Question

Check please!! vv
3x-[5-4(2x-7)]
my answer was x-7

Answer 1

also I have...\[\sqrt{16x^2}\times \sqrt{3x^3}\]
my answer was...\[4x^2\sqrt{3x}\]

Answer 2

then the first answer is x+7 ?

Answer 3

first I did parentheses 5-4 = 1 then keeping in parentheses distribute -1 to 2x -7
then 3x-2x =x and (-1)(-7) = 7....right??

Answer 4

3x - [5 - 4 (2x - 7)]
The most inside parentheses first so distribute the 4
3x - [5 - 8x - 28]

Answer 5

if you are distributing -4 times -7 wouldn't it be +28 ?

Answer 6

Yup.

Answer 7

then would you just add/subtract like terms?

Answer 8

3x - [5 - 8x + 28]
combine like terms inside parentheses
3x - [ - 8x + 33]
distribute negative
3x + 8x - 33

Answer 9

okay thank you!! Can you also check my work for my second problem way up there?

Answer 10

Is that sqrt (16x^2) * sqrt (3x^2)? I can't really read it.

Answer 11

If it's sqrt (16x^2) * sqrt (3x^3) then it's good.

Answer 12

yes it's sqrt 3x^3
thank you both!

Answer 13

:)