Answer: 6 x 10^12
Step-by-step explanation:
Move the decimal to the left so there is one non-zero digit number left of the decimal point. This will be the exponent placed on 10
16x^2 + 25y^2 + 160x - 200y + 400 = 0 Rearrange and regroup.
(16x^2 + 160x) + (25y^2 - 200y ) = 0-400. Group the xs together and the ys together.
16(X^2 + 10x) + 25(y^2-8y) = -400. Factorising.
We are going to use completing the square method.
Coefficient of x in the first expression = 10.
Half of it = 1/2 * 10 = 5. (Note this value)
Square it = 5^2 = 25. (Note this value)
Coefficient of y in the second expression = -8.
Half of it = 1/2 * -8 = -4. (Note this value)
Square it = (-4)^2 = 16. (Note this value)
We are going to carry out a manipulation of completing the square with the values
25 and 16. By adding and substracting it.
16(X^2 + 10x) + 25(y^2-8y) = -400
16(X^2 + 10x + 25 -25) + 25(y^2-8y + 16 -16) = -400
Note that +25 - 25 = 0. +16 -16 = 0. So the equation is not altered.
16(X^2 + 10x + 25) -16(25) + 25(y^2-8y + 16) -25(16) = -400
16(X^2 + 10x + 25) + 25(y^2-8y + 16) = -400 +16(25) + 25(16) Transferring the terms -16(25) and -25(16)
to other side of equation. And 16*25 = 400
16(X^2 + 10x + 25) + 25(y^2-8y + 16) = 25(16)
16(X^2 + 10x + 25) + 25(y^2-8y + 16) = 400
We now complete the square by using the value when coefficient was halved.
16(x-5)^2 + 25(y-4)^2 = 400
Divide both sides of the equation by 400
(16(x-5)^2)/400 + (25(y-4)^2)/400 = 400/400 Note also that, 16*25 = 400.
((x-5)^2)/25 + ((y-4)^2)/16 = 1
((x-5)^2)/(5^2) + ((y-4)^2)/(4^2) = 1
Comparing to the general format of an ellipse.
((x-h)^2)/(a^2) + ((y-k)^2)/(b^2) = 1
Coordinates of the center = (h,k).
Comparing with above (x-5) = (x - h) , h = 5.
Comparing with above (y-k) = (y - k) , k = 4.
Therefore center = (h,k) = (5,4).
Sorry the answer came a little late. Cheers.
Set up an algebraic equation. This is what it looks like based on what the question tells you:
Let a = area of pentagon
a+27 = 4a-15
3a-15 = 27
3a = 42
a = 14
Another triple integral. We're integrating over the interior of the sphere
Let's do the outer integral over z. z stays within the sphere so it goes from -2 to 2.
For the middle integral we have
x is the inner integral so at this point we conservatively say its zero. That means y goes from 
and 
Similarly the inner integral x goes between 
So we rewrite the integral
Let's work on the inner one,
There's no z in the integrand, so we treat it as a constant.
So the middle integral is
I gotta go so I'll stop here, sorry.
(7,2) is the answer.
There you go.
