Speed because speed is key
The first thing we are going to do to find (g°f)(3), is find (g°f)(x). To do that, we are going to evaluate g(x) at f(x):
(g°f)(x)=
![g(f(x))=g(3x-2)=(3x-2)^2=9x^2-12x+4](https://tex.z-dn.net/?f=g%28f%28x%29%29%3Dg%283x-2%29%3D%283x-2%29%5E2%3D9x%5E2-12x%2B4)
Now, we can evaluate (g°f) at 3:
![g(f(3))=9(3)^2-12(3)+4](https://tex.z-dn.net/?f=g%28f%283%29%29%3D9%283%29%5E2-12%283%29%2B4)
![g(f(3))=9(9)-36+4](https://tex.z-dn.net/?f=g%28f%283%29%29%3D9%289%29-36%2B4)
![g(f(3))=81-32](https://tex.z-dn.net/?f=g%28f%283%29%29%3D81-32)
![g(f(3))=49](https://tex.z-dn.net/?f=g%28f%283%29%29%3D49)
We can conclude that the correct answer is
c. 49
9514 1404 393
Answer:
Step-by-step explanation:
Let x and y represent the weights of the large and small boxes, respectively. The problem statement gives rise to the system of equations ...
x + y = 85 . . . . . combined weight of a large and small box
70x +50y = 5350 . . . . combined weight of 70 large and 50 small boxes
We can subtract 50 times the first equation from the second to find the weight of a large box.
(70x +50y) -50(x +y) = (5350) -50(85)
20x = 1100 . . . . simplify
x = 55 . . . . . . . divide by 20
Using this in the first equation, we can find the weight of a small box.
55 +y = 85
y = 30 . . . . . . . subtract 55
A large box weighs 55 pounds; a small box weighs 30 pounds.
Answer:
54 cm^2
Step-by-step explanation:
side ( a ) = 3 cm
Formula : -
Surface area ( cube ) = 6a^2
Surface area ( cube )
= 6 x ( 3 )^2
= 6 x 9
= 54 cm^2