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3 years ago

Please solve these for 20 points : solving formulas for a variable

Mathematics

1 answer:

Digiron [165]3 years ago

6 0

6. (A/pi = r^2)
7. [(P - 2l)/2 = w]
8. [C/(2pi)= r]
9. (2A/h = b)
10. (E/c^2 = m)

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Mrac [35]

Answer:

2(c+1)

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Step-by-step explanation:

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3 years ago

Rationalize the denominator of $\displaystyle \frac{1}{\sqrt{2} + \sqrt{3} + \sqrt{7}}$, and write your answer in the form\[

Butoxors [25]

9514 1404 393

Answer:

Step-by-step explanation:

Apparently, you want to simplify ...

$\displaystyle \frac{1}{\sqrt{2} + \sqrt{3} + \sqrt{7}}$

so the denominator is rational. It looks like the form you want is ...

$\dfrac{A\sqrt{2} + B\sqrt{3} + C\sqrt{7} + D\sqrt{E}}{F}$

And you want to know the sum A+B+C+D+E+F.

We can start by multiplying numerator and denominator by a conjugate of the denominator. Then we can multiply numerator and denominator by a conjugate of the resulting denominator.

$\displaystyle =\frac{1}{\sqrt{2} + \sqrt{3} + \sqrt{7}}\cdot\frac{\sqrt{2} + \sqrt{3} - \sqrt{7}}{\sqrt{2} + \sqrt{3} - \sqrt{7}}=\frac{\sqrt{2} + \sqrt{3} - \sqrt{7}}{2\sqrt{6}-2}\\\\=\frac{\sqrt{2} + \sqrt{3} - \sqrt{7}}{2\sqrt{6}-2}\cdot\frac{\sqrt{6}+1}{\sqrt{6}+1}=\frac{(1+\sqrt{6})(\sqrt{2}+\sqrt{3}-\sqrt{7})}{10}\\\\=\frac{\sqrt{2}+\sqrt{3}-\sqrt{7}+2\sqrt{3}+3\sqrt{2}-\sqrt{42}}{10}=\frac{4\sqrt{2}+3\sqrt{3}-\sqrt{7}-\sqrt{42}}{10}$

Comparing this to the desired form we have ...

A = 4, B = 3, C = -1, D = -1, E = 42, F = 10

Then the sum is ...

A +B +C +D +E +F = 4 + 3 -1 -1 +42 +10 = 59 -2 = 57

The sum of interest is 57.

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Answer:

Just substitute the values in!
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