Question

1. in jay’s business, the annual cost of operating a car, c, is a linear function of the number of kilometers the car is driven, k. the annual cost of operating a car that has been driven 19 375km is approximately $3875. the annual cost of operating a car that has been driven 20 000km is approximately $3900.

a) write an equation in slope - point form to represent this function
b) use the equation in part a to determine how many kilometers a car has been driven if the annual operating cost is approximately $4350.

1. students at tahayghen secondary school sell punch during the school carnival. the number of cups sold, n, is a linear function of the temperature in degrees celsius. the students sold 458 cups when the temperature was 25°c. they sold 534 cups when the temperature was 29°c.

a) write an equation in slope - point form to represent this function
b) use the equation in part a to determine the approximate temperature when

Explanation:

Part 17a

Step 1: Identify two points

We have two points: \((k_1, c_1)=(19375, 3875)\) and \((k_2, c_2)=(20000, 3900)\).

Step 2: Calculate the slope \(m\)

The slope formula is \(m = \frac{c_2 - c_1}{k_2 - k_1}\).
Substitute the values: \(m=\frac{3900 - 3875}{20000 - 19375}=\frac{25}{625}=\frac{1}{25}\).

Step 3: Write the slope - point form

Using the point \((20000, 3900)\) (we could also use the other point), the slope - point form of a linear equation is \(c - c_1=m(k - k_1)\).
Substitute \(m = \frac{1}{25}\), \(c_1 = 3900\) and \(k_1=20000\) into the formula: \(c - 3900=\frac{1}{25}(k - 20000)\).

Part 17b

Step 1: Substitute \(c = 4350\) into the equation

We have the equation \(c - 3900=\frac{1}{25}(k - 20000)\). Substitute \(c = 4350\):
\(4350-3900=\frac{1}{25}(k - 20000)\)

Step 2: Simplify the left - hand side

\(450=\frac{1}{25}(k - 20000)\)

Step 3: Multiply both sides by 25

\(450\times25=k - 20000\)
\(11250=k - 20000\)

Step 4: Solve for \(k\)

Add 20000 to both sides: \(k=11250 + 20000=31250\).

Part 18a

Step 1: Identify two points

We have two points: \((t_1, n_1)=(25, 458)\) and \((t_2, n_2)=(29, 534)\).

Step 2: Calculate the slope \(m\)

The slope formula is \(m=\frac{n_2 - n_1}{t_2 - t_1}\).
Substitute the values: \(m=\frac{534 - 458}{29 - 25}=\frac{76}{4} = 19\).

Step 3: Write the slope - point form

Using the point \((29, 534)\) (we could also use the other point), the slope - point form of a linear equation is \(n - n_1=m(t - t_1)\).
Substitute \(m = 19\), \(n_1 = 534\) and \(t_1 = 29\) into the formula: \(n - 534=19(t - 29)\).

Part 18b (assuming we want to find the temperature when, for example, if we assume we want to find the temperature when \(n\) is a certain value, but since the problem statement for 18b is incomplete. If we assume we want to find the temperature when \(n\) is a given value, say if we want to find the temperature when \(n\) is some value \(n_0\), we would substitute \(n = n_0\) into the equation \(n - 534=19(t - 29)\) and solve for \(t\). But since the problem statement for 18b is not fully given, we can't provide a complete solution for it now. However, we can show the general method.

If we assume we want to find the temperature when \(n\) is a particular value, say \(n = n_0\):

Step 1: Substitute \(n=n_0\) into the equation \(n - 534=19(t - 29)\)

\(n_0-534 = 19(t - 29)\)

Step 2: Divide both sides by 19

\(\frac{n_0 - 534}{19}=t - 29\)

Step 3: Solve for \(t\)

\(t=\frac{n_0 - 534}{19}+29\)

Answer:

s:
17a) \(c - 3900=\frac{1}{25}(k - 20000)\)
17b) \(31250\) kilometers
18a) \(n - 534=19(t - 29)\)
(For 18b, the solution depends on the complete problem statement. If you provide the missing information about the number of cups or other relevant data, we can solve it completely.)