<h3>
Answer: B) Only the first equation is an identity</h3>
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I'm using x in place of theta. For each equation, I'm only altering the left hand side.
Part 1
cos(270+x) = sin(x)
cos(270)cos(x) - sin(270)sin(x) = sin(x)
0*cos(x) - (-1)*sin(x) = sin(x)
0 + sin(x) = sin(x)
sin(x) = sin(x) ... equation is true
Identity is confirmed
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Part 2
sin(270+x) = -sin(x)
sin(270)cos(x) + cos(270)sin(x) = -sin(x)
-1*cos(x) + 0*sin(x) = -sin(x)
-cos(x) = -sin(x)
We don't have an identity. If the right hand side was -cos(x), instead of -sin(x), then we would have an identity.
Answer:
Solve for
K
by simplifying both sides of the equation, then isolati.ng the variable.
K
=
m
v
2 i.f yo.u wa.nt th.e re.al answ.er g.o h.ere: >>>>https://www.math.way.com/po.pular-prob.lems/Alg.ebra/229798
2
Step-by-step explanation:
The cross section would be a semicircle because a cross section of the whole thing would be a circle
Hope this helps
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