37. Emma is storing 432 ounces of soup into small 6-ounces containers and medium 8-ounces
containers. If there must be at least 50 small
containers, what is the least possible number
of containers needed to store all the soup
without any leftover soup remaining?

15 8 ounce containers and 52 6 ounce containers

Step-by-step explanation:

First figure out how much soup must be in the 50 small containers

50 * 6 = 300

Subtract that from the total amount of soup

432 - 300 = 132

We have to put 132 ounces of soup into 6 ounce and 8 ounce containers

Let x be the number of 6 ounce and y be the number of 8 ounce containers

6x+8y = 132

x+y = minimum

We want to use as many 8 ounce containers as possible

132/8 = 16.5

16*8 = 128 r4  we cannot use 16 because we do not have a 4ounce container

15*8 = 120  r12  12/6 =2  we can do this because we can use 2 6 ounce containers

We need 15 8 ounce containers and 2 6 ounce for the 132 ounces left

We have 50 for the 300 ounces

For a total of

15 8 ounce containers and 52 6 ounce containers

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

the factor is 2

step-by-step explanation:

let f(x)=8(3^(x−2)+2)

to stretch a graph by 3, it would be 3f(x)

g(x) =3f(x)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

x     −2     0     2     4     6

f(x) 1/64 1/16   1/4     1     4

1/64 * z * z = 1/16

i multiplies by z twice because it its 2 steps from -2 to 0

1/64 z^2 = 1/16

multiply by 64 to clear the fraction on the left side

1/4 * 64 z^2 =1/16 *64

z^2 = 4

take the square root of each side

sqrt(z^2) = ±sqrt(4)

z = ±2  

we are multiplying by 2 each time.