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tiny-mole [99]
3 years ago
5

Can someone explain how to do question 3?

Mathematics
1 answer:
konstantin123 [22]3 years ago
5 0
2x^2+2y^2-4x+12y+6=0
you want it in the form
(x-a)^2+(y-b)^2=r^2 where (a,b) is the center and r is the radius hence the name standard center radius form.
dividing by 2 and rearranging
x^2-2x+y^2+6y=-3
by completing the square
x^2-2x+1+y^2+6y+9=-3+9+1
(x-1)^2+(y+3)^2=7
this is the standard radius center form (part b)
to get part a we multipy the equation by 2 to get
2(x-1)^2+2(y+3)^2=14 which is choice (2)
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sleet_krkn [62]
<h3>Answer:  7/13</h3>

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Explanation:

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Step-by-step explanation:

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Evaluate the integral. 3 2 t3i t t − 2 j t sin(πt)k dt
sveta [45]

∫(t = 2 to 3) t^3 dt

= (1/4)t^4 {for t = 2 to 3}

= 65/4.

----

∫(t = 2 to 3) t √(t - 2) dt

= ∫(u = 0 to 1) (u + 2) √u du, letting u = t - 2

= ∫(u = 0 to 1) (u^(3/2) + 2u^(1/2)) du

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For the k-entry, use integration by parts with

u = t, dv = sin(πt) dt

du = 1 dt, v = (-1/π) cos(πt).


So, ∫(t = 2 to 3) t sin(πt) dt

= (-1/π) t cos(πt) {for t = 2 to 3} - ∫(t = 2 to 3) (-1/π) cos(πt) dt

= (-1/π) (3 * -1 - 2 * 1) + [(1/π^2) sin(πt) {for t = 2 to 3}]

= 5/π + 0

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∫(t = 2 to 3) <t^3, t√(t - 2), t sin(πt)> dt = <65/4, 26/15, 5/π>.

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hram777 [196]

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