Maggy started a savings account in march of 2003. on march 2007, she had $4,200. on march 2015, she had $10,400. if maggy's saving is modeled by a linear function, what was her initial deposit?

First find the slope of the linear equation...

slope=m=(y2-y1)/(x2-x1)

m=(10400-4200)/(2015-2007)=6200/8=775 so we so far have a line of:

s=775(y-2003)+b, using either original point we can solve for b, the y-intercept which is the initial value as well...I'll use point (2007, 4200) and we get:

4200=775(2007-2003)+b

4200=3100+b

b=$1100

s(y)=775(y-2003)+1100

The initial deposit was $1100.00.

My B-day is March 2007!

jhfdergmllm: $"$& (& #@$& () 9887¥€¢£`~¥°℅™℅℅℅=

step-by-step explanation:

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option (b) is correct

step-by-step explanation:

to find the solution of this system, we need to convert it into reduced row-echelon form. we have,

after applying row operations r2 -> r2/2 and r3 -> r3/3, we get,