Question

Hello I need help with this equation
simplify (2 rad.8) (3 rad.6) + 19 rad.48

Answer 1

23997878253756106444997335735942416102218125380085922630175553 :)

Answer 2

what does rad.8 mean

Answer 3

to the power of 8

Answer 4

Okay (\[(2\sqrt{8}) (2\sqrt{6}) + 19 \sqrt{48}\]

Answer 5

3 \[sorry (3\sqrt{6 })\]

Answer 6

46\[\sqrt{12}\]

Answer 7

for the first part, you can change the order of the multiplies
\( 2 \cdot 3 \sqrt{8}\cdot \sqrt{6} \)

and use the rule
\[ \sqrt{a} \sqrt{b} = \sqrt{a b} \]

Answer 8

should i simplify

Answer 9

its 46\[\sqrt{12}\]

Answer 10

or 92\[\sqrt{3}\]

Answer 11

if there are any options i can choose

Answer 12

so would it be like \[2 \sqrt{2}\sqrt{4} and 3 \sqrt{2}\sqrt{3} + 19 \sqrt{2}\sqrt{14}\]

Answer 13

ya likely

Answer 14

sorry rathan reddy the answers are 1) 25 rad.3 2) 19 rad. 6 3) 19 rad 12 + 2rad 48 4) 100 rad 3 5) 100 rad 12 + 3 rad 48

Answer 15

yes you could do that (but the "and" is really multiply)

but I was thinking just write
the first part as
\[ 2 \cdot 3 \sqrt{8}\cdot \sqrt{6} = 6 \sqrt{48}\]

so that the problem is
\[ 6 \sqrt{48} + 19 \sqrt{48} \]

you have 6 square roots (of 48) and add 19 more,
how many do you have ?

Answer 16

you should get 25 sqrt(48)

now simplify sqrt(48)

can you do that ?

Answer 17

for 48 I got 1*48 2*14 4*12 6*8

Answer 18

i think its option 3.

Answer 19

that way I simplify square roots is break them into "prime factors" and look for PAIRs of numbers:
I start by dividing by little numbers:
48 = 2* 24 = 2* 2* 12 = 2*2*2*6 = 2*2*2*2*3
we have two pairs of twos: (2*2) * (2*2) * 3
pull each pair outside the square root and put just one of the pair on the outside.
you get
\[ \sqrt{ 2\cdot 2\cdot 2\cdot 2\cdot 3} = 2\cdot 2\sqrt{3}\]

Answer 20

which is 4 sqrt(3)

so you have
25 * 4 * sqrt(3)
which simplifies

Answer 21

sorry its 100 rad.3
i will simplyfy it now

Answer 22

I don't understand the

pull each pair outside the square root and put just one of the pair on the outside.
you get
2⋅2⋅2⋅2⋅3 − − − − − − − − − − √ =2⋅23 √

Answer 23

(2 rad.8)(3 rad.6)+19rad.48
6 rad.48+19 rad.48
25 rad.48
25 rad.16*3
25*4 rad.3
100 rad.3

Answer 24

did you get it

Answer 25

btw, when you said
\[ 2 \sqrt{2}\sqrt{4} and 3 \sqrt{2}\sqrt{3} + 19 \sqrt{2}\sqrt{14} \]
that should be
\[ 2 \sqrt{2}\sqrt{4} \cdot 3 \sqrt{2}\sqrt{3} + 19 \sqrt{2}\sqrt{24} \]
you could continue by saying
\[ 2 \cdot 3 \sqrt{2}\cdot \sqrt{2}\cdot \sqrt{4}\sqrt{3} + 19 \sqrt{2}\sqrt{24} \]
and the left part is 6 * 2 * 2 * sqrt(3)
you would have to simplify the right part (the sqrt(24) ) to finish

Answer 26

Rathanreddy I need you to draw the last 3 parts of your answer please

Answer 27

Phi I am confused with the