Question

from unit 2, lesson 3
the table shows the amount of money, a, in a savings account after m months.
select all the equations that represent the relationship between the amount of money, a, and the number of months, m.
a a = 100m
b a = 100(m - 5)
c a - 700 = 100m
d a - 1,200 = 100m
e a = 700 + 100m
f a = 1,200 + 100m
g a = 1,200 + 100(m - 5)
+ learning targets
+ i can use graphing technology to graph linear equations and identify solutions to the equations.
+ i understand how the coordinates of the points on the graph of a linear equation are related to the equation.
+ when given the graph of a linear equation, i can explain the meaning of the points on the graph in terms of the situation it represents.

Explanation:

Step1: Find the slope

The change in amount $\Delta A=1300 - 1200=100$ when $\Delta m=6 - 5 = 1$. So the slope of the linear - relationship is $100$.

Step2: Find the y - intercept

Using the point - slope form with the point $(m = 5,A = 1200)$. The general form of a linear equation is $A=A_0 + k(m - m_0)$. Substituting $m_0 = 5$, $A_0 = 1200$ and $k = 100$ gives $A=1200+100(m - 5)$. Expanding $A=1200+100m-500=700 + 100m$.

Step3: Check each option

• Option A: When $m = 5$, $A=100\times5 = 500

eq1200$, so A is wrong.

• Option B: When $m = 5$, $A=100(5 - 5)=0

eq1200$, so B is wrong.

• Option C: Rearranging gives $A=100m + 700$, when $m = 5$, $A=100\times5+700 = 1200$. But the form is not in the most intuitive way for this problem.
• Option D: When $m = 5$, $A=100\times5+1200=1700

eq1200$, so D is wrong.

• Option E: $A = 700+100m$ is correct as derived above.
• Option F: When $m = 5$, $A=1200+100\times5=1700

eq1200$, so F is wrong.

• Option G: $A = 1200+100(m - 5)$ is correct as derived above.

Answer:

E. $A = 700+100m$, G. $A = 1200 + 100(m - 5)$