The length of the DEFG rectangle
is 6/4 of the length of the ABCD rectangle.
6/4=3/2 = 1,5
How many times greater than the surface of ABCD is the DEFG area?
1,5 times
He would need to play 12 average scoring games.
900 (goal) - 300 (has) = 600
and 600/50 is 12
hope this helps :)
Answer:11 days
Step-by-step explanation:
y =2x+1
7 divided by 3= 2 remainder 3
Don't worry about the three use the 2 times 5 to get 10.
Then 10 plus 1 to get 11. If you don't believe me then do this: 3 times 2 to get 6 then plus 1 to get 7
Let 1993 = time 0 = 0.
Let 1999 = time 6 = 6
Let 2012 = time 19 = 19
So, a = 171 (million). First solve for k.
176 = 171 e^k6
176/171 = e^(k*6)
ln (176/171) = 6k
k = 1/6 ln (176/171)
So, in 2012 we have: P(19) = 171 e^(19k), where k = 1/6 ln (176/171)
Hope this helped!
I think C would be for number 11
A is for 12.
x/2 < - 18
