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emmainna [20.7K]
3 years ago
11

Rearrange the formula A = Θr2 2 for Θ.

Mathematics
2 answers:
In-s [12.5K]3 years ago
5 0

Answer:

\theta=\frac{2A}{r^{2} }

Step-by-step explanation:

The given formula is

A=\frac{\theta r^{2} }{2}

We have to solve the formula for \theta. To do that, we first need to past the divisor 2 to the other side, as follows

2A=\theta r^{2}

Then, we have to move the squared radius to the other side with the opposite operation, that is, dividing

\frac{2A}{r^{2} } =\theta \\\theta=\frac{2A}{r^{2} }

Threfore, the given expression solved for \theta is

\theta=\frac{2A}{r^{2} }

Lina20 [59]3 years ago
3 0

Answer:

\theta=\frac{2A}{r^2}

Step-by-step explanation:

Given : The formula A=\frac{\theta r^2}{2}

We have to rearrange the given formula for \theta

Consider the given formula A=\frac{\theta r^2}{2}

Multiply both side by 2, we have,

2A={\theta r^2

Divide both side by r^2 , we have,

\frac{2A}{r^2}= \frac{\theta r^2}{r^2}

Simplify, we get,

\frac{2A}{r^2}=\theta

Thus, \theta=\frac{2A}{r^2}  

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Step-by-step explanation:

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3 years ago
4. Using the geometric sum formulas, evaluate each of the following sums and express your answer in Cartesian form.
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Answer:

\sum_{n=0}^9cos(\frac{\pi n}{2})=1

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Step-by-step explanation:

\sum_{n=0}^9cos(\frac{\pi n}{2})=\frac{1}{2}(\sum_{n=0}^9 (e^{\frac{i\pi n}{2}}+ e^{\frac{i\pi n}{2}}))

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2nd

\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=\frac{1-e^{\frac{i2\pi N}{N}}}{1-e^{\frac{i2\pi}{N}}}

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3th

\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})==\frac{1}{2}(\sum_{n=0}^\infty ((\frac{e^{\frac{i\pi n}{2}}}{2})^n+ (\frac{e^{-\frac{i\pi n}{2}}}{2})^n))

=\frac{1}{2}(\frac{1-0}{1-i}+\frac{1-0}{1+i})=\frac{1}{2}

What we use?

We use that

e^{i\pi n}=cos(\pi n)+i sin(\pi n)

and

\sum_{n=0}^k r^k=\frac{1-r^{k+1}}{1-r}

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