Cathy is dressing a store mannequin and can choose from an assortment of clothes consisting of 5 summer dresses, 5 hats, and 2 pairs of shoes. How many different ways can Cathy dress the mannequin with a dress, a hat, and a pair of shoes? **Show your work is for scrap paper**​

50 ways

Step-by-step explanation:

Given

Cathy has an assortment of clothes consisting

5 dresses, 5 hats and 2 pair of shoes

Cathy has 5 ways to select a dress, 5 ways to select a hat, and 2 ways to select the pair of shoes

Therefore no of the ways she can dress the mannequin is

x -

median = 1.25

range = 12-14.5

interquartile = q12-q14.5

y -

median = 1

range = 15-17

interquartile = q15-q17

hello from mrbilldoesmath!

answer:   (7x + 6y) ( 49x^2 - 42xy + 36y^2)

discussion:

in general, the factorization of a^3 + b^3 is

a^3 + b^ 3 = (a + b) ( a^2 - ab + b^2)

in our case note that 343 = 7^3 and 216 = 6^3 so the equation becomes:

343x^3 + 216y^3 = (7x)^3 + (6y)^3

if you think of a (above) as 7x and b as 6y the factorization becomes

343x^3 + 216y^3 =

((7x + 6y) (   (7x)^2 - 7x*6y + (6y)^2)   =

(7x + 6y) ( 49x^2 - 42xy + 36y^2)

you,

mrb

(a + b) (a^2 - a b + b^2)