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Zielflug [23.3K]
3 years ago
10

Which inequality correctly compares Four-fifths, One-fourth, and Four-ninths?

Mathematics
2 answers:
tatiyna3 years ago
8 0

Answer:

a

Step-by-step explanation:

TiliK225 [7]3 years ago
5 0

Answer:

\frac{4}{5} > \frac{4}{9} > \frac{1}{4} or \frac{1}{4} < \frac{4}{9} < \frac{4}{5}

Step-by-step explanation:

Lets find the decimal equivalent of each fraction, and then compare them.

Four-fifths:

4 is lower than 5, so the integer result is 0.

The decimal part is given by 40/5 = 8.

So 0.8

One-fourth

1 is lower than 4, so the integer result is 0.

The decimal part is given by 10/4 = 2.5.

So 0.25

Four-ninths

4 is lower than 5, so the integer result is 0.

The decimal part is given by 40/9 = 4.444.

So 0.4444

These following inequalities can combate them:

\frac{4}{5} > \frac{4}{9} > \frac{1}{4} or \frac{1}{4} < \frac{4}{9} < \frac{4}{5}

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====================================================

Work Shown:

\displaystyle \lim_{\Delta x \to 0^{+}} \frac{\frac{9}{x+\Delta x} - \frac{9}{x}}{\Delta x}\\\\\\\displaystyle \lim_{\Delta x \to 0^{+}} \frac{\frac{9x}{x(x+\Delta x)} - \frac{9(x+\Delta x)}{x(x+\Delta x)}}{\Delta x}\\\\\\\displaystyle \lim_{\Delta x \to 0^{+}} \frac{\frac{9x-9(x+\Delta x)}{x(x+\Delta x)}}{\Delta x}\\\\\\\displaystyle \lim_{\Delta x \to 0^{+}} \frac{9x-9x-9\Delta x}{x\Delta x(x+\Delta x)}\\\\\\

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