Date: Page 3) If n Cu)=40, n(A) =25,0(B) = 10 and CA n BJ = 5 find: in (AUB) ii) n(A) iii) OCA is no (B) ng ?​

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)

answer: can you show the options ?

step-by-step explanation:

192/15 = 12.8.

12.8

step-by-step explanation:

192/15 is 12.8.

however, it says to round to the nearest foot, which would equal 13.

first, let's see what the pattern is.

58 - x = 47

46 - x = 36

x = 9

so this means that the term after 25 is 25 - 9:

the next term in the sequence is 16.