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Amiraneli [1.4K]
2 years ago
14

What is the constant rate of change for this equation y = 1/9x-20 ?​

Mathematics
2 answers:
padilas [110]2 years ago
7 0

Answer: The constant rate change would be \frac{1}{9}.

Step-by-step explanation: The general rate of change can be found by using the difference quotient formula. To find the average rate of change over an interval, enter a function with an interval: f (x) = x^{2} , [2,3]

Write y = \frac{1}{9}x - 20 as a function which is f (x) = \frac{1}{9}x - 20

Consider the difference quotient formula which is \frac{f (x +h) - f (x) }{h}.

Find the components of the definition. \frac{f (x+h) = \frac{h}{9} + \frac{x}{9} - 20}simplify then it would be \frac{f (x) = \frac{x}{9} - 20 }.

Lastly plug in all the components.

\frac{f (x + h) - f (x)}{h} = \frac{\frac{h}{9} +\frac{x}{9} - 20 - (\frac{x}{9} - 20) }{h}

After solving all this the answer would be \frac{1}{9}

k0ka [10]2 years ago
3 0

Answer:

like:-

1/9x20 = <u>2 2/9</u>

1/9-20 = <u>19 8/9 </u>

Step-by-step explanation:

<h3><u>I'm not sure</u></h3>

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