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Kobotan [32]
3 years ago
14

How do you convert 60 miles per hour is how many feet per second? can iny one help

Mathematics
1 answer:
DiKsa [7]3 years ago
8 0
That's two questions.

The answer to the second question is:  Yes.

The answer to the first question is:

Multiply 60 miles per hour by ' 1 ' a few times.  Use fractions that have
the same thing on top and bottom, since those are always equal to ' 1 '.
Then cancel units if the same unit is on the top and the bottom.
Here's how it goes:

(60 mi/hr) x (5280 ft/mile) x (1hr/3600 sec) = (60 x 5280 / 3600) ft/sec = <u>88 ft/sec</u>

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3 years ago
If g (x) = 1/x then [g (x+h) - g (x)] /h
lys-0071 [83]

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\dfrac{-1}{x(x+h)}, h\ne 0

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  \begin{aligned} \\\frac{g(x+h)-g(x)}{h} &= \frac{1}{h} \cdot [g(x+h) - g(x)] \\&= \frac{1}{h} \left( \frac{1}{x+h} - \frac{1}{x} \right)\end{aligned}

Technically we are done, but some more simplification can be made. We can get a common denominator between 1/(x+h) and 1/x.

  \begin{aligned} \\\frac{g(x+h)-g(x)}{h} &= \frac{1}{h} \left( \frac{1}{x+h} - \frac{1}{x} \right)\\&=\frac{1}{h} \left(\frac{x}{x(x+h)} - \frac{x+h}{x(x+h)} \right) \\ &=\frac{1}{h} \left(\frac{x-(x+h)}{x(x+h)}\right) \\ &=\frac{1}{h} \left(\frac{x-x-h}{x(x+h)}\right) \\ &=\frac{1}{h} \left(\frac{-h}{x(x+h)}\right) \end{aligned}

Now we can cancel the h in the numerator and denominator under the assumption that h is not 0.

  = \dfrac{-1}{x(x+h)}, h\ne 0

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