Answer:
radius 4
center (3,6)
Step-by-step explanation:
(x - h)^2 + (y - k)^2 = r^2
coordinates of the center (h, k) and the radius is (r)
x² + y²- 6x - 12y +29=0
x² - 6x + y²- 12y +29=0
(x² - 6x) + (y²- 12y) +29=0
complete the square
(x² - 6x) + 9 + (y²- 12y) +36 +29= + 9 +36
(x² - 6x + 9) + (y²- 12y + 36) = + 9 +36 -29
(x-3)^2 + (y-6)^2 = 16
(x - h)^2 + (y - k)^2 = r^2
coordinates of the center (h, k) and the radius is (r)
center (3,6)
radius 4
cuemathcom
varsitytutors
Answer:
≥ 999,999,999
Step-by-step explanation:
that is the less than or equal to sign
hope this helps
now, for a rational expression, the domain, or "values that x can safely take", applies to the denominator NOT becoming 0, because if the denominator is 0, then the rational turns to
undefined.
now, what value of "x" makes this denominator turn to 0, let's check by setting it to 0 then.
![\bf 2-x^{12}=0\implies 2=x^{12}\implies \pm\sqrt[12]{2}=x\\\\ -------------------------------\\\\ \cfrac{x^2-9}{2-x^{12}}\qquad \boxed{x=\pm \sqrt[12]{2}}\qquad \cfrac{x^2-9}{2-(\pm\sqrt[12]{2})^{12}}\implies \cfrac{x^2-9}{2-\boxed{2}}\implies \stackrel{und efined}{\cfrac{x^2-9}{0}}](https://tex.z-dn.net/?f=%5Cbf%202-x%5E%7B12%7D%3D0%5Cimplies%202%3Dx%5E%7B12%7D%5Cimplies%20%5Cpm%5Csqrt%5B12%5D%7B2%7D%3Dx%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A%5Ccfrac%7Bx%5E2-9%7D%7B2-x%5E%7B12%7D%7D%5Cqquad%20%5Cboxed%7Bx%3D%5Cpm%20%5Csqrt%5B12%5D%7B2%7D%7D%5Cqquad%20%5Ccfrac%7Bx%5E2-9%7D%7B2-%28%5Cpm%5Csqrt%5B12%5D%7B2%7D%29%5E%7B12%7D%7D%5Cimplies%20%5Ccfrac%7Bx%5E2-9%7D%7B2-%5Cboxed%7B2%7D%7D%5Cimplies%20%5Cstackrel%7Bund%20efined%7D%7B%5Ccfrac%7Bx%5E2-9%7D%7B0%7D%7D)
so, the domain is all real numbers EXCEPT that one.
Answer:
24
Step-by-step explanation:
PEMDAS.
