- #1

thomas49th

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A = πr² + r²root(k²-1)

i need to rearrange it to find r

i go it to

r³ = (2A/π+root(k²-1)

to get just r (with no powers) what will the final equation look like and why.

Thanks

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- Thread starter thomas49th
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- #1

thomas49th

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A = πr² + r²root(k²-1)

i need to rearrange it to find r

i go it to

r³ = (2A/π+root(k²-1)

to get just r (with no powers) what will the final equation look like and why.

Thanks

- #2

arildno

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Eeeh?

Your original equation is:

[tex]A=\pi{r}^{2}+r^{2}\sqrt{k^{2}-1}[/tex]

Agreed?

Your original equation is:

[tex]A=\pi{r}^{2}+r^{2}\sqrt{k^{2}-1}[/tex]

Agreed?

- #3

thomas49th

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yeh, that's the one

- #4

arildno

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So, what is a common factor between the two terms on the right-hand side?

- #5

thomas49th

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r² is the common factor

- #6

arildno

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So, you may rewrite your equation as:

[tex]r^{2}(\pi+\sqrt{k^{2}-1})=A[/tex]

What would you do next?

[tex]r^{2}(\pi+\sqrt{k^{2}-1})=A[/tex]

What would you do next?

- #7

thomas49th

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cross multiply?

[tex]r^{2} = A/(\pi+\sqrt{k^{2}-1})[/tex]

[tex]r^{2} = A/(\pi+\sqrt{k^{2}-1})[/tex]

- #8

arildno

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1. Now, does this equal what you posted before?

2. Since you now know the SQUARE of a number, how do we get what the number itself is?

- #9

thomas49th

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do you square the whole RHS?

- #10

HallsofIvy

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You want to **get rid of** the square on r.

What is the**opposite** of squaring?

What is the

- #11

thomas49th

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sorry, my bad

square root the whole RHS

square root the whole RHS

- #12

arildno

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Do you know what an equation is, and what is allowed to do with one?

- #13

arildno

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sorry, my bad

square root the whole RHS

Wrong.

WHAT must you take the square root of?

- #14

thomas49th

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isn't the answer:

[tex]r= \sqrt{A/(\pi+\sqrt{k^{2}-1})}[/tex]

[tex]r= \sqrt{A/(\pi+\sqrt{k^{2}-1})}[/tex]

- #15

arildno

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2. Put parentheses about the correct radicand.

- #16

thomas49th

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So here's another one:

v² - av² = u²

v²(1 - a) = u²

v² = [tex]u²\1-a[/text]

v = [tex]\sqrt{u²/1-a}[/tex]

am I right?

- #17

thomas49th

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So here's another one:

v² - av² = u²

v²(1 - a) = u²

v² = [tex]u^{2}/1-a[/tex]

v = [tex]\sqrt{u^{2}/1-a}[/tex]

am I right?

- #18

arildno

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Is it that hard to get?

Secondly, in the prior exercise I assumed that "r" was a radius, and hence necessarily a non-negative quantity (you didn't say).

Now, must "v" be a non-negative quantity?

- #19

thomas49th

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v is meaningless, I'm just praticing rearranging the formula

- #20

arildno

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Is it "meaningless"?

Is it not even a number?

Is it not even a number?

- #21

thomas49th

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Did i get the question right?

[tex]v = \sqrt{u^{2}/(1-a)}[/tex]

- #22

arildno

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v can be either of the two numbers:

[tex]v=\pm\sqrt{\frac{u^{2}}{1-a}}[/tex]

[tex]v=\pm\sqrt{\frac{u^{2}}{1-a}}[/tex]

- #23

thomas49th

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would changing it to

[tex]v=\frac{u}\sqrt{1-a}}[/tex]

be simplfying?

[tex]v=\frac{u}\sqrt{1-a}}[/tex]

be simplfying?

- #24

arildno

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would changing it to

[tex]v=\frac{u^{2}}\sqrt{{1-a}}[/tex]

How would that come about?

- #25

thomas49th

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Would this be considered simplyfying?

[tex]v=\frac{u}\sqrt{(1-a)}[/tex]

- #26

arildno

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Almost; but you forget you have TWO solutions for v:

[tex]v=\pm\frac{u}{\sqrt{1-a}}[/tex]

[tex]v=\pm\frac{u}{\sqrt{1-a}}[/tex]

- #27

thomas49th

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- #28

arildno

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That's a matter of taste, mostly.

The first expression is about as simple; however, most would regard the square root of a square (i.e, your numerator) as a non-simplified expression.

- #29

thomas49th

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How would this equation go then...

[tex]v^{2} = u^{2} + a^{2\5}[/tex] find x

[tex]v^{2} = u^{2} + a^{2\5}[/tex] find x

Last edited:

- #30

arildno

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Eeh, what x?

The one under the table, or the one NOT appearing in your equation??

The one under the table, or the one NOT appearing in your equation??

- #31

thomas49th

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[tex]v^{2} = u^{2} + av^{\frac{x}5}

1. Rearrange to x

2. Rearrange to get v

- #32

arildno

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1. Easy

2. Forget it.

2. Forget it.

- #33

thomas49th

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okay. If you say it's easy ill believe you.

Just, somthing about surds.

Simplify

http://www.bbc.co.uk/schools/gcsebitesize/img/ma_surd25.gif" [Broken]

can you show me the steps you did it in

Just, somthing about surds.

Simplify

http://www.bbc.co.uk/schools/gcsebitesize/img/ma_surd25.gif" [Broken]

can you show me the steps you did it in

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- #34

arildno

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I don't see that any rewriting of that expression is any simpler than the one given.

- #35

thomas49th

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youre not suppost to leave surds on the bottom. Apparently the answer should be

http://www.bbc.co.uk/schools/gcsebitesize/img/ma_surd28.gif" [Broken]

But I looked at their method and it looked dogdy. Is the answer

[tex]\frac{3(\sqrt{6} - \sqrt{2})}4[/tex]

as the link above looks like it's 3/4 multiplied by root 6 - root 2

http://www.bbc.co.uk/schools/gcsebitesize/img/ma_surd28.gif" [Broken]

But I looked at their method and it looked dogdy. Is the answer

[tex]\frac{3(\sqrt{6} - \sqrt{2})}4[/tex]

as the link above looks like it's 3/4 multiplied by root 6 - root 2

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