Chris withdraws $7.50 per week from his savings account. After 15 weeks, the balance in the account is $585.
A. Write an equation in point -slope form to model this.
B. Write the equation in slope-intercept form.
C. How much is in the account after 1 year (52 weeks)?
;Details or extra information would be appreciated .
A. Ao = 7.5*15 + 585 = $697.50 = Initial amount.
Y = 697.5 - 7.5x
Y = -7.5x + 697.5.
P1(X1,Y1) = (15,585).
Y-Y1 = m(X-X1). Point-slope form.
Y-585 = -7.5(X-15).
B. Y = mx + b. Slope-intercept form.
Y = -7.5x + 697.5.
C. Y = -7.5*52 + 697.5.
To write the equation in point-slope form, we need to find the slope and a point on the line.
We know that Chris withdraws $7.50 per week from his savings account. This means that for each week, the balance decreases by $7.50. So the slope of the line is -7.50.
Also, after 15 weeks, the balance in the account is $585. This gives us a point on the line, (15, 585).
Using the point-slope form of a linear equation, y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line, we can substitute the values to get the equation:
y - 585 = -7.50(x - 15)
Now, let's simplify it to slope-intercept form.
To convert the equation to slope-intercept form (y = mx + b), we need to solve for y:
y - 585 = -7.50x + 112.50
Adding 585 to both sides:
y = -7.50x + 697.50
So, the equation in slope-intercept form is y = -7.50x + 697.50.
Now, let's move on to part C.
To find out how much is in the account after 1 year (52 weeks), we can substitute x = 52 into the equation and solve for y:
y = -7.50(52) + 697.50
Calculating this gives us:
y = -390 + 697.50
y = 307.50
Therefore, after 1 year, the balance in the account is $307.50.
Note: The solution assumes that no deposits or other transactions have been made besides the weekly withdrawals.