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Math  /  Algebra

QuestionThe table represents a linear function. The rate of change between the points (−5,10)(-5,10) and (−4,5)(-4,5) is -5 . What is the rate of change between the points (−3,0)(-3,0) and (−2,−5)(-2,-5) ? \begin{tabular}{|c|c|} \hlinexx & yy \\ \hline-5 & 10 \\ \hline-4 & 5 \\ \hline-3 & 0 \\ \hline-2 & -5 \\ \hline \hline \end{tabular} −5-5 −15-\frac{1}{5} 15\frac{1}{5} 5 Mark this and return Save and Exit Next Submit

Studdy Solution

STEP 1

1. The table given represents a linear function, which means the rate of change (slope) between any two points on the line is constant.
2. The rate of change (slope) between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is calculated using the formula: $ m = \frac{y_2 - y_1}{x_2 - x_1} \]

STEP 2

1. Verify the rate of change between the points (−5,10)(-5, 10) and (−4,5)(-4, 5).
2. Calculate the rate of change between the points (−3,0)(-3, 0) and (−2,−5)(-2, -5).

STEP 3

Verify the rate of change between the points (−5,10)(-5, 10) and (−4,5)(-4, 5):
m=5−10−4−(−5)=−51=−5m = \frac{5 - 10}{-4 - (-5)} = \frac{-5}{1} = -5
This confirms the given rate of change is correct.

STEP 4

Calculate the rate of change between the points (−3,0)(-3, 0) and (−2,−5)(-2, -5):
m=−5−0−2−(−3)=−51=−5m = \frac{-5 - 0}{-2 - (-3)} = \frac{-5}{1} = -5
The rate of change between the points (−3,0)(-3, 0) and (−2,−5)(-2, -5) is −5-5.
−5\boxed{-5}

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