Chapter 8.1, Problem 45IP

Solution

(a)

An equation that represents the distance d the butterfly will travel in t days

The equation is $d=52.5t$

Given:

The North American Monarch butterfly migrates up to 3000 miles to California and butterfly travels on average 52.5 miles per day

Concept Used:

  $\text{Speed}=\frac{\text{Distance}}{\text{Time}}$

Rules of Addition/ Subtraction:

• Two numbers with similar sign always get added and the resulting number will carry the similar sign.
• Two numbers with opposite signs always get subtracted and the resulting number will carry the sign of larger number.

Rules of Multiplication/ Division:

• The product/quotient of two similar sign numbers is always positive.
• The product/quotient of two numbers with opposite signs is always negative.

Calculation:

In order to find the equation the distance d the butterfly will travel in t days with the average speed of 52.5 miles is below:

  $\begin{array}{c}\text{Speed}=\frac{\text{Distance}}{\text{Time}}\\ 52.5=\frac{d}{t}\\ 52.5t=\frac{d}{t}\times t\\ 52.5t=d\\ \Rightarrow d=52.5t\end{array}$

Thus, the equation is $d=52.5t$

(b)

Complete the table

The table is below

Time (days)	1	2	3	4	5	6
Distance (miles) $d=52.5t$	52.5	105	157.5	210	262.5	315

Given:

  

Concept Used:

  $d=52.5t$

Rules of Addition/ Subtraction:

• Two numbers with similar sign always get added and the resulting number will carry the similar sign.
• Two numbers with opposite signs always get subtracted and the resulting number will carry the sign of larger number.

Rules of Multiplication/ Division:

• The product/quotient of two similar sign numbers is always positive.
• The product/quotient of two numbers with opposite signs is always negative.

Calculation:

In order to complete the table by using equation $d=52.5t$ is below:

Time (days)	1	2	3	4	5	6
Distance (miles) $d=52.5t$	$\begin{array}{l}d=52.5t\\ d=52.5\end{array}$	$\begin{array}{c}d=52.5t\\ =52.5\times 2\\ d=105\end{array}$	$\begin{array}{c}d=52.5t\\ =52.5\times 3\\ d=157.5\end{array}$	$\begin{array}{c}d=52.5t\\ =52.5\times 4\\ d=210\end{array}$	$\begin{array}{c}d=52.5t\\ =52.5\times 5\\ d=262.5\end{array}$	$\begin{array}{c}d=52.5t\\ =52.5\times 6\\ d=315\end{array}$

Thus, the table is below

Time (days)	1	2	3	4	5	6
Distance (miles) $d=52.5t$	52.5	105	157.5	210	262.5	315

(c)

The graph of points in the table

  

Given:

The table is below

Time (days)	1	2	3	4	5	6
Distance (miles) $d=52.5t$	52.5	105	157.5	210	262.5	315

Concept Used:

In rectangular coordinate system any point be in the form of $\left(x,y\right)$ .

In Quadrant I, both the coordinates of ordered pair are positive, that is, $\left(+,+\right)$ .

In Quadrant II, first number is negative and second is positive, that is, $\left(-,+\right)$ .

In Quadrant III, both the coordinates of ordered pair are negative, that is, $\left(-,-\right)$ .

In Quadrant IV, first point is positive and second is negative, that is, $\left(+,-\right)$ .

Explanation: In order to plot the point $\left(1,52.5\right),\left(2,105\right),\left(3,157.5\right),\left(4,210\right),\left(5,262.5\right),\left(6,315\right)$ , observe that both numbers of the ordered pair are positive, so the point will be in first quadrant.

To plot the graph of the point $\left(1,52.5\right)$ , first move 1units to the right starting from origin and then move 52.5 units up.

To plot the graph of the point $\left(1,52.5\right)$ , first move 1units to the right starting from origin and then move 52.5 units up.

To plot the graph of the point $\left(2,105\right)$ , first move 2 units to the right starting from origin and then move 105 units up.

To plot the graph of the point $\left(3,175.5\right)$ , first move 3 units to the right starting from origin and then move 175.5 units up.

To plot the graph of the point $\left(4,210\right)$ , first move 4 units to the right starting from origin and then move 210 units up.

To plot the graph of the point $\left(5,262.5\right)$ , first move 5 units to the right starting from origin and then move 262.5 units up.

To plot the graph of the point $\left(6,315\right)$ , first move 6 units to the right starting from origin and then move 315 units up.

The graph of the point is shown below:

  

(d)

The number of days it will take the butterfly to travel 475 days

The time is $9days$ .

Given:

Butterfly travels 475 miles

Concept Used:

  $\text{Speed}=\frac{\text{Distance}}{\text{Time}}$

Rules of Addition/ Subtraction:

• Two numbers with similar sign always get added and the resulting number will carry the similar sign.
• Two numbers with opposite signs always get subtracted and the resulting number will carry the sign of larger number.

Rules of Multiplication/ Division:

• The product/quotient of two similar sign numbers is always positive.
• The product/quotient of two numbers with opposite signs is always negative.

Calculation:

In order to find the time is below:

  $\begin{array}{c}d=52.5t\\ 475=52.5t\\ \frac{475}{52.5}=\frac{52.5t}{52.5}\\ 9=t\\ \Rightarrow t=9\end{array}$

Thus, the time is $9days$ .

(e)

How many days will it take the butterfly to travel 2100 miles

The days that the butterfly to travel 2100 miles is $40days$

Given:

  $\text{d}=\text{52}\text{.5t}$

Concept Used:

  $\text{Speed}=\frac{\text{Distance}}{\text{Time}}$

Rules of Addition/ Subtraction:

• Two numbers with similar sign always get added and the resulting number will carry the similar sign.
• Two numbers with opposite signs always get subtracted and the resulting number will carry the sign of larger number.

Rules of Multiplication/ Division:

• The product/quotient of two similar sign numbers is always positive.
• The product/quotient of two numbers with opposite signs is always negative.

Calculation:

In order to find the days is below:

  $\begin{array}{c}\text{d}=\text{52}\text{.5t}\\ 2100=52.5t\\ \frac{2100}{52.5}=\frac{52.5t}{52.5}\\ 40=t\\ \Rightarrow t=40\end{array}$

Thus, the days that the butterfly to travel 2100 miles is $40days$