Initially, there are 5 red beans and 9 black beans in a bag. Then, 3 red beans and 3 black beans are added to the bag. If one bean is randomly chosen from the bag,
what is the probability that this bean is red?
First correct answer will get marked brainiest

the answer is 890898 beans find this by 500 x 900

Step-by-step explanation:

70

step-by-step explanation:

2*5=10

4*15=60

60+10=

70

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)