Question

6 from unit 2 lesson 3 a basketball coach purchases bananas for the players on his team. the table shows total price in dollars, p, of n bananas. which equation could represent the total price in dollars for bananas? number of bananas total price in dollars 7 4.13 8 4.72 9 5.31 10 5.90 a. p = 0.59n b. p = 5.90 - 0.59n c. p = \frac{5.90}{n} d. p = n + 0.59 7 from unit 2 lesson 4 kiran is collecting dimes and quarters in a jar. he has collected $10.00 so far and has d dimes and q quarters. the relationship between the numbers of dimes and quarters, and the amount of money in dollars is represented by the equation 0.1d + 0.25q = 10. select all the values (d, q) that could be solutions to the equation. a (100, 0) b. (20, 50) c (50, 20) d. (0, 100) e. (10, 36) algebra 1 unit 2 practice problems cc by nc 2023

Explanation:

Question 6

Step1: Find the unit - price

To find the equation representing the total price $P$ of $n$ bananas, we first find the unit - price. We can use the formula for unit - price $u=\frac{P}{n}$. For example, when $n = 7$ and $P=4.13$, $u=\frac{4.13}{7}=0.59$. When $n = 8$ and $P = 4.72$, $u=\frac{4.72}{8}=0.59$. When $n = 9$ and $P=5.31$, $u=\frac{5.31}{9}=0.59$. When $n = 10$ and $P = 5.90$, $u=\frac{5.90}{10}=0.59$.

Step2: Determine the equation

The total price $P$ is the unit - price times the number of bananas. So the equation is $P = 0.59n$.

Step1: Substitute values into the equation

We have the equation $0.1d+0.25q = 10$, where $d$ is the number of dimes and $q$ is the number of quarters.

• For option A: When $d = 100$ and $q = 0$, $0.1\times100+0.25\times0=10 + 0=10$.
• For option B: When $d = 20$ and $q = 50$, $0.1\times20+0.25\times50=2 + 12.5 = 14.5

eq10$.

• For option C: When $d = 50$ and $q = 20$, $0.1\times50+0.25\times20=5+5 = 10$.
• For option D: When $d = 0$ and $q = 100$, $0.1\times0+0.25\times100=0 + 25=25

eq10$.

• For option E: When $d = 10$ and $q = 36$, $0.1\times10+0.25\times36=1+9 = 10$.

Answer:

A. $P = 0.59n$

Question 7