Answer:
Step-by-step explanation:a = m + (p-1)*d
b = m + (q-1)*d
c = m + (r-1)*d
p(b-c) = p*(q-r)*d
q(c-a) = q*(r-p)*d
r(a-b) = r*(p-q)*d
p(b-c)+q(c-a)+r(a-b)
= p*(q-r)*d + q*(r-p)*d +r*(p-q)*d
= (pq-pr+qr-pq+rp-qr)*d
= 0*d = 0
So i prove p(b-c)+q(c-a)+r(a-b)=0 hope this is helpfull
-1, 0, 1/2, 6, 9 hope that helps
Answer: No solution
Step-by-step explanation: -12x-14=-62-12x
You would have to simplify the -12x but you can't so this would be a no solution.
The standard equation for the circle is:
(x-h)^2+(y-k)^2=r^2, where (h,k) is the center of the circle and r is the radius.
We are given that the center is (5,1) and r=10 so
(x-5)^2+(y-1)^2=100
