Question

9 from unit 2, lesson 8
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 = 100 m )
b ( a = 100 ( m - 5 ) )
c ( a - 700 = 100 m )
d ( a - 1,200 = 100 m )
e ( a = 700 + 100 m )
f ( a = 1,200 + 100 m )
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 slope \(m\) (rate of change) is calculated as \(\frac{\Delta A}{\Delta m}\). For \(m = 5,A = 1200\) and \(m=6,A = 1300\), the slope \(m=\frac{1300 - 1200}{6 - 5}=100\).

Step2: Use the point - slope form

Using the point - slope form \(A - A_1=m(m - m_1)\). Taking the point \((m_1,A_1)=(5,1200)\), we have \(A-1200 = 100(m - 5)\), which simplifies to \(A=1200+100(m - 5)=1200 + 100m-500=700 + 100m\).

Step3: Check each option

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

eq1200\).

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

eq1200\).

• Option C: When \(m = 5\), \(A-700=100\times5\Rightarrow A = 1200\). But when \(m = 6\), \(A-700=100\times6\Rightarrow A=1300\). However, \(A-700 = 100m\Rightarrow A=700 + 100m\) (correct form).
• Option D: When \(m = 5\), \(A-1200=100\times5\Rightarrow A=1700

eq1200\).

• Option E: \(A = 700+100m\). When \(m = 5\), \(A=700 + 100\times5=1200\); when \(m = 6\), \(A=700+100\times6 = 1300\).
• Option F: When \(m = 5\), \(A=1200+100\times5=1700

eq1200\).

• Option G: \(A=1200+100(m - 5)=1200+100m-500=700 + 100m\). When \(m = 5\), \(A=1200+100(5 - 5)=1200\); when \(m = 6\), \(A=1200+100(6 - 5)=1300\).

Answer:

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