How much larger is 3 than 0.3?
1 answer:
You might be interested in
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
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:
Ratio = 3 : 2 and value of m = 5.
Step-by-step explanation:
We are given the end points ( -3,-1 ) and ( -8,9 ) of a line and a point P = ( -6,m ) divides this line in a particular ratio.
Let us assume that it cuts the line in k : 1 ratio.
Then, the co-ordinates of P = .
But, = -6
i.e. -8k-3 = -6k-6
i.e. -2k = -3
i.e.
So, the ratio is k : 1 i.e i.e. 3 : 2.
Hence, the ratio in which P divides the line is 3 : 2.
Also, = m where
i.e. m =
i.e. m =
i.e. m =
i.e. m = 5.
Hence, the value of m is 5.