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.
Answer:
idk
Step-by-step explanation:
Answer:
Tito have
textbooks in terms of x
Step-by-step explanation:
Rudy has 6 textbooks
We are given that Tito and Maria each have y textbooks
So, Tito has no. of books = y
Maria has no. of books = y
So, total number of textbooks = 6+y+y=6+2y
We are given that Rudy, Maria, and Tito have a total of x textbooks
So,6+2y=x
So, 2y=x-6
![y=\frac{x-6}{2}](https://tex.z-dn.net/?f=y%3D%5Cfrac%7Bx-6%7D%7B2%7D)
Hence Tito have
textbooks in terms of x
For this case we have that by definition, the median is the number of half in a set of numbers. To find it, the numbers are placed in order and the middle number is found.
If we order the given data we have:
6,7,10,12,14
The middle number is 10. So the median is 10.
Answer:
10
Answer :
![\boxed{\textsf{ The final answer is \textbf{n}$^{\textbf{3}}$ .}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Ctextsf%7B%20The%20final%20answer%20is%20%5Ctextbf%7Bn%7D%24%5E%7B%5Ctextbf%7B3%7D%7D%24%20.%7D%7D)
Step-by-step explanation:
Its given that n is the middle out of the three consecutive integers . So ,
<u>The </u><u>last</u><u> </u><u>integer</u><u> </u><u>will</u><u> </u><u>be</u><u> </u><u>:</u><u>-</u><u> </u>
<u>The </u><u>next</u><u> </u><u>Integer</u><u> </u><u>will</u><u> be</u><u> </u><u>:</u><u>-</u>
Now the Question says that the three integers are multipled to give a product . So that would be.
![\sf\implies Product_{(three\ consecutive\ integers)}= (n-1)n(n+1) = (n^2-1)(n) = \pink{n^3-n}](https://tex.z-dn.net/?f=%5Csf%5Cimplies%20Product_%7B%28three%5C%20consecutive%5C%20integers%29%7D%3D%20%28n-1%29n%28n%2B1%29%20%3D%20%28n%5E2-1%29%28n%29%20%3D%20%5Cpink%7Bn%5E3-n%7D)
Now thirdly it's given that n is added to the given integer . That would be ,
Here - n and +n gets cancelled. So we are ultimately left out with n³.
Hence the final number is a cube of some number.