Question

The points \((-1.3, 2.6)\) and \((3.3, 9.5)\) are on the graph of a linear relationship between two variables, $x$ and $y$. a. What is the constant rate of change of $y$ with respect to $x$? b. Write a formula that expresses $y$ in terms of $x$. c. What is the value of $y$ when $x = 110$? $y = $

          The points \((-1.3, 2.6)\) and \((3.3, 9.5)\) are on the graph of a linear relationship between two variables, $x$ and $y$.
a. What is the constant rate of change of $y$ with respect to $x$?
b. Write a formula that expresses $y$ in terms of $x$.
c. What is the value of $y$ when $x = 110$?
$y = $
        
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The points (-1.3, 2.6) and (3.3, 9.5) are on the graph of a linear relationship between two variables, x and y.
a. What is the constant rate of change of y with respect to x?
b. Write a formula that expresses y in terms of x.
c. What is the value of y when x = 110?
y =

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Precalculus with Limits
Precalculus with Limits
Ron Larson 2nd Edition
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The points (-1.3, 2.6) and (3.3, 9.5) are on the graph of a linear relationship between two variables, x and y. a. What is the constant rate of change of y with respect to x? b. Write a formula that expresses y in terms of x. c. What is the value of y when x = 110? y =
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Transcript

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00:01 So given points we have here that is minus 1 .8 comma 2 .6 and second point we have here that is and second point is 2 .8 comma 10 .42.
00:19 So now we need to find here the rate of change of y with respect to x.
00:26 So it becomes then x is equal to y2 minus y1 upon x2 minus x1.
00:40 So let's say this is our x1 point, this is our y1 point, this is our x2 point and this is our y2 point.
00:48 This is 10 .42.
00:50 So after putting here value we will get that is 10 .42 minus 2 .6 upon 2 .8 plus 1 .8.
01:11 So x become from here that is 7 .82 upon 4 .6.
01:21 So after simplify this it becomes 1 .7.
01:25 So rate of change here that is 1 .7.
01:32 So this will be our answer...
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