Can you help me in my recent
50
64=1, 2, 4, 8, 16, 32, 64
80
The answer is 27a^6
Hope this helps.
9514 1404 393
Answer:
x = 10·cos(θ) -4·cot(θ)
Step-by-step explanation:
Apparently, we are to assume that the horizontal lines are parallel to each other.
The relevant trig relations are ...
Sin = Opposite/Hypotenuse
Cos = Adjacent/Hypotenuse
If the junction point in the middle of AB is labeled X, then we have ...
sin(θ) = 4/BX ⇒ BX = 4/sin(θ)
cos(θ) = x/XA ⇒ XA = x/cos(θ)
Then ...
BX +XA = AB = 10
Substituting for BX and XA using the above relations, we get
4/sin(θ) +x/cos(θ) = 10
Solving for x gives ...
x = (10 -4/sin(θ))·cos(θ)
x = 10·cos(θ) -4·cot(θ) . . . . . simplify
_____
We used the identity ...
cot(θ) = cos(θ)/sin(θ)
