Recall the double angle identity for cosine:
It follows that
Since 0° < 22° < 90°, we know that sin(22°) must be positive, so csc(22°) is also positive. Let x = 22°; then the closest answer would be C,
but the problem is that none of these claims are true; cot(32°) ≠ 4/3, cos(44°) ≠ 5/13, and csc(22°) ≠ √13/2...
<em><u>There is no solution!</u></em>
Find the area for each shape.
For the triangle face you would do 4 times 6 divided by 2. Then times it by 2 because there are two triangle faces.
For the rectangle face do 5 times 12, then times 3 because there are 3 triangle faces.
Finally add both numbers together to get the surface area.
I don’t know if you can see this but I hope it does if u don’t understand let me know so I can explain!
If y = cos(kt), then its first two derivatives are
y' = -k sin(kt)
y'' = -k² cos(kt)
Substituting y and y'' into 49y'' = -16y gives
-49k² cos(kt) = -15 cos(kt)
⇒ 49k² = 15
⇒ k² = 15/49
⇒ k = ±√15/7
Note that both values of k give the same solution y = cos(√15/7 t) since cosine is even.
