4. It cannot be a constant or unit rate relationship because the value of k = y/x vary across table values.
5. The relationship in Tuesday's table can be described by a constant unit rate.
<h3>What is a Constant Unit rate Relationship?</h3>
A constant unit rate relationship can be described as a relationship between two input and output variables, x and y, in such a way that, k = y/x. "k" is the constant of proportionality of the relationship or the unit rate between the two variables.
4. We have the following:
5/20 = 1/4
10/40 = 1/4
12/60 = 1/5
14/80 = 7/40
16/100 = 4/25
Thus, this cannot be a constant or unit rate relationship because the value of k = y/x vary across table values.
5. 4/20 = 1/5
8/40 = 1/5
12/60 = 1/5
16/80 = 1/5
20/100 = 1/5
Thus, the value of k is the same across all table values, therefore, the relationship in Tuesday's table can be described by a constant unit rate.
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Answer:
D
Step-by-step explanation:
eq. of given line is x+y=10
eq. of any line parallel to given line is x+y=a
if it passes through (-1,5) then -1+5=a
or a=4
reqd. eq of line is x+y=4
or y=-x+4
58,927 - move decimal 4 places to the left
5.8927
Answer:
Three ways to find the slope of a line: You may have two points #(x_1,y_1)# and #(x_2,y_2)# (often one or both of these points may be intercepts of the #x# and/or #y# axes). The slope is given by the equation. #m=(y_2-y_1)/(x_2-x_1)#. You may have a linear equation that is either in the form or can be manipulated into the form. #y = mx + b#.
Step-by-step explanation:
