step(i):-</em></u>
Given equation
y = -2 x² + 5 x + 7 ...(i)
Differentiating equation (i) with respective to 'x' , we get

Equating zero

⇒ -4 x +5 =0
⇒ -4 x = -5
⇒
<em> The rocket hit the ground at </em>
<em></em>
<u><em>Step(ii):</em></u>-
...(ii)
Again differentiating equation (ii) with respective to 'x' , we get

The maximum height at x = 
y = -2 x² + 5 x + 7



<em>The maximum height of the rocket = 12.468 feet</em>
Answer:
By AA similarity
Step-by-step explanation:
We have been given that ABCD is a parallelogram
So, by the property of parallelogram AB ||CD and FD is cutting the line BC
Hence, FD is transverse line. In transverse line alternate angles are equal.
Therefore, ∠AFD=∠EDC (alternate interior angles)
And ∠FAD=∠ECD (opposite angles in parallelogram)
Therefore, by AA similarity △ADF∼△CDE