The height at t seconds after launch is
s(t) = - 16t² + V₀t
where V₀ = initial launch velocity.
Part a:
When s = 192 ft, and V₀ = 112 ft/s, then
-16t² + 112t = 192
16t² - 112t + 192 = 0
t² - 7t + 12 = 0
(t - 3)(t - 4) = 0
t = 3 s, or t = 4 s
The projectile reaches a height of 192 ft at 3 s on the way up, and at 4 s on the way down.
Part b:
When the projectile reaches the ground, s = 0.
Therefore
-16t² + 112t = 0
-16t(t - 7) = 0
t = 0 or t = 7 s
When t=0, the projectile is launched.
When t = 7 s, the projectile returns to the ground.
Answer: 7 s
Answer:
8 or 9
Step-by-step explanation:
So, you first need to take 4/5 of 400 (320). Then, you can count by 75s for a bit
75students; 2chaperones
150; 4
225; 6
300; 8
now, you have 20 students left over...since one chaperone would watch about 37 kids, you can either divide the remaining kids up among the other chaperones, or you can add one more chaperone to watch the 20 kids. i don't know what the teacher wants, so either 8 or 9 chaperones
Mean: 52.2 because its the average of all the numbers
Range: 2 because (60-58)
Median: 53 because its in the middle of all the numbers
Mode : None
Answer:
m = 10 , n = 5
Step-by-step explanation:
using the cosine and tangent ratio in the right triangle and the exact values
cos45° =
and tan45° = 1 , then
cos45° =
=
=
( cross- multiply )
m = 5
×
= 5 × 2 = 10
-----------------------------------------
tan45° =
=
= 1 , then
n = 5
Answer:
a. 14 is a coefficient, and b is a variable.
This can be expanded to 14*b
b. 30 is a coefficient, and j and k are variables.
This can be expanded to 30*j*k