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Fynjy0 [20]
3 years ago
6

Part A: Sam rented a boat at $225 for 2 days. If he rents the same boat for 5 days, he has to pay a total rent of $480.

Mathematics
1 answer:
miss Akunina [59]3 years ago
4 0

Answer:

Here's what I get.

Step-by-step explanation:

Part A. Equation in standard form

The question is asking you to find the equation of a straight line that passes through two points

Let x = the number of days

and y = the cost

Then the coordinates of the two points are (2,225) and (5,480).

(i) Calculate the slope of the line

\begin{array}{rcl}m & = & \dfrac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\& = & \dfrac{480 - 225}{5 - 2}\\\\& = & \dfrac{255}{3}\\\\& = & 85\\\end{array}

In other words, the daily rental is $85/day.

(ii) Calculate the y-intercept

\begin{array}{rcl}y & = & mx + b\\480 & = & 85 \times 5 + b\\480 & = & 425 + b\\b & = & 55\\\end{array}

(iii) Write the equation for the line

y = 85x + 55

That is, the cost is $55  plus $85/day

Part B. Equation in function notation

Replace y with ƒ(x)

ƒ(x) = 85x + 55

Part C. Graphing

Let's say you want to plot a graph of the rental cost for up to ten days.

(i) Calculate two points on the graph.

When x = 0, y = 85; when x = 10, y = 905.

(ii) Scale your axes

A good number of intervals is about ten.

Your x-axis should have tick marks at 1-day intervals.

Your largest y-value is 905. Ten intervals would make about $90/interval. However, you should round that up to $100/interval for easy interpolation.

Your y-axis will run from 0 to $1000 in $100 intervals.

Plot your two points and draw a straight line through them.

(iii) Axis labels

x represents the number of days, so the label on the x-axis could be "No. of days."

y represents the cost of renting the boat, so the label on the y-axis could be "Rental cost."

Your graph should resemble the one below.

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Answer:

1) tanФ = 0.8392 ⇒ 2nd answer

2) cosФ = 20/29 ⇒ 3rd answer

3) cosФ = -√2/2 , cotФ = -1 ⇒ first answer

Step-by-step explanation:

* At first check the quadrant of the angle

∵ 180° < Ф < 270°

∴ Ф lies on the third quadrant

* Remember in the 3rd quadrant tanФ and cotФ only positive

∵ sinФ ≅ -0.7660

∵ tan²Ф = sec²Ф - 1

∵ secФ = 1/sinФ

∵ secФ = 1/-0.7660

∴ sec²Ф = (1/-0.7660)²

* Substitute this value in the equation

∴ tan²Ф = (1/-0.7660)² - 1 = 0.70428594 ⇒ take √ for both sides

∴ tanФ ≅ ± 0.8392

∵ Ф lies on the 3rd quadrant

∴ tanФ = 0.8392

* At first check the quadrant of the angle

∵ 0° < Ф < 90°

∴ Ф lies on the first quadrant

* Remember in the 1st quadrant all are positive

∵ sinФ = 21/29

∵ sin²Ф + cos²Ф = 1

∴ (21/29)² + cos²Ф = 1 ⇒ subtract (21/29)² from both sides

∴ cos²Ф = 1 - (21/29)² = 400/841 ⇒ take √ in both sides

∴ cosФ = ± 20/29

* Because Ф lies in the 1st quadrant

∴ cosФ = 20/29

* Remember that when the point is at the terminal side of angle Ф

∴ Its x-coordinate is cosФ

∴ Its y-coordinate is sinФ

* The point is (-√2/2 , √2/2)

∵ x-coordinate is negative ad y-coordinate is positive

∴ the point is on the 2nd quadrant

∴ 90° < Ф < 180°

∴ The values of sinФ and cscФ only positive

* From previous we know that:

∴ cosФ = -√2/2

∴ sinФ = √2/2

∵ cotФ = cosФ/sinФ

∴ cotФ = -√2/2 ÷ √2/2 = -√2/2 × 2/√2 = -√2/√2 = -1

* cosФ = -√2/2 , cotФ = -1

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