70 pls

maia is comparing the cost for one months usage of two cell phone companies. company a charges $40 per month plus $8.50 per gigabyte of data used. company b charges $30 per month plus $10.50 per gigabyte of data used.

part a
graph these costs as a system of equations on the coordinate plane below.

for each line in the system of equations, select two points on the coordinate plane. a line will be drawn through the points you selected.
(graph above)

part b

how many gigabytes of data would have to be used for company a and b to cost the same amount?

company A:  y= 40+8.5*(x)

company B: y=30+10.5*(x)

5 gigabytes of data would have to be used for company A and B to cost the same amount.

Step-by-step explanation:

[common monthly charges are $40 and let x represent the number of gigabytes of data used. so, 8.5*x gives cost due to to total gigabytes. hence, i have added 40 and  8.5*x to represent the total cost for first month]

y=30+10.5*(x)

[obtain this also in the above mentioned manner]

40+8.5*(x)=30+10.5*(x)

subtracting 30 from both sides,

10+8.5x=10.5x

subtracting 8.5x from both sides,

10=2x

dividing both the sides by 2.5,

5=x [since, 10/2 = 5)

40% off the regular price is the correct answer.

step-by-step explanation:

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)