3 years ago

Estimate length using compensation

Mathematics

1 answer:

Murljashka [212]3 years ago

5 0

What do you have to estimate?

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Y=f(x)=2^x find f(x) when x=1

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$f(x) = 2^x$

When x = 1

$f(1) = 2^{1} = 2$

Answer : f(1) = 2

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3 years ago

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Step-by-step explanation:

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2 years ago

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3 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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3 years ago