What is the center of the equation X^2+y^2-4x-8y-16=0
1 answer:
x² + y² - 4x - 8y - 16 = 0
+ 16 + 16
x² + y² - 4x - 8y = 16
x² - 4x + y² - 8y = 16
(x² - 4x + 4) + (y² - 8y + 16) = 16 + 4 + 16
(x² - 2x - 2x + 4) + (y² - 4y - 4y + 16) = 20 + 16
[x(x) - x(2) - 2(x) - 2(-2)] + [y(y) - y(4) - 4(y) - 4(4)] = 36
[x(x - 2) - 2(x - 2)] + [y(y - 4) - 4(y - 4)] = 36
(x - 2)(x - 2) + (y - 4)(y - 4) = 36
(x - 2)² + (y - 4)² = 36
36 36 36
¹/₃₆(x - 2)² + ¹/₃₆(y - 4)² = 1
Center: (2, 4)
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Step-by-step explanation:
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Answer:b
Step-by-step explanation:9 multiply 8 multiply 6
We split [2, 4] into 
subintervals of length 
,
![[2,4]=\left[2,2+\dfrac2n\right]\cup\left[2+\dfrac2n,2+\dfrac4n\right]\cup\left[2+\dfrac4n,2+\dfrac6n\right]\cup\cdots\cup\left[2+\dfrac{2(n-1)}n,4\right]](https://tex.z-dn.net/?f=%5B2%2C4%5D%3D%5Cleft%5B2%2C2%2B%5Cdfrac2n%5Cright%5D%5Ccup%5Cleft%5B2%2B%5Cdfrac2n%2C2%2B%5Cdfrac4n%5Cright%5D%5Ccup%5Cleft%5B2%2B%5Cdfrac4n%2C2%2B%5Cdfrac6n%5Cright%5D%5Ccup%5Ccdots%5Ccup%5Cleft%5B2%2B%5Cdfrac%7B2%28n-1%29%7Dn%2C4%5Cright%5D)
so that the right endpoints are given by the sequence
for 
. Then the Riemann sum approximating
is
The integral is given exactly as 
, for which we get
To check: we have
