<span>A. cos(x) = cos(-x) [correct, since cos(x) is an even function]
B. Since the cosine function is even, reflection over the x-axis [y-axis] does not change the graph. [false]
C. cos(x) = -cos(x) [ false]
D. The cosine function is odd [even], so it is symmetrical across the origin. [false]</span>
Answer:
8 hours
Step-by-step explanation:
rental fee = 12 + 4/hr
spent = 44
44 = 12+4(hr)
32 = 4(hr)
32/4 = hr
hr = 8
We have that
<span>tan(theta)sin(theta)+cos(theta)=sec(theta)
</span><span>[sin(theta)/cos(theta)] sin(theta)+cos(theta)=sec(theta)
</span>[sin²<span>(theta)/cos(theta)]+cos(theta)=sec(theta)
</span><span>the next step in this proof
is </span>write cos(theta)=cos²<span>(theta)/cos(theta) to find a common denominator
so
</span>[sin²(theta)/cos(theta)]+[cos²(theta)/cos(theta)]=sec(theta)<span>
</span>{[sin²(theta)+cos²(theta)]/cos(theta)}=sec(theta)<span>
remember that
</span>sin²(theta)+cos²(theta)=1
{[sin²(theta)+cos²(theta)]/cos(theta)}------------> 1/cos(theta)
and
1/cos(theta)=sec(theta)-------------> is ok
the answer is the option <span>B.)
He should write cos(theta)=cos^2(theta)/cos(theta) to find a common denominator.</span>
Answer:
-7
Step-by-step explanation:
given f(x)=4x+k
therfore, f(2)=4*2+k
we know that f(2) =1
so, 4*2+k=1
8+k=1
k=1-8
k=-7
