2cos(x) - 4sin(x) = 3
use identity [cos(x)]^2 +[ sin(x)]^2 = 1 => cos(x) = √[1 - (sin(x))^2]
2√[1 - (sin(x))^2] - 4 sin(x) = 3
2√[1 - (sin(x))^2] = 3 + 4 sin(x)
square both sides
4[1 - (sin(x))^2] = 9 + 24 sin(x) + 16 (sin(x))^2
expand, reagrup and add like terms
4 - 4[sin(x)]^2 = 9 + 24sin(x) + 16sin^2(x)
20[sin(x)]^2 + 24sin(x) +5 = 0
use quadratic formula and you get sin(x) = -0.93166 and sin(x) = -0.26834
Now use the inverse functions to find x:
arcsin(-0.93166) = 76.33 degrees
arcsin(-0.26834) = 17.30 degrees
Answer:
x ≥ 1
Step-by-step explanation:
- The circle is over 1
- The circle is closed (filled in), which means it will be either ≤ or ≥.
- The arrow is pointing right, so it's ≥.
- The number can be represented by x
- Put it together: x ≥ 1
I hope this helps!
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
.20 or 20. one of those i cant give you the answer because how are u going to learn <span />