<span>since sin and cos = each other at pi/4; take your integrals from 0 to pi/4
</span><span>[S] cos(t) dt - [S] sin(t) dt ;[0,pi/4]
</span>
<span>to revolve it around the x axis;
we do a sum of areas
[S] 2pi [f(x)]^2 dx
</span>
<span>take the cos first and subtract out the sin next; like cutting a hole out of a donuts.
</span><span>pi [S] cos(x)^2 dx - [S] sin(x)^2 dx ; [0,pi/4]
</span>
<span>cos(2t) = 2cos^2 - 1
cos^2 = (1+cos(2t))/2
</span>
<span>1/sqrt(2) - (-1/sqrt(2) +1)
1/sqrt(2) + 1/sqrt(2) -1
(2sqrt(2) - sqrt(2))/sqrt(2) = sqrt(2)/sqrt(2) = 1</span>
Second one is the correct answer
The solution to the problem is as follows:
cot θ = 1 / tanθ
<span>cot θ tanθ = tanθ / tanθ = 1
</span>
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Answer:
Perpendicular lines have opposite-reciprocal slopes, so the slope of the line we want to find is 1/2. Plugging in the point given into the equation y = 1/2x + b and solving for b, we get b = 6. Thus, the equation of the line is y = ½x + 6.
The answer is 6
Explanation: 7.50•6+2=44 and 6•6+8=44
