Answer: I think its (B) Reflection over the x-axis
Step-by-step explanation: Hope this helps!
Answer:
The input is x−2, therefore
Step-by-step explanation:
f(x−2)=(x−2)2−3(x−2)+1
What we have so far:
Kinetic energy = 0. The reason behind that is because the beam is not moving at a height of 40m.
Gavity, g = 9.8m/s²
Height = 40m
Potential energy = mgh; this is equal to 0 because m, stands for mass and in this problem, we do not have a value for the mass of the beam. Hence, 0 x 9.8m/s² x 40m = 0. Potential energy = 0.
Solution:
We will use the equation of Total energy:
TE = potential energy + kinetic energy
TE = 0 + 0
∴ TE = 0
The answer is: Assuming no air resistance, the total energy of the beam as it hits the ground is 0.
(15p⁺⁴.q⁻⁶) /(-20p⁻¹².q⁻³).
Remember that a⁻ⁿ = 1/aⁿ and 1/a⁻ⁿ = aⁿ
(-15/4).(p⁻⁴.q⁻⁶)(p⁺¹².q⁺³).
(-15/20).(p⁻⁴.p¹².q⁻⁶.q³)
Remember aⁿ.aˣ = aⁿ⁺ˣ
(-15/20).(p⁸.q⁻³)
-3/5(p⁸.q⁻³)