Apassword is 4 characters long and must consist of 3 letters and 1 of 10 special characters. if letters can be repeated and the special character is at the end of the password, how many possibilities are there? a. 175,760 b. 456,976 c. 703,040 d. 1,679,616

there are a total of 26 letters that can be used

so you have 26 x 26 x 26 x 10 = 175,760 possibilities

Answer is A

please mark brainliest

b i think

step-by-step explanation:

not really sure

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)