Answer:
Choice B
Step-by-step explanation:
Given radical expression:
![\sqrt[4]{1296 {x}^{16} {y}^{12} }](https://tex.z-dn.net/?f=%20%5Csqrt%5B4%5D%7B1296%20%7Bx%7D%5E%7B16%7D%20%20%7By%7D%5E%7B12%7D%20%7D%20)
To Find:
The Simpler form of this expression
Soln:
![= \sqrt[4]{1296 {x}^{16} {y}^{12} }](https://tex.z-dn.net/?f=%20%3D%20%20%5Csqrt%5B4%5D%7B1296%20%7Bx%7D%5E%7B16%7D%20%20%7By%7D%5E%7B12%7D%20%7D%20)
We could re-write the given expression, according to the law of exponents:
![= \tt \sqrt[4]{(6x {}^{4}y {}^{3}) {}^{4} }](https://tex.z-dn.net/?f=%20%3D%20%20%5Ctt%20%5Csqrt%5B4%5D%7B%286x%20%7B%7D%5E%7B4%7Dy%20%7B%7D%5E%7B3%7D%29%20%20%7B%7D%5E%7B4%7D%20%20%7D%20)
Now we need to bring terms out of the radical as:

Bring out 6x^4 from the absolute & put y^3 only in it:

Choice B is accurate.
Answer:
D and E because they are bigger than 28. C is 28 but there isn't a greater OR EQUAL sign next to it
Answer:

Step-by-step explanation:
The vertical component of the initial launch can be found using basic trigonometry. In any right triangle, the sine of an angle is equal to its opposite side divided by the hypotenuse. Let the vertical component at launch be
. The magnitude of
will be the hypotenuse, and the relevant angle is the angle to the horizontal at launch. Since we're given that the angle of elevation is
, we have:
(Recall that
)
Now that we've found the vertical component of the velocity and launch, we can use kinematics equation
to solve this problem, where
is final and initial velocity, respectively,
is acceleration, and
is distance travelled. The only acceleration is acceleration due to gravity, which is approximately
. However, since the projectile is moving up and gravity is pulling down, acceleration should be negative, represent the direction of the acceleration.
What we know:
Solving for
:

Answer:
2880
Step-by-step explanation:
V=whl=12·12·20=2880