Question

PLEASE HELP!!! 2 Algebra questions. Will fan and medal

Answer 1

Simplify the radical show all your steps please \sqrt{363} -3\sqrt{27}\]

Answer 2

\[\sqrt{363}-3\sqrt{27}\]

Answer 3

@mathmale @Kainui @samigupta8

Answer 4

Please somebody

Answer 5

Okay so first, lets simplify these radicals by thinking how we can rewrite them

For example \(\large \sqrt{27} = \sqrt{9 \times 3}\)

We can now look at this as
\[\large \sqrt{9} \sqrt{3}\]

And what do we know about \(\large \sqrt{9}\) ?

Answer 6

It's square root is 3?

Answer 7

Correct so we go from

\[\large \sqrt{27}\]
to
\[\large \sqrt{9 \times 3} = \sqrt{9} \times \sqrt{3}\]
to
\[\large 3\sqrt{3}\]

The reasoning we went through that process is because...we know 27 does not have a square root...so we need to simplify it into smaller factors that we can then work with that WILL have square roots

Answer 8

Let me know if that makes sense before we move on :)

Answer 9

It does :)

Answer 10

Great! Okay stay with me

The original problem was

\[\large \sqrt{363} - 3\sqrt{27}\]

What we JUST did was simplify \(\large \sqrt{27}\) into \(\large 3\sqrt{3}\) right?

So I'm rewriting this to make sure we don't forget the original problem had \(\large 3\sqrt{37}\)

so that answer we just found needs to be multiplied by 3 right? so we have

\(\large 3\sqrt{27} = 9\sqrt{3}\) right?

Answer 11

Typo up there...that 37 should be 27 >.<

Answer 12

Ok

Answer 13

Alright so...now we need to tackle the first radical

\[\large \sqrt{363}\]

What numbers *think of one we know the square root of* can be multiplied to make this number?

Answer 14

Um...3? I do not know honestly. Also I have to leave somewhere in a couple minutes..

Answer 15

I'm guessing 3 would be one

Answer 16

You are absolutely correct....3 is going to be involved...but WHAT times 3 = 363?

\[\large \sqrt{363} = \sqrt{121 \times 3} = \sqrt{121} \times \sqrt{3}\]

Right? well guess what? \(\large \sqrt{121} = 11\) right?

Answer 17

Yeah!

Answer 18

So now we have, Altogether
\[\large 11\sqrt{3} - 9\sqrt{3}\]
Which we know is....?

Answer 19

\[2\sqrt{3}\]

Answer 20

Exactly! :)

Answer 21

Tysm!!! Last one, I will go ahead and fan you incase i don't get the answer immediately :(
Which functions is best shown on the graph below,
a. y=2sqrtx-6-1
b. y=2sqrtx+6+1
c. y=2sqrtx-6+1
d. y=2sqrtx+6-1

Answer 22

Oh no huge typo hold on

Answer 23

a. y=sqrt2x-6-1
b. y=sqrt2x+6+1
c. y=sqrt2x-6+1
d. y=sqrt2x+6-1

Answer 24

sqrt as in square root as in a *insert radical sign*

Answer 25

Alright so it is
a) \(\large y=\sqrt{2x -6} - 1\)
b) \(\large y=\sqrt{2x + 6} + 1\)
c) \(\large y=\sqrt{2x -6} + 1\)
d) \(\large y=\sqrt{2x +6} - 1\)

Right?

Answer 26

So here, we start with the BIG overall function and work forward

Think about what the graph of \(\large y = \sqrt{x}\) looks like

Answer 27

Yes

Answer 28

Now think about what would happen if we multiplied each of those 'x' by 2 to give us the graph of \(\large y=\sqrt{2x}\)



*Still looks basically the same..just a little higher in amplitude

Answer 29

Now comes the parts you will SEE

If we add or subtract a number from that 'x' ***NOT FROM THE WHOLE FUNCTION*** we shift the graph Left or Right Respectively

Example...if I add 2 to the 'x' and I now have the graph of \(\large y = \sqrt{2x + 2}\) it would look like

Answer 30

Add a number = shift to the left
Subtract a number = shift to the right

So HERE we can clearly see that the graph you want has been shifted to the right correct? So are we adding or subtracting 6 under the radical?

Answer 31

substrcct

Answer 32

Exactly...so now, so far, we have the equation as \(\large y = \sqrt{2x - 6}\)

and that looks like

Answer 33

Finally, your graph has been shifted UP 1 unit...*it begins at y = 1 instead of y = 0

This vertical shift comes from adding or subtracting a number ***FROM THE FUNCTION THIS TIME*** which will shift the graph up or down respectively

Adding a number = shift UP
Subtracting a number = shift DOWN

So are we to add or subtract the 1 from the function?

Answer 34

add

Answer 35

Anddd that allows us to say the function graphed is \(\large y = \sqrt{2x - 6} + 1\)

Answer 36

TY!!

Answer 37

Not a problem!

Answer 38

Hi! I noticed your comment on another post saying you needed help. Do you still need any or are you good? (: