Answer:
No, because the ratio of pay to hours is not the same for each pair of value.
Answer:
h(t) = -16t2 + 144
h(1) = -16(12) + 144 = 128 ft
h(2) = -16(22) + 144 = 80 ft
h(2) - h(1) = 80 - 128 = -48 ft
It fell 48 ft between t = 1 and t = 2 seconds.
It reaches the ground when h(t) = 0
0 = -16t2 + 144
t = √(144/16) s = 3s
It reaches the ground 3s after being dropped.
Step-by-step explanation:
Assuming 5a3 equals 5a^3, and the same for everything else, you would plug in 4 for a, so 5(4)^3 - 2(4)^2 + 4 - 45, which equals 247.
Answer:
Step-by-step explanation:
Given the formula for calculating the distance travelled expressed as;
s=1/2at^2
Given
a = 3
t = 10
Required
lower bound of s
Substitute the given values into the equation;
s=1/2at^2
S = 1/2(3)(10)^2
S = 1/2 * 3 * 100
S = 3 * 50
S = 150
Hence the lower bound of distance S is 150
