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
Answer:
are you ok?
Step-by-step explanation:
why did you need help lol
If m ACD = 30 => m DCB = 60.
In triangle ACD:
AC^{2}+CB^{2}=AB^{2} => AB=
=32.
=> P=16+32+
=48 +
.
60,000,000,000 in scientific notation is 6 x 10^10 so you move the decimal 10 times.
