Colin predicted whether he got answers right or wrong in his 50 question exam. He identified the 11 questions he thought he got wrong. It turns out that Colin got 5 questions right that he thought he got wrong. Colin also got a total score of 34 out of 50 in the test. What is the percentage accuracy he had with predicting his scores?

54.54..%

Step-by-step explanation:

I'm not 100 percent sure but I believe this is the right answer

if you read over the problem it tells you he got 6 out of 11 of the questions right. the questions says that Colon thought 11 were incorrect but 5 were actually correct so 11-5 is how I got 6 out 11

and to make the fraction 6/11 a percentage you divide 6 by 11 and I got 0.545454..

Next multiple by 100 l, so 54% with a repeating decimal of 54.

my guess would be b and d, but i didn’t actually get out the pen and paper. lol

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)