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?
A) $1,100
B) $1,300
C) $1,500
D) $700

A) $1,100

Step-by-step explanation:

Let S be savings and t he number of years

S = mt + c

In 2007, t = 4 and S = 4200

4200 = 4m + c (1)

In 2015, t = 12 and S = 10400

10400 = 12m + c (2)

Equation (2) - Equation (1)

6200 = 8m

m = 6200/8

m = 775

(1) 4200 = 4m + c

4200 = 4(775) + c

4200 = 3100 + c

c = 4200 - 3100

c = 1100

S = 775t + 1100

Initial deposit is when t = 0,

S = $1100

answer: answer is c!

ok i got to organize the

3/15 and 12/55

3*

aha

3*4 equals 12 but not

a: no

b:

8/24 12/35

no

gautie

it's c

5/18 = 25/90

5*5=25

18*5 = 90

hope that you!

one more happy customer!

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step-by-step explanation: