Answer:
(x-3)^2 + (y+2)^2 = 9^2
Step-by-step explanation:
x^2 -6x+y^2+4y-68 = 0
Complete the square
x^2 -6x+y^2+4y-68+68 = 0+68
x^2 -6x+y^2+4y = 68
Find the term to add for x
-6 /2 = -3 -3^2 = 9
Find the term to add for y
4/2 =2 2^2 = 4
Add 9 and 4
x^2 -6x+9+y^2+4y+4 = 68+9+4
(x-3)^2 + (y+2)^2 = 81
(x-3)^2 + (y+2)^2 = 9^2
The standard form is
(x-h)^2 + (y-k)^2 = r^2 where (h,k) is the center and r is the radius
Answer:
b > -16
Step-by-step explanation:
The inequality is given as:
b - 1 > -17
To solve this problem, add +1 to both sides of the expression:
b - 1 + 1 > -17 + 1
Now;
b > -16
So, all values for which b is more than -16 satisfy the solution for this problem.
you did a question not an answer pal sorry
Recall Euler's theorem: if , then
where is Euler's totient function.
We have - in fact, for any since and share no common divisors - as well as .
Now,
where the are positive integer coefficients from the binomial expansion. By Euler's theorem,
so that