Given
s=16t^2
where
s=distance in feet travelled (downwards) since airborne with zero vertical velocity and zero air-resistance
t=time in seconds after release
Here we're given
s=144 feet
=>
s=144=16t^2
=>
t^2=144/16=9
so
t=3
Ans. after 3 seconds, the object hits the ground 144 ft. below.
The simplified version of this expression is 110g^4h^16v^6. first, remove the parenthesis. the expression becomes 10g^3h^8v^6 times 11gh^8. then, calculate the product.
im pretty sure the answer is 30
cos4x = cos2x
We know that:
cos2x = 1-2cos^2 x
==> cos4x = 1-2cos^2 (2x)
Now substitute:
==> 1-2cos^2 (2x) = cos2x
==> 2cos^2 (2x) + cos2x - 1 = 0
Now factor:
==> (2cos2x -1)(cos2x + 1) = 0
==> 2cos2x -1 = 0 ==> cos2x =1/2 ==> 2x= pi/3
==> x1= pi/6 , 7pi/6
==> x1= pi/6 + 2npi
==> x2= 7pi/6 + 2npi
==> cos2x = -1 ==> 2x= pi ==> x3 = pi/2 + 2npi.
<span>==> x= { pi/6+2npi, 7pi/6+2npi, pi/2+2npi}</span>
Answer:
(0, -3), (2, 1)
Step-by-step explanation:
Solve this by solving for the first variable in one of the equations, then substitute the result into the other equation.
