100 Points last one i promise! helpp!
2 answers:
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Now, the cosecant of θ is -6, or namely -6/1.
however, the cosecant is really the hypotenuse/opposite, but the hypotenuse is never negative, since is just a distance unit from the center of the circle, so in the fraction -6/1, the negative must be the 1, or 6/-1 then.
we know the cosine is positive, and we know the opposite side is -1, or negative, the only happens in the IV quadrant, so θ is in the IV quadrant, now
recall that
therefore, let's just plug that on the remaining ones,
now, let's rationalize the denominator on tangent and secant,
however, the cosecant is really the hypotenuse/opposite, but the hypotenuse is never negative, since is just a distance unit from the center of the circle, so in the fraction -6/1, the negative must be the 1, or 6/-1 then.
we know the cosine is positive, and we know the opposite side is -1, or negative, the only happens in the IV quadrant, so θ is in the IV quadrant, now
recall that
therefore, let's just plug that on the remaining ones,
now, let's rationalize the denominator on tangent and secant,
Answer:
For w = 18 units perimeter is minimum
P = 2(18 + w)
Step-by-step explanation:
Given;
Area of the rectangle = 324 units²
P is the perimeter
w is the width
Let L be the length of the rectangle
therefore,
P = 2(L + w) ............(1)
also,
Lw = 324
or
L = ..........(2)
substituting 2 in 1
P =
now,
for minimizing the perimeter
= 0
or
= 0
or
= 0
or
= -1
or
w² = 324
or
w = 18 units
For w = 18 units perimeter is minimum
therefore,
from 2
L =
or
L = 18 units
objective function for P is:
P = 2(18 + w)