Solve.
-An investment is worth $\$ 3358$ in 2006. By 2011 it has grown to $\$ 6193$. Let y be the value of the investment in the year $x$, where $x = 0$ represents 2006 . Write a linear equation that relates value $y$ to year x.
A) $y = \frac { 1 } { 567 } x + 3358$
B) $y = - 567 x + 9028$
C) $y = - 567 x + 3358$
D) $y = 567 x + 3358$

Question

Quizplus · Accepted Answer

The increase in investment value from 2006 to 2011 (5 years) is $6193 - $3358 = $2835. The annual increase is $2835 / 5 = $567. Thus, the equation is $y = 567x + 3358$, where $x = 0$ in 2006.

Solve -An Investment Is Worth $3358\$ 3358$3358 In 2006 by 2011 It Has Grown To In 2006

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-An Investment Is Worth $\$ 3358$ In 2006 by 2011 It Has Grown To In 2006