A house was purchased for $67,000 After 7 years the value of the house was $130,000. Express the house's value V(t) in terms of time t in years.
if linear (doubt that really by the way)
V(t) = 67,000 + (130,000-67,000) t/7
= 67,000 + 9 t
what is t that's what I'm trying to find
I have a typo, forgot thousands
V(t) = 67,000 + 9,000 t
It asked you to write V(t) as a function of t where t is the years since you bought the house.
when t = 0 , V = 67,000 + 9,000 * 1
when t = 2 , V = 67,000 + 9,000 * 2
etc
another typo
when t = 0 , v = 67,000 + 9,000 * 0
when t = 1 , V = 67,000 + 9,000 * 1
when t = 2 , V = 67,000 + 9,000 * 2
etc
To express the house's value V(t) in terms of time t in years, we can use the concept of linear growth.
Linear growth is represented by the equation of a straight line: y = mx + b, where y is the dependent variable (in this case, the house's value), x is the independent variable (in this case, time), m is the slope (rate of change), and b is the y-intercept (initial value).
In this case, we know that the value of the house increased from $67,000 to $130,000 over a period of 7 years. Let's use this information to find the slope (m) and y-intercept (b).
First, we calculate the rate of change (slope):
slope (m) = (change in value)/(change in time)
slope (m) = ($130,000 - $67,000)/(7 years)
slope (m) = $63,000/7 years
slope (m) = $9,000/year
Now that we have the slope, we can find the y-intercept (b). We can use either of the given data points (the initial or final value and time) to find the y-intercept.
Using the initial value ($67,000) at time t = 0:
V(0) = (slope × t) + b
$67,000 = ($9,000/year × 0 years) + b
$67,000 = b
Therefore, the y-intercept (b) is $67,000.
Now we can write the equation for the house's value V(t) in terms of time t:
V(t) = (slope × t) + b
V(t) = ($9,000/year × t) + $67,000
Therefore, the expression for the house's value V(t) in terms of time t is:
V(t) = $9,000t + $67,000