Mel opened a savings account in april with an initial deposit of $250. each month after april, mel deposited an additional $60. find the point–slope form of the linear equation that expresses mel's total deposits (y) as a function of the month (x). let x = 1 represent january, x = 2 represent february, and so on. what are mel's total deposits through the month of november?
=
total deposits through november

(y - 310) = 60(x - 1)

$670

Step-by-step explanation:

Mel opened a savings account in April with an initial deposit of $250. Each month after April, Mel deposited an additional $60.

If we express Mel's total deposit (y, in dollars) after x months since the month of April, then the slope of the linear function m will be 60 and (1, 250 + 60) = (1,310) is a point on the line graph.

Now, in point-slope form of a straight line equation gives if the slope of a straight line is m and (h,k) is a point on the line, then its equation can be written as (y - k) = m(x - h).

Now, in our case, the equation of the expression for Mel's total deposit will be given by

(y - 310) = 60(x - 1) (1) (Answer)

Now, at the month of November, the x value will be 7, then from equation (1), Mel's total deposits through the month of November will be  

y - 310 = 60(7 - 1)

⇒ y = 310 + 360 = $670 (Answer)

step-by-step explanation:

compare

you can first factor out the n to get n(1+5). we know that 1+5 is 6, so we just have 6n.