Question

\[2x \sqrt{36-x ^{2}} \] Solve the area equation for x=3

HELP ME SOLVE!!!

Answer 1

write
\[2\times 3\times \sqrt{36-3^2}\] and compute

Answer 2

\[6\sqrt{36+9}\]

Answer 3

@satellite73 ^ ?

Answer 4

no

Answer 5

\[36-3^2=36-9=27\]

Answer 6

so would it be 12\[\sqrt{5}\]

Answer 7

lets go slow

Answer 8

12\[\sqrt{5}\]

Answer 9

the first step is \[2\times 3\times \sqrt{36-3^2}\]

Answer 10

then you compute and get
\[6\sqrt{27}\] because \(36-9=27\)

Answer 11

Ok, this helps going step by step.

Answer 12

since \(27=9\times 3\) you have \(\sqrt{27}=\sqrt{9}\times \sqrt{3}\) giving \[\sqrt{27}=3\sqrt{3}\]

Answer 13

Would you then multiply the 3 and 6?

Answer 14

so now you have
\[6\times 3\sqrt{3}=18\sqrt{3}\] as a final answer

Answer 15

yes

Answer 16

Blah! Now the next part is awfully tricky.

Answer 17

A(x)=2x\[\sqrt{36-x ^{2}}\]

Answer 18

x=\[\sqrt{3}\]

Answer 19

@satellite73 What about this?

Answer 20

\[A(x)=2x\sqrt{36-x^2}\]
\[A(\sqrt3)=2\sqrt{3}\times \sqrt{36-(\sqrt{3})^2}\]

Answer 21

you get
\[A(\sqrt{3}=2\sqrt{3}\sqrt{36-3}=2\sqrt{3}\sqrt{33}=2\sqrt{66}\]

Answer 22

okay i made a mistake

Answer 23

How did you end up with 66 inside the square root? I'm new to all of this. :-/

Answer 24

it is
\[2\sqrt{3}\sqrt{33}=2\sqrt{99}\]

Answer 25

yeah i was wrong

Answer 26

\[\sqrt{a}\sqrt{b}=\sqrt{ab}\]

Answer 27

How did you get the 99, just multiply 3 and 33?

Answer 28

and \(3\times 33=99\)

Answer 29

yes

Answer 30

but you are still not done

Answer 31

Ok, let's move forward.

Answer 32

since \(99=9\times 11\) you get
\[\sqrt{99}=\sqrt{9}\sqrt{11}=3\sqrt{11}\]

Answer 33

for a final answer of
\[6\sqrt{11}\]

Answer 34

This material is complicated.

Answer 35

Thank you for helping. :)

Answer 36

yw