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stich3 [128]
3 years ago
13

Witch is greater 4 cups or 2 quarts

Mathematics
1 answer:
mario62 [17]3 years ago
8 0
2 quarts because 4 cups only goes into 1 quart and 2 quarts is a greater number then 1 so 2 quarts
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(-8q^3r^4s^2)^2 =

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Answer:
Area of the shaded region is 200 mm squared

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What estimate can you make for 1/2 and 4/6?
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Find f(a), f(a+h), and<br> 71. f(x) = 7x - 3<br> f(a+h)-f(a)<br> h<br> if h = 0.<br> 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 <em>h</em> 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

4 0
2 years ago
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