Question

Pretty please help me? :)

Answer 1

All ou have to do is apply the sqr rt to all!
4c^2 - c + 3c + 3c!!
Like rt a^2 = a
Like that OKAY??

Answer 2

@AkashdeepDeb Please solve it step by step... I can't get the answer

Answer 3

\[\sqrt (16c^4) - \sqrt c^2 + 3 \sqrt c^2 + \sqrt (9c^2)\]
\[4c^2 - c + 3c + 3c\]
\[4c^2 + 5c\]
That is how you do it!
Find the sqrt.

Answer 4

Do you get it?

Answer 5

@AkashdeepDeb Yes! Will you help me with another problem?

Answer 6

Kudos? pleaseee
Yeah SURE!

Answer 7

These types confuse me. Perhaps because of the way its written. Step by step please? I don't just want answers I want to learn :)

Answer 8

\[\large \sqrt{16c^4} - \sqrt{c^2}+3\sqrt{c^2}+ \sqrt{9c^2}\]\[\large \sqrt{\color{green}{4\cdot 4 \cdot c \cdot c \cdot c \cdot c}} - \sqrt{\color{green}{c \cdot c}} +3\sqrt{\color{green}{c \cdot c}} + \sqrt{\color{green}{3 \cdot 3 \cdot c \cdot c}}\] Now pull out all the pairs of numbers that are highlighted, for each pair, 1 will come outofthe square root. \[\large 4c^2 -c+3c+3c\] Simplify

Answer 9

Remember rt a * rt b = rt. (ab) [KEEPING THAT IN MIND]
That answer to this is
8 rt. (2t) - 7 rt. (t)
Are you getting it?

Answer 10

\[\large \sqrt{t}(8\sqrt{2}-7)\] Distributive property . multiplying 2 numbers under a square root results in them both being under a square root. \[\large 8\sqrt{2t} -7\sqrt{t}\]

Answer 11

Yes, okay. Those kind of problems have always been weird to me. I know you both are probably tired on me but I still need help :( Are you willing to help more? It's on different kinds of problems now

Answer 12

Its easier if you post new problems on a new post

Answer 13

I've been doing that... I don't get good answers or people who can explain it as well as you both did :\ Thanks for helping me x

Answer 14

No problem :)