- #1

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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

- Thread starter thomas49th
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- #1

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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

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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

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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

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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

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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

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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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Wrong.sorry, my bad

square root the whole RHS

WHAT must you take the square root of?

- #14

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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

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So heres 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

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So heres 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

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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

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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

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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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How would that come about?would changing it to

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

- #25

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Would this be considered simplyfying?

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

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