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:
see explanation
Step-by-step explanation:
(5)
Since ∠EBA = 90° then ∠ABD = 90° ( straight angle ) and
∠ABC + ∠CBD = ∠ABD, that is
2x + 3x - 10 = 90, simplifying
5x - 10 = 90 ( add 10 to both sides )
5x = 100 ( divide both sides by 5 )
x = 20, thus
∠ABC = 2x = 2 × 20 = 40°
∠CBD = 3x - 10 = (3 × 20) - 10 = 60 - 10 = 50°
(6)
4x - 18 = 3x + 7 ( vertical angles are congruent )
Subtract 3x from both sides
x - 18 = 7 ( add 18 to both sides )
x = 25
7y = 5y + 28 ( vertical angles are congruent )
Subtract 5y from both sides
2y = 28 ( divide both sides by 2 )
y = 14
Answer:

Step-by-step explanation:






The answer is A. 0
i hope this helps!
Answer:
x = 3
Step-by-step explanation:
Here, we want to find the value of x
By mathematical convention;
NM = NL + LM
Now, substitute individual values
3x + 13 = (6x-5) + (2x + 3)
3x + 13 = 6x-5 + 2x + 3
3x + 13 = 8x -2
Collect like terms
8x -3x = 13 + 2
5x = 15
x = 15/5
x = 3