Answer:
Question 1;
∴ The y values are 1.49 , 2.98 , 4.47 and 5.96
Question 2;
∴ The y values are 2.00 , 2.8 , 3.6 and 4.4
Question 3;
∴ The y values are -9 , -3 , 3 and 9
Question 4;
∴ y values are[tex] \frac{1}{2}[/tex] , [tex]\frac{3}{4}[/tex] , 1 and [tex]\frac{5}{4}[/tex]
Step-by-step explanation:
Question 1;
Given data,
Apples cost [tex]\$1.49[/tex] per pound.
[tex]x[/tex]= weight in pound
[tex]y[/tex]= Cost in [tex]\$[/tex]
Here the relation between [tex]x[/tex] and [tex]y[/tex] comes;
[tex]y=x\times 1.49[/tex] (equation 1)
Let, Plug all x value in equation 1;
when [tex]x=1[/tex] then [tex]y=1\times 1.49=1.49[/tex],
when [tex]x=2[/tex] then [tex]y=2\times 1.49=2.98[/tex],
when [tex]x=3[/tex] then [tex]y=3\times 1.49=4.47[/tex],
when [tex]x=4[/tex] then [tex]y=4\times 1.49=5.96[/tex],
∴ The y values are 1.49 , 2.98 , 4.47 and 5.96
Question 2;
From question,
Initial fees of cab is [tex]\$[/tex]2.00 and Passengers are charged [tex]\$[/tex]0.80 for every minute.
Let,
[tex]x[/tex]= Time in minutes
[tex]y[/tex]= Amount in [tex]\$[/tex]
Here the relation between x and y comes;
[tex]y=x\times 0.80 +2.00[/tex] (equation 2)
Let,By plug all x value in equation 2;
When [tex]x=0[/tex] then [tex]y=0\times 0.80+2.00=2.00[/tex],
When [tex]x=1[/tex] then [tex]y=1\times 0.80+2.00=2.8[/tex],
When [tex]x=2[/tex] then [tex]y=2\times 0.80+2.00=3.6[/tex],
When [tex]x=3[/tex] then [tex]y=3\times 0.80+2.00=4.4[/tex],
∴ The y values are 2.00 , 2.8 , 3.6 and 4.4
Question 3;
Given relation between [tex]x[/tex] and [tex]y[/tex];
[tex]y=6\times x +3[/tex]
Plug all the values of x in above equation,
When [tex]x=-2[/tex] then [tex]y=(6\times -2) +3=-9[/tex],
When [tex]x=-1[/tex] then [tex]y=(6\times --1) +3=-3[/tex],
When [tex]x=0[/tex] then [tex]y=(6\times 0) +3=3[/tex],
When [tex]x=1[/tex] then [tex]y=(6\times 1) +3=9[/tex],
∴ The y values are -9 , -3 , 3 and 9
Question 4;
Given relation between [tex]x[/tex] and [tex]y[/tex];
[tex]y=(\frac{1}{4} \times x)+1[/tex]
Plug all the values of x in above equation,
When [tex]x=-2[/tex] then [tex]y=(\frac{1}{4} \times -2)+1=\frac{1}{2}[/tex],
When [tex]x=-1[/tex] then [tex]y=(\frac{1}{4} \times -1)+1=\frac{3}{4}[/tex],
When [tex]x=0[/tex] then [tex]y=(\frac{1}{4} \times 0)+1=1[/tex],
When [tex]x=1[/tex] then [tex]y=(\frac{1}{4} \times 1)+1=\frac{5}{4}[/tex],
∴ y values are [tex]\frac{1}{2}[/tex] , [tex]\frac{3}{4}[/tex] , 1 and [tex]\frac{5}{4}[/tex]
