I'm sorry, do you have any more detail. i'm not sure what this means?
Distribute the 3a^n to all the other values then solve...
(3a^n*a^n)+(3a^n*a^n)+(3a^n*-1)
4a^2n+4a^2n-3a^n=
8a^2n-3a^n
# of numbers between 1 & 103: 102
Half of that would be even numbers. So, 51 even numbers.
Between 4 and x, there are 102 numbers. 4+102=106. 4+106=x. 106=x
Answer:
(0 , -1)
Step-by-step explanation:
WR = 4 + 2 = 6
WS = 1/3 x WR = 2 .... the point of 1:2 of WR
S (-2 + 2 , -4) i.e. (0,4)
The same reason: RQ = 1/3 x RY = 1/3 x (5 + 4) = 3
Q (-4 , -4 + 3) i.e. (-4, -1)
QP // RW
p (0 , -1)