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2 years ago

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 = 2 2/9

1/9-20 = 19 8/9

Step-by-step explanation:

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

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