Question

PLEASE HELP D: I really need help with about 8 questions????

Answer 1

I can try and help with some but I don't know...

Answer 2

yay! Okay so I'll give you the first 3.......

Answer 3

Ok for the first question here:
Answer: (A)

-> [ 98h^3 ] ^(1/2)

-> [ (49h^2) x (2h)]^ (1/2)

the square root of 49 =7 and the square root of h^2 = h

-> 7h x [ (2h) ]^(1/2)

Answer 4

THANK YOU :D

Answer 5

Can you answer any more?

Answer 6

Yes...I'm working on it now

Answer 7

Yay!

Answer 8

on 11 just combine like terms
you can add (3-9)sqrt11 and sqrt6 stays as it is

Answer 9

The answer to second question: B.
Just combine like terms and I did this on my calculator too

Answer 10

okay!

Answer 11

That's all I can do for now...but I'll tag some people...maybe they can help you, cause' I got to go @BulletWithButterflyWings @chaylacryder @Kitt020912 @kyrsten17 @uri there ya go Bye!

Answer 12

okay thanks :)

Answer 13

Those are the questions :)

Answer 14

I got 10.11 answered

Answer 15

All I need now is 12,13,15,16, and 20 :)

Answer 16

for question 12, did you try factoring 52 and 117 ?

Answer 17

Do you mean 13?

Answer 18

yes, I mean Q13.

Answer 19

what is factoring again?

Answer 20

breaking up a number into 2 numbers that multiply together. Hint: one of the numbers will be 13 (if this will simplify)

Answer 21

ohh okay!

Answer 22

so 4*13 and 9*13?

Answer 23

yes. now use the idea
\[ \sqrt{4 \cdot 13} = \sqrt{4} \sqrt{13} \]

Answer 24

is √13√4 the answer?

Answer 25

it's on the way. Are you saying you don't know the square root of 4 ?

Answer 26

its 2.

Answer 27

so sqrt(4) sqrt(13) is the same as 2 sqrt(13)

what about \[ \sqrt{9} \sqrt{13} \]

Answer 28

3 sqrt (13)

Answer 29

so your problem can be written as
\[ \sqrt{52} - \sqrt{13} + \sqrt{117} \\ 2\sqrt{13} - 1 \sqrt{13} + 3 \sqrt{13} \]

if you have 2 square roots of 13, take away 1, and add 3, how many square roots of 13 do you have ?

Answer 30

4!

Answer 31

so the answer is \[ 4 \sqrt{13} \]

Answer 32

Yay! Thanks!

Answer 33

for Q15, you can change the order of the multiplies
\[ 2 \sqrt{15 k} \cdot 6k\sqrt{3} \\ 2 \cdot 6\cdot k \cdot \sqrt{15k} \sqrt{3} \]
you should simplify 2*6*k
you should factor the 15.

Answer 34

1*3*k. Get rid of the 1.
15= 5*3

Answer 35

2*6*k is not 1*3*k

write down what you have so far

Answer 36

so 12*k?

Answer 37

yes, or just 12k (in algebra people leave out the multiply sign when you have a number and a letter.... of course this won't work for two numbers because 2*2 written as 22 is wrong)

but write down what you have, including the radicals.

Answer 38

Okay I am just going to guess on the rest of these questions... Thanks for the help though!! :)

Answer 39

It is one thing to not know this... it is another to be lazy

Answer 40

I know, I just have too much to do and I started late.