Question

7 from unit 2, lesson 2
a student on the cross - country team runs 30 minutes a day as a part of her training.
write an equation to describe the relationship between the distance she runs in miles, d,
and her running speed, in miles per hour, when she runs:
a. at a constant speed of 4 miles per hour for the entire 30 minutes.
b. at a constant speed of 5 miles per hour the first 20 minutes, and then at 4 miles per
hour the last 10 minutes.
c. at a constant speed of 6 miles per hour the first 15 minutes, and then at 5.5 miles per
hour for the remaining 15 minutes.
d. at a constant speed of a miles per hour the first 6 minutes, and then at 6.5 miles per
hour for the remaining 24 minutes.
e. at a constant speed of 5.4 miles per hour for m minutes, and then at b miles per hour
for n minutes.
8 from unit 2, lesson 3
in the 21st century, people measure length in feet and meters. at various points in history,
people measured length in hands, cubits, and paces. there are 9 hands in 2 cubits. there
are 5 cubits in 3 paces.
a. write an equation to express the relationship between hands, h, and cubits, c.
b. write an equation to express the relationship between hands, h, and paces, p.

Explanation:

7.

Step1: Recall distance - speed - time formula

The formula is $d = s\times t$, where $d$ is distance, $s$ is speed and $t$ is time. We need to convert time from minutes to hours since speed is in miles per hour ($1$ hour = 60 minutes, so $t$ (in hours)=$\frac{\text{minutes}}{60}$).

Step2: Solve part a

The student runs at a constant speed $s = 4$ miles per hour for $t = 30$ minutes or $t=\frac{30}{60}=0.5$ hours. The equation for distance $d$ is $d = 4\times0.5$.

Step3: Solve part b

The student runs at 5 miles per hour for 20 minutes ($t_1=\frac{20}{60}=\frac{1}{3}$ hours) and 4 miles per hour for 10 minutes ($t_2=\frac{10}{60}=\frac{1}{6}$ hours). The distance $d = 5\times\frac{1}{3}+4\times\frac{1}{6}$.

Step4: Solve part c

The student runs at 6 miles per hour for 15 minutes ($t_1=\frac{15}{60}=\frac{1}{4}$ hours) and 5.5 miles per hour for 15 minutes ($t_2=\frac{15}{60}=\frac{1}{4}$ hours). The distance $d=6\times\frac{1}{4}+5.5\times\frac{1}{4}$.

Step5: Solve part d

The student runs at $a$ miles per hour for 6 minutes ($t_1=\frac{6}{60}=\frac{1}{10}$ hours) and 6.5 miles per hour for 24 minutes ($t_2=\frac{24}{60}=\frac{2}{5}$ hours). The distance $d=a\times\frac{1}{10}+6.5\times\frac{2}{5}$.

Step6: Solve part e

The student runs at 5.4 miles per hour for $m$ minutes ($t_1 = \frac{m}{60}$ hours) and $b$ miles per hour for $n$ minutes ($t_2=\frac{n}{60}$ hours). Since $m + n=30$, the distance $d = 5.4\times\frac{m}{60}+b\times\frac{n}{60}$.

Step1: Find the relationship between hands and cubits

Given that there are 9 hands in 2 cubits. Let $h$ be the number of hands and $c$ be the number of cubits. The ratio of hands to cubits is $\frac{h}{c}=\frac{9}{2}$, so $h=\frac{9}{2}c$.

Step2: Find the relationship between hands and paces

First, from "5 cubits in 3 paces", we have $c=\frac{5}{3}p$. Substitute $c$ into $h=\frac{9}{2}c$. Then $h=\frac{9}{2}\times\frac{5}{3}p$. Simplify $\frac{9}{2}\times\frac{5}{3}=\frac{15}{2}$, so $h=\frac{15}{2}p$.

Answer:

a. $d = 4\times0.5$
b. $d = 5\times\frac{1}{3}+4\times\frac{1}{6}$
c. $d=6\times\frac{1}{4}+5.5\times\frac{1}{4}$
d. $d=a\times\frac{1}{10}+6.5\times\frac{2}{5}$
e. $d = 5.4\times\frac{m}{60}+b\times\frac{n}{60}$ ($m + n = 30$)

8.