Sure, let's go through each problem step-by-step and write down the corresponding equations in the form [tex]\( y = mx + b \)[/tex] where applicable, and then we will solve each problem.
### Problem 1
Situation: A sales associate is given a \[tex]$500 hiring bonus and earns an average commission of \$[/tex]250 per week for the next 12 weeks.
Equation: [tex]\( y = 250x + 500 \)[/tex]
- [tex]\( y \)[/tex] represents total earnings.
- [tex]\( x \)[/tex] represents the number of weeks.
- 250 is the weekly commission (slope [tex]\( m \)[/tex]).
- 500 is the hiring bonus (y-intercept [tex]\( b \)[/tex]).
Solution:
[tex]\[
\begin{align*}
x & = 12 \\
y & = 250(12) + 500 \\
y & = 3000 + 500 \\
y & = 3500
\end{align*}
\][/tex]
So, she earns \$3500 after 12 weeks.
### Problem 2
Situation: A farm has 75 acres of wheat, and the farmer can harvest 12 acres per day. How many days to harvest all the fields?
Equation: [tex]\( y = \frac{75}{12} \)[/tex]
- [tex]\( y \)[/tex] represents the total days required to harvest.
- 75 is the total acres of wheat.
- 12 is the acres harvested per day.
Solution:
[tex]\[
\begin{align*}
y & = \frac{75}{12} \\
y & = 6.25
\end{align*}
\][/tex]
So, it will take 6.25 days to harvest all the fields.
### Problem 3
Situation: A contractor's crew can frame 3 houses in a week. How long will it take them to frame 54 houses if they frame the same number each week?
Equation: [tex]\( y = \frac{54}{3} \)[/tex]
- [tex]\( y \)[/tex] represents the total weeks required.
- 54 is the total number of houses.
- 3 is the number of houses framed per week.
Solution:
[tex]\[
\begin{align*}
y & = \frac{54}{3} \\
y & = 18
\end{align*}
\][/tex]
So, it will take 18 weeks to frame all 54 houses.
### Problem 4
Situation: A water tank holds 18,000 gallons. How long will it take for the water level to reach 6,000 gallons if water is used at an average rate of 450 gallons per day?
Equation: [tex]\( y = \frac{(18000 - 6000)}{450} \)[/tex]
- [tex]\( y \)[/tex] represents the total days required to reach the required level.
- 18,000 is the tank capacity.
- 6,000 is the required water level.
- 450 is the usage rate per day.
Solution:
[tex]\[
\begin{align*}
y & = \frac{18000 - 6000}{450} \\
y & = \frac{12000}{450} \\
y & = 26.\overline{6}
\end{align*}
\][/tex]
So, it will take approximately 26.67 days for the water level to reach 6,000 gallons.
Each problem has been carefully solved to give us the respective answers.
