Seth’s parents gave him $5000 to invest for his 16th birthday. He is considering two investment options. Option A will pay him 4.5% interest compounded annually. Option B will pay him 4.6% compounded quarterly. 1) Write a function of option A and option B that calculates the value of each account after n years. 2) Seth plans to use the money after he graduates from college in 6 years. Determine how much more money option B will earn than option A to the nearest cent. 3) Algebraically determine, to the nearest tenth of a year, how long it would take for option B to double Seth’s initial investment.

1) Function of option

option .

2)After 6 years, how much more money option B will earn than option A is   67.57 $.

3)It  would take 15 years for option B to double Seth’s initial investment.

Step-by-step explanation:

Given:

Initial Investment=5000$

Option A(rate)= 4.5% .....annually

Option B(rate)=4.6 %..........Quarterly

To Find:

1)Write a function of option A and option B

2)After 6 years, how much more money option B will earn than option A

3) how long it would take for option B to double Seth’s initial investment.

Solution:

To write the function use formula of compound interest as ,

For option A ,P=5000$ r=4.5 % annually

For Option B ,P=5000$ r=4.6 % Quarterly

2)After 6 years, how much more money option B will earn than option A,

Here t=6 so Above equation will be ,

$

For Option B

B=5000(1.0115)^ 4*6

$

B will earn more money as

therefore B -A

=67.57 $

3)how long it would take for option B to double Seth’s initial investment

By doubling the invest i.e  for 10000 $  how much time will required.

So B=10000$ , P=5000$  and r= 4.5 % Quarterly

10000=5000(1.0115)^4t

2=(1.0115)^4t

Using the definition of the logarithm as ,

4t=Log2 with base 1.0115............. use this in calculator

4t=60.62

t=15.155 years

i.e, t=15 years.

step-by-step explanation:

we are given a graph for system of inequalities.

in order to write the system of inequalities for the given graph, we need to find the inequality for each of the line separately.

let us find the slope of dotted line first.

taking two coordinates on the graph.

let us say (0,0) and (3,1).

let us find the slope of the line now.

slope of the dotted line would be =.

y-intercept of dotted line is y=0 only.

plugging slope m= and b=0 in slope-intercept form y=mx+b, we get

.

the line is shaded up of that dotted line.

therefore, for shading up we would take greater than inequality symbol > .

and because it's a dotted line we would take only > symbol in place of = sign in above equation of line.

therefore, inequality for dotted line would be

let us take two points (3,1) and (0,4) on the solid line.

slope of the solid line would be =.

for the solid line we have y-intercept b= 4.

plugging slope m= -1 and b=4 in slope-intercept form y=mx+b, we get

y = -1x + 4.

the line is shaded down of that solid line.

therefore, for shading down we would take less than inequality symbol < .

and because it's a solid line we would take ≤ symbol in place of = sign in above equation of line.

therefore, inequality for solid line would be