You can use photomath it will give you the exact steps
1+7 and 7+1 are the same equations. The numbers are just switched around .
Example:
1+2=3
2+1+3
<span>They add up to the same answer no matter where they are placed, therefore knowing 1+7 helps you find the sum of 7+1 (again, because they are the same) </span>
Answer:
Some number less than 5
Step-by-step explanation:
If it was more we wouldn't need to subtract to find n.
First, we should calculate the areas separately.
4x8=32
5.6 x 4 = 22.4/2 = 11.2
4.5x4=22 22/2=11
4x8=32
32+11+32+11.2
64+22.2=86.2
So 86.2 ft2
Hope this helped. ^^
Answer:
a) 72.25sec
b) 6.25secs
c) after 10.5secs and 2 secs
Step-by-step explanation:
Given the height reached by the rocket expressed as;
s(t)= -4t^2 + 50t - 84
At maximum height, the velocity of the rocket is zero i.e ds/dt = 0
ds/dt = -8t + 50
0 = -8t + 50
8t = 50
t = 50/8
t = 6.25secs
Hence it will reach the maximum height after 6.25secs
To get the maximum height, you will substitute t - 6.25s into the given expression
s(t)= -4t^2 + 50t - 84
s(6.25) = -4(6.25)^2 + 50(6.25) - 84
s(6.25) = -156.25 + 312.5 - 84
s(6.25) = 72.25feet
Hence the maximum height reached by the rocket is 72.25feet
The rocket will reach the ground when s(t) = 0
Substitute into the expression
s(t)= -4t^2 + 50t - 84
0 = -4t^2 + 50t - 84
4t^2 - 50t + 84 = 0
2t^2 - 25t + 42 = 0
2t^2 - 4t - 21t + 42 = 0
2t(t-2)-21(t-2) = 0
(2t - 21) (t - 2) = 0
2t - 21 = 0 and t - 2 = 0
2t = 21 and t = 2
t = 10.5 and 2
Hence the time the rocket will reach the ground are after 10.5secs and 2 secs
