It is the gradient of the straight line joining the origin to any point on the graph. Thus, if A = (p,q) is any point on the graph, the average unit rate between the origin and A is q/p (provided p is non-zero).
you can compare two measurements using ratios to find the unit rate.
the unit rate of this problem that we need to find our deals with 1
To find the unit rate, divide the price by the number of items. $5.60 / 7 = $0.80. The unit rate is 80 cents.
If you are given a rate of x to y then the equivalent unit rate is x/y to 1.
It is the gradient of the straight line joining the origin to any point on the graph. Thus, if A = (p,q) is any point on the graph, the average unit rate between the origin and A is q/p (provided p is non-zero).
run 2.3km in 7min find unit rate
Divide the ordinate (y-coord) of any point on the graph by its abscissa (x-coord).
Find the slope of the tangent to the graph at the point of interest.
you can compare two measurements using ratios to find the unit rate.
the unit rate of this problem that we need to find our deals with 1
To find the unit rate, divide the price by the number of items. $5.60 / 7 = $0.80. The unit rate is 80 cents.
If you are given a rate of x to y then the equivalent unit rate is x/y to 1.
You find the average rate of change of the function. That gives you the derivative on different points of the graph.
Differentiate the graph with respect to time.
It is the gradient of the line. Pick any two points on the line. Measure the RISE = difference in their heights (distances from the x-axis), and the RUN = difference in their horizontal displacements (distances from the y-axis). The unit rate is RISE/RUN.
Division.