A rental car company charges $20.59 per day to rent a car and $0.06 for every mile driven. Abigail wants to rent a car, knowing that: She plans to drive 150 miles.
She has at most $420 to spend.

Which inequality can be used to determine xx, the maximum number of days Abigail can afford to rent for while staying within her budget?
a. 9x+20.59≤420
b. 9+20.59x≤420
c. 9x+20.59≥420
d. 9+20.59x≥420

step-by-step explanation:

actually your concept is wrong

step-by-step explanation:

150% of x =300

150x=300×100

x=(300×100)÷150

x=200

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)

step-by-step explanation: