Question

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Answer 1

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$\frac{1}{2}$ , $\frac{3}{4}$ , 1 and $\frac{5}{4}$

Step-by-step explanation:

Question 1;

Given data,

Apples cost $\ 1.49$ per pound.

$x$= weight in pound

$y$= Cost in

Here the relation between $x$ and $y$ comes;

$y=x\times 1.49$   (equation 1)

Let, Plug all x value in equation 1;

when $x=1$ then $y=1\times 1.49=1.49$,

when $x=2$ then $y=2\times 1.49=2.98$,

when $x=3$ then $y=3\times 1.49=4.47$,

when $x=4$ then $y=4\times 1.49=5.96$,

∴ The y values are 1.49 , 2.98 , 4.47 and 5.96

Question 2;

From question,

Initial fees of cab is 2.00 and Passengers are charged  0.80 for every minute.

Let,

$x$= Time in minutes

$y$= Amount in

Here the relation between x and y comes;

$y=x\times 0.80 +2.00$    (equation 2)

Let,By plug all x value in equation 2;

When $x=0$ then $y=0\times 0.80+2.00=2.00$,

When $x=1$ then $y=1\times 0.80+2.00=2.8$,

When $x=2$ then  $y=2\times 0.80+2.00=3.6$,

When $x=3$ then  $y=3\times 0.80+2.00=4.4$,

∴ The y values are 2.00 , 2.8 , 3.6 and 4.4

Question 3;

Given relation between $x$ and $y$;

$y=6\times x +3$

Plug all the values of x in above equation,

When $x=-2$ then $y=(6\times -2) +3=-9$,

When $x=-1$ then  $y=(6\times --1) +3=-3$,

When $x=0$ then $y=(6\times 0) +3=3$,

When $x=1$ then $y=(6\times 1) +3=9$,

∴ The y values are -9 , -3 , 3 and 9

Question 4;

Given relation between $x$ and $y$;

$y=(\frac{1}{4} \times x)+1$

Plug all the values of x in above equation,

When $x=-2$ then $y=(\frac{1}{4} \times -2)+1=\frac{1}{2}$,

When $x=-1$ then $y=(\frac{1}{4} \times -1)+1=\frac{3}{4}$,

When $x=0$ then $y=(\frac{1}{4} \times 0)+1=1$,

When $x=1$ then $y=(\frac{1}{4} \times 1)+1=\frac{5}{4}$,

∴ y values are $\frac{1}{2}$ , $\frac{3}{4}$ , 1 and $\frac{5}{4}$