Hannah was preparing to make 'goody bags' for tricker-treaters on halloween. she had large bags of snack pack candies: one bag held 48 pieces of chocolate, a second held 64 pieces lollipops, and the third held 80 pieces of bubblegum . how many bags can she prepare if she does not want any candy left over. how many of each type will go into the bags?

There are 8 bags she can prepare if she does not want any candy left over.

There are 6 pieces of chocolates , 8 pieces of lollipops and 10 pieces of bubblegum into bags.

Step-by-step explanation:

Since we have given that

Number of pieces of chocolate in one bag = 48

Number of pieces of lollipops in second bag = 64

Number of pieces of bubblegum in third bag = 80

So, we need a large bag of snack pack candies.

So, for Number of bags she can prepare if she does not want any candy left over , we will do "H.C.F. of 48, 64, 80" which is equal to 8.

So, Number of bags she need to prepare = 8

Number of chocolates in bag would be

Number of lollipops in bag would be

Number of bubblegum in bag would be

Hence, there are 6 pieces of chocolates , 8 pieces of lollipops and 10 pieces of bubblegum in 8 bags.

c. y = 1/4 x + 5

step-by-step explanation:

first, put the equation of the line on the graph into slope-intercept form

(y = mx + b):   y = -4x + 8

second, a perpendicular line has to have "an opposite/reciprocal slope" to the original slope. so, find the "m" value in your original equation (it's -4, right? )

original slope: m = -4   perpendicular slope: change the sign to its opposite and find the reciprocal ("flip" the number)

new slope is now positive (b/c the original slope was negative 4), and it's 1/4 (not the original slope of 4 or 4/1)