With this. joe runs 8.25 times around a track in 1,119.803 seconds. if one lap around the track is 402.3 meters, which is the best estimate of the runners average speed in meters per second (m/s)?

a. 30 m/s

b. 0.3 m/s

c. 3 m/s

d. 0.4 m/s

i am unable to solve this problem

step-by-step explanation:

if you are saying purchasing 3 social media ads and one newspaper ad counts as one purchase then there are:

6 purchases

if you are asking how many of each then:

18 social media ads

6 newspaper ads

i need more details sorry

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)