i can't comment so someone else is going to have to answer
but they need the whole page to answer this question.
Consider a circle with radius
centered at some point
on the
-axis. This circle has equation

Revolve the region bounded by this circle across the
-axis to get a torus. Using the shell method, the volume of the resulting torus is

where
.
So the volume is

Substitute

and the integral becomes

Notice that
is an odd function, so the integral over
is 0. This leaves us with

Write

so the volume is

Answer:
See below ~
Step-by-step explanation:
Finding t₁₂ :
⇒ t₁₂ = t₁ + 11d
⇒ t₁₂ = 6 + 11(2)
⇒ t₁₂ = 6 + 22
⇒ t₁₂ = 28
=============================================================
Finding S₁₂ :
⇒ S₁₂ = 12/2 × 2t₁ + (12 - 1)d
⇒ S₁₂ = 6 × 2(6) + 11(2)
⇒ S₁₂ = 6 × 12 + 22
⇒ S₁₂ = 6 × 34
⇒ S₁₂ = 204
So basically you are going to line up your equations (already done). Next, you will just add each term. So x + 3x = 4x, then 2y - 2y = 0 (we wouldn't put anything) then 7 - 3 = 4
So then we have all of our terms figured out and will have 4x = 4 then divide 4x by 4 to get x alone and then you will also divide the four that is by itself by four to equal one. So, your answer is x = 1. I hope that helps!
Answer:
Solve the equation for x by finding a
, b
, and c
of the quadratic then applying the quadratic formula.
Exact Form:
x
=
−
4
±
√
5
Decimal Form:
x
=
−
1.76393202
…
,
−
6.23606797
…