To find the rate of change of the savings account in this context, we need to identify how much the amount of money (in dollars) increases for each month (time in months).
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Look at the values given:
- When \( t = 0 \), \( y = 0 \)
- When \( t = 1 \), \( y = 12.5 \)
- When \( t = 2 \), \( y = 25 \)
- When \( t = 3 \), \( y = 37.5 \)
- When \( t = 4 \), \( y = 50 \)
- When \( t = 5 \), \( y = 62.5 \)
- When \( t = 6 \), \( y = 75 \)
- When \( t = 7 \), \( y = 87.5 \)
- When \( t = 8 \), \( y = 100 \)
- When \( t = 9 \), \( y = 112.5 \)
- When \( t = 10 \), \( y = 125 \)
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Calculate the change in y (dollars in the savings account) for each 1-month increase in t:
The increase from one month to the next is:
- From \( t = 0 \) to \( t = 1 \): \( 12.5 - 0 = 12.5 \)
- From \( t = 1 \) to \( t = 2 \): \( 25 - 12.5 = 12.5 \)
- From \( t = 2 \) to \( t = 3 \): \( 37.5 - 25 = 12.5 \)
- From \( t = 3 \) to \( t = 4 \): \( 50 - 37.5 = 12.5 \)
- From \( t = 4 \) to \( t = 5 \): \( 62.5 - 50 = 12.5 \)
- From \( t = 5 \) to \( t = 6 \): \( 75 - 62.5 = 12.5 \)
- From \( t = 6 \) to \( t = 7 \): \( 87.5 - 75 = 12.5 \)
- From \( t = 7 \) to \( t = 8 \): \( 100 - 87.5 = 12.5 \)
- From \( t = 8 \) to \( t = 9 \): \( 112.5 - 100 = 12.5 \)
- From \( t = 9 \) to \( t = 10 \): \( 125 - 112.5 = 12.5 \)
The rate of change is consistent throughout, specifically \( 12.5 \) dollars per month.
Therefore, the rate of change is:
Rate of change: 12.5 dollars per month.
With this rate, we can also establish the equation of the linear function, where \( y \) (the amount in the account) is dependent on \( t \) (time in months):
The formula for a linear function can be given by:
\[ y = mx + b \]
where:
- \( m \) is the rate of change (slope) and equals \( 12.5 \)
- \( b \) is the initial amount in the account (the value of \( y \) when \( t = 0 \)), which is \( 0 \).
Plugging in these values, we get:
\[ y = 12.5t + 0 \] or simply: \[ y = 12.5t \]
This is the equation representing the linear function for the savings account.
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