Answer:
it's C.
Step-by-step explanation:
Hope that helped :D
Answer:
3
Step-by-step explanation:
I think its 3 because when we add 2 to x the graphic must be continue two step to right
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:
The factors of 2q²-5pq-2q+5p are (2q-5p) (q-1)....
Step-by-step explanation:
The given expression is:
2q²-5pq-2q+5p
Make a pair of first two terms and last two terms:
(2q²-5pq) - (2q-5p)
Now factor out the common factor from each group.
Note that there is no common factor in second group. So we will take 1 as a common factor.
q(2q-5p) -1(2q-5p)
Now factor the polynomial by factoring out the G.C.F, 2q-5p
(2q-5p) (q-1)
Thus the factors of 2q²-5pq-2q+5p are (2q-5p) (q-1)....
Answer:
definitely 4
Step-by-step explanation:
15/3 is 5 if u add 4 packets then I should he 20
