18. How can you tell if a fraction will have a repeating decimal

1 answer:

7 0

Terminating and Repeating Decimals. Any rational number (that is, a fraction in lowest terms) can be written as either a terminating decimal or a repeating decimal . Just divide the numerator by the denominator . If you end up with a remainder of , then you have a terminating decimal.

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Find f(a), f(a+h), and 71. f(x) = 7x - 3 f(a+h)-f(a) h if h = 0. 72. f(x) = 5x²

Leni [432]

Answer:

$71. \ \ \ f(a) \ = \ 7a \ - \ 3; \ f(a+h) \ = \ 7a \ + \ 7h \ - \ 3; \ \displaystyle\frac{f(a+h) \ - \ f(a)}{h} \ = \ 7$

$72. \ \ \ f(a) \ = \ 5a^{2}; \ f(a+h) \ = \ {5a}^{2} \ + \ 10ah \ + \ {5h}^{2}; \ \displaystyle\frac{f(a+h) \ - \ f(a)}{h} \ = \ 10a \ + \ 5h$

Step-by-step explanation:

In single-variable calculus, the difference quotient is the expression

$\displaystyle\frac{f(x+h) \ - \ f(x)}{h}$ ,

which its name comes from the fact that it is the quotient of the difference of the evaluated values of the function by the difference of its corresponding input values (as shown in the figure below).

This expression looks similar to the method of evaluating the slope of a line. Indeed, the difference quotient provides the slope of a secant line (in blue) that passes through two coordinate points on a curve.

$m \ \ = \ \ \displaystyle\frac{\Delta y}{\Delta x} \ \ = \ \ \displaystyle\frac{rise}{run}$ .

Similarly, the difference quotient is a measure of the average rate of change of the function over an interval. When the limit of the difference quotient is taken as h approaches 0 gives the instantaneous rate of change (rate of change in an instant) or the derivative of the function.

Therefore,

$71. \ \ \ \ \ \displaystyle\frac{f(a \ + \ h) \ - \ f(a)}{h} \ \ = \ \ \displaystyle\frac{(7a \ + \ 7h \ - \ 3) \ - \ (7a \ - \ 3)}{h} \\ \\ \-\hspace{4.25cm} = \ \ \displaystyle\frac{7h}{h} \\ \\ \-\hspace{4.25cm} = \ \ 7$

$72. \ \ \ \ \ \displaystyle\frac{f(a \ + \ h) \ - \ f(a)}{h} \ \ = \ \ \displaystyle\frac{{5(a \ + \ h)}^{2} \ - \ {5(a)}^{2}}{h} \\ \\ \-\hspace{4.25cm} = \ \ \displaystyle\frac{{5a}^{2} \ + \ 10ah \ + \ {5h}^{2} \ - \ {5a}^{2}}{h} \\ \\ \-\hspace{4.25cm} = \ \ \displaystyle\frac{h(10a \ + \ 5h)}{h} \\ \\ \-\hspace{4.25cm} = \ \ 10a \ + \ 5h$