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2 years ago

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Mathematics

1 answer:

Andrej [43]2 years ago

7 0

Oh so easy!

The answer is 4!

The question is

WHAT IS 2+2?

It's written in morse code! I used to study it! :)

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2 years ago

Simplify . (1/c + 1/h)/(1/(c ^ 2) - 1/(r ^ 2))

Marrrta [24]

Answer:

$\frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}$

Step-by-step explanation:

$\frac{\frac{1}{c}+\frac{1}{h}}{\frac{1}{c^2}-\frac{1}{r^2}}$

Combine $\frac{1}{c} + \frac{1}{h}$

$\frac{\frac{h+c}{ch}}{\frac{1}{c^2}-\frac{1}{r^2}}$

Combine the bottom, too.

$=\frac{\frac{h+c}{ch}}{\frac{r^2-c^2}{c^2r^2}}$

Apply the fraction rule

$=\frac{\left(h+c\right)c^2r^2}{ch\left(r^2-c^2\right)}$

Cancel

$=\frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}$

Therefore, $\frac{\left(\frac{1}{c}+\frac{1}{h}\right)}{\left(\frac{1}{\left(c^2\right)}-\frac{1}{\left(r^2\right)}\right)}:\quad \frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}$

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2 years ago

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son4ous [18]

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In fact, every polynomial has real and/or complex solutions. If all solutions are real, we're good. But if not all of them are real, then the complex ones come in couple of conjugate solutions. Since $1 + 13i$ and $1 - 13i$ are conjugate complex numbers, if one of them is a solution, the other must be as well.

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2 years ago

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