Question

http://prntscr.com/de940u
Am I correct?

Answer 1

@pooja195

Answer 2

@Nnesha

Answer 3

for some reason i can't open that link.

Answer 4

@Nnesha There :)

Answer 5

thanks

Answer 6

can i see your work? how did you get 41sqrt{3} ?

Answer 7

Umm... :/ lol
Ehhhhh...

Wait. Let me actually try and do the work bleh haha
Cuz tbh, the teacher went over this last week with us but I didn't do that hw so she gave it back to me... >.>
Can you show me how to do it? :)

Answer 8

first simplify the sqrt{} \[\sqrt{27}\]
factor 27
what are the factor of 27 ? (where one number sholuld be the perfect square root)

Answer 9

9

Answer 10

9 and 3
\[7 \sqrt{9 \cdot 3}\]
which can be written as \[7\cdot \sqrt{9}\cdot \sqrt{ 3}\]
take the square root of 9

Answer 11

3

Answer 12

\[\sqrt{48}\]
same with this
what are the factors of 48 ?

Answer 13

let me switch laptops real quick brb

Answer 14

yes right
\[7 \color{Red}{\sqrt{9}} \cdot \sqrt{3}\]\[7 \cdot \color{Red}{3} \cdot \sqrt{3}\]

Answer 15

factors of 48... 6 and 8

Answer 16

correct.
but 6 & 8 aren't the perfect square root

Answer 17

Hmmmmm... 12 and 4

Answer 18

ok. what are the factors of 12 ???

Answer 19

3 and 4

Answer 20

\[5\sqrt {4 \cdot 3 \cdot 4 }\]simplify

Answer 21

How? :/ That's basically 5sqroot48 again? :/

Answer 22

Oh wait!! i think it's 20 sqroot 12

Answer 23

\[7 \sqrt { 9 \cdot 3} \rightarrow 7 \sqrt{9} \sqrt{3} \rightarrow 7 \cdot 3 \sqrt{3}\]

Answer 24

\[\sqrt{ a \cdot b } \rightarrow \sqrt{a}{ \cdot \sqrt{b}}\]

Answer 25

Okay so \[60\sqrt{4}\]

Answer 26

factors of 48 = 12 , 4
factors of 12 = 4, 3
so we can rewrite 12 ,4 as \[5\sqrt {4 \cdot 3 \cdot 4 }\]
4 is perfect square
so take the square root of each 4
remember you can rewrite that radical as \[5 \cdot \sqrt{4} \cdot \sqrt{3}\cdot \sqrt{4}\]

Answer 27

So 5 x 2 x 3 x 2?

Answer 28

3?

Answer 29

\[\[5\times2\times \sqrt{3}\times2\]

Answer 30

yes looks good

Answer 31

So then 5 x 2 = 10 then \[10\sqrt{5}\]

Answer 32

\[7 \cdot 3 \color{Red}{\sqrt{3}} + 5 \cdot 2 \cdot 2 \color{Red}{\sqrt{3}}\]

Answer 33

well here is easy way to find the factor whenever you have to simplify radical
take the highest perfect square factor

Answer 34

\[62\sqrt{3~or~6}\]

Answer 35

1, 2, 3, 4, 6, 8, 12, 16, 24 and 48 (Factors of 48)
here HIGHEST perfect square root factor is 16
16 times 3 = 48
\[5 \sqrt{16 \cdot 3} \rightarrow 5 \cdot \sqrt{16} \cdot \sqrt{3} \rightarrow 5 \cdot 4 \cdot \sqrt{3}\]

Answer 36

Okay got that

Answer 37

so it's 5 times 4 not 2

Answer 38

\[7 \times 3 \sqrt{3}+ 5 \times 4 \sqrt{3}\]
multiply the the numbers and then combine like terms

Answer 39

41

Answer 40

just 41?

Answer 41

\[41\sqrt{3}\]

Answer 42

oh yes!!

Answer 43

Oh wait.. We're adding the square roots so it's \[41\sqrt{6}\]

Answer 44

\[ \color{blue}{7 \cdot 3}\color{Red}{ \sqrt{3}} + \color{blue}{ 5 \cdot 4}\color{Red}{ \sqrt{3}}\]
\[ \color{blue}{21}\color{Red}{ \sqrt{3}} + \color{blue}{ 20}\color{Red}{ \sqrt{3}}\]
sqrt{3} is just like variable
2x+2x=4x( just add the coefficient )
\[ \color{blue}({21+20)}\color{Red}{ \sqrt{3}}\]

Answer 45

So it is sqrt 3 or 6? :/

Answer 46

no we don't combine the sqrt
it would stay the same

Answer 47

Okay... Wow these are like very hard!
Thank you for helping me. :)

Answer 48

So basically we just simplified the square roots and then added the bases?

Answer 49

yeah harder than polar stuff

Answer 50

polar stuff is soooo hard :))

Answer 51

sorry.. i meant to say *yes sir, it is hard* o^_^o

Answer 52

-_-

Answer 53

poolar coordinates*

Answer 54

polar equations *