Aaron estimated that 148% of 333 is 495. is his estimate reasonable?
a. no, because his calculation should be 140% of 340.
b. no, because his calculation should be 150% of 340.
c. yes, because his calculation should be 140% of 330.
d. yes, because his calculation should be 150% of 330.

D. Yes, because his calculation should be 150% of 330.

Step-by-step explanation:

Here, the given phrase,

148% of 333 is 495,

Since, a number ending with digit 5 or more than 5 is rounded to next 10,

While a number ending with a digit less than 5 is rounded to previous 10,

That is,

148 ≈ 150,

333 ≈ 330

Also, 150% of 330 =

So 148% of 333 ≈ 495,

Hence, option 'D' is correct.

D. Yes, because his calculation should be 150% of 330.

Step-by-step explanation:

Here, the given phrase,

148% of 333 is 495,

Since, a number ending with digit 5 or more than 5 is rounded to next 10,

While a number ending with a digit less than 5 is rounded to previous 10,

That is,

148 ≈ 150,

333 ≈ 330

Also, 150% of 330 =

So 148% of 333 ≈ 495,

Hence, option 'D' is correct.

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)