The measure of angle DBE is (0.1x - 22) and the measure of angle CBE is (0.3x - 54). The value of x is 415
Step-by-step explanation:
Given that,
The measure of angle DBE (0.1x - 22)
The measure of angle CBE (0.3x - 54)
<u>Step 1</u>
As per the Angle Addition Postulate,
m∠DBE+m∠CBE=m∠DBC
<u>
Step 2</u>
Since∠ DBC is a right angle as indicated by the square symbol, then m∠DBC=90°.
Hence,
(0.1x-22)+(0.3x-54)= 90
<u>Step 3</u>
Lets solve the equation for x
We know that 0.4x-76= 90
0.4x= 166
x=166/0.4
x=415
<u>Step 4</u>
<u>Answer :</u> The value of x is 415
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