Answer:
General form of a linear equation: [tex]y=mx+b[/tex]
(where m is the slope and b is the y-intercept)
A linear equation has a constant rate of change which is represented by the slope (m). It is the rate at which one quantity is changing with respect to another quantity.
Question 3
Part (a)
Given equation: [tex]y=-9x+55000[/tex]
where:
- y = value of car (in dollars)
- x = number of miles driven
The rate of change (slope) is -9 and represents that for every 1 mile driven, the value of the car decreases by $9.
The initial value is the y-intercept (b) as this is the value when x = 0.
Therefore, the initial value of car is $55,000.
Part (b)
To find the number of miles at which the car be valued at $35,002, set the equation to $35,002 and solve for x:
[tex]\implies -9x+55000=35002[/tex]
[tex]\implies -9x=35002-55000[/tex]
[tex]\implies -9x=-19998[/tex]
[tex]\implies x=\dfrac{-19998}{-9}[/tex]
[tex]\implies x=2222[/tex]
Therefore, the car will be valued at $35,002 after 2,222 miles.
Question 4
Given equation: [tex]y=0.39x+2.55[/tex]
where:
- y = price of carrots (in dollars)
- x = weight (in oz)
The rate of change (slope) is 0.39 and represents that for every increase of 1 oz of weight in carrots, the price of the carrots increases by $0.39
The initial value is the y-intercept (b) as this is the value when x = 0.
Therefore, the initial cost of the carrots is $2.55.
Part (b)
To find the weight of carrots you can buy for $6.06, set the equation to $6.06 and solve for x:
[tex]\implies 0.39x+2.55=6.06[/tex]
[tex]\implies 0.39x=6.06-2.55[/tex]
[tex]\implies 0.39x=3.51[/tex]
[tex]\implies x=\dfrac{3.51}{0.39}[/tex]
[tex]\implies x=9[/tex]
Therefore, you can buy 9 oz of carrots for $6.06
Part (c)
To find out how much it costs to buy 21 oz of carrots, substitute x = 21 into the equation and solve for y:
[tex]\implies y=0.39(21)+2.55[/tex]
[tex]\implies y=10.74[/tex]
Therefore, it costs $10.74 to buy 21 oz of carrots.
