<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>
Answer:
22
Step-by-step explanation:
QS =2×11 = 22
__________
Answer:
x = 1, y = 2, z = -1
Step-by-step explanation:
2x + 4y = 10
2x = -4y + 10
x = -2y + 5
now sub -2y + 5 in for x, back into the other 2 equations
2x + 2y + 3z = 3 -3x + y + 2z = -3
2(-2y + 5) + 2y + 3z = 3 -3(-2y + 5) + y + 2z = -3
-4y + 10 + 2y + 3z = 3 6y - 15 + y + 2z = -3
-2y + 3z = 3 - 10 7y + 2z = - 3 + 15
-2y + 3z = - 7 7y + 2z = 12
-2y + 3z = -7....multiply by 2
7y + 2z = 12...multiply by -3
--------------------------
-4y + 6z = -14 (result of multiplying by 2)
-21y - 6z = -36 (result of multiplying by -3)
---------------------------add
-25y = - 50
y = -50/-25
y = 2 <===
2x + 4y = 10 2x + 2y + 3z = 3
2x + 4(2) = 10 2(1) + 2(2) + 3z = 3
2x + 8 = 10 2 + 4 + 3z = 3
2x = 10 - 8 6 + 3z = 3
2x = 2 3z = 3 - 6
x = 2/2 3z = -3
x = 1 <== z = -3/3
z = -1 <===
Value of x = 19
(8x - 20) = (5x + 37)
3x = 57
x = 19
37=10+4.50x
x=number of students needed to make $37
37=10+4.50x
-10 -10
27=4.50x
/4.50 /4.50
6=x
She needs to 6 students to make $37 for an hour.
Hope it helps!
