I assume that there is an operato ^ missing in each function and that the right functions are:
f(m) = (m+40)^2 + 10 and
g(m) = (m+12)^2 - 50
C(m) = f(m) + g(m)
To perform that sum you need to expand the two square parentheses, this way:
f(m) = (m+40)^2 + 10 = m^2 + 80m + 1600 + 10 = m^2 + 80m + 1610
g(m) = (m+12)^2 - 50 = m^2 + 24m + 144 - 50 = m^2 +24m + 94
Now you can add f(m) + g(m) = 2m^2 + 104m + 1704
Answer: c(m) = 2m^2 + 104m + 1704
Answer:
10c3 =(10)!/(3!*7!) = (10*9*8)/(3*2*1) = 10*3*4 =120
Answer:
awesome! thanks for sharing!
Step-by-step explanation:
Answer:
(4,10) is the solution
Step-by-step explanation:
2x + y = 18
x – y = –6
Using elimination
2x + y = 18
x – y = –6
----------------------
3x = 12
Dividing each side by 3
3x/3 = 12/3
x =4
Now we can find y
2x+y = 18
2(4) +y =18
8+y = 18
Subtracting 8 from each side
8-8+y = 18-8
y=10
(4,10) is the solution