John, Sally, and Natalie would all like to save some money. John decides that it
would be best to save money in a jar in his closet every single month. He decides
to start with $300, and then save $100 each month. Sally has $6000 and decides
to put her money in the bank in an account that has a 7% interest rate that is
compounded annually. Natalie has $5000 and decides to put her money in the
bank in an account that has a 10% interest rate that is compounded continuously.

How much money have after 2 years?

How much money will sally have in 10 years?

What type of exponential model is Natalie’s situation?

Write the model equation for Natalie’s situation

How much money will Natalie have after 2 years?

How much money will Natalie have after 10 years

Part 1) John’s situation is modeled by a linear equation (see the explanation)

Part 2)  

Part 3)

Part 4)

Part 5) Is a exponential growth function

Part 6)

Part 7)

Part 8)  

Part 9) Is a exponential growth function

Part 10)    or  

Part 11)  

Part 12)

Part 13)  Natalie has the most money after 10 years

Part 14)  Sally has the most money after 2 years

Step-by-step explanation:

Part 1) What type of equation models John’s situation?

Let

y ----> the total money saved in a jar

x ---> the time in months

The linear equation in slope intercept form

y=mx+b

The slope is equal to

The y-intercept or initial value is

so

therefore

John’s situation is modeled by a linear equation

Part 2) Write the model equation for John’s situation

see part 1)

Part 3) How much money will John have after 10 years?

Remember that

1 year is equal to 12 months

so

For x=120 months

substitute in the linear equation

Part 4) How much money will John have after 2 years?

Remember that

1 year is equal to 12 months

so

For x=24 months

substitute in the linear equation

Part 5) What type of exponential model is Sally’s situation?

we know that    

The compound interest formula is equal to  

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest  in decimal

t is Number of Time Periods  

n is the number of times interest is compounded per year

in this problem we have  

substitute in the formula above

therefore

Is a exponential growth function

Part 6) Write the model equation for Sally’s situation

see the Part 5)

Part 7) How much money will Sally have after 10 years?

For t=10 years

substitute  the value of t in the exponential growth function

Part 8) How much money will Sally have after 2 years?

For t=2 years

substitute  the value of t in the exponential growth function

Part 9) What type of exponential model is Natalie’s situation?

we know that

The formula to calculate continuously compounded interest is equal to

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest in decimal  

t is Number of Time Periods  

e is the mathematical constant number

we have  

substitute in the formula above

Applying property of exponents

 therefore

Is a exponential growth function

Part 10) Write the model equation for Natalie’s situation

   or  

see Part 9)

Part 11) How much money will Natalie have after 10 years?

For t=10 years

substitute

Part 12) How much money will Natalie have after 2 years?

For t=2 years

substitute

Part 13) Who will have the most money after 10 years?

Compare the final investment after 10 years of John, Sally, and Natalie

Natalie has the most money after 10 years

Part 14) Who will have the most money after 2 years?

Compare the final investment after 2 years of John, Sally, and Natalie

Sally has the most money after 2 years

step-by-step explanation:

a.   we have two

lines:   y = 4-x   and  

y = 8-x^-1

given two simultaneous equations that are both to be

true, then the solution is the points where the lines cross. the intersection is

where the two equations are equal. therefore the solution that works for both

equations is when

4-x = 8-x^-1

this is where the two lines will cross and that is the

common point that satisfies both equations.

b.   4-x = 8-x^-1

  x           4-x       8-x^-1

-3           7        

8.33

-2           6         8.5

-1           5         9

  0           4        

-

  1           3        

7

  2           2        

7.5

  3           1         7.67

the table shows that none of the x values from -3 to 3 is

the solution because in no case does

4-x = 8-x^-1

to find the solution we need to rearrange the equation to

find for x:

4-x = 8-x^-1

multiply both sides with x:

4x-x^2 = 8x-1

x^2+4x-1=0

x= -4.236, 0.236

therefore there are two points that satisfies the

equation.

find y:  

x=-4.236

y = 4-x   = 4 – (-4.236)

= 8.236

y = 8-x^-1 =   .236)^-1

= 8.236

x=0.236

  y = 4-x   = 4 – (0.236) = 3.764

y = 8-x^-1 =   8-(0.236)^-1

= 3.764

thus the two lines cross at 2 points:

(-4.236, 8.236) & (0.236, 3.764)

c.   to solve

graphically the equation 4-x = 8-x^-1

we would graph both lines: y = 4-x   and   y

= 8-x^-1

the point on the graph where the lines cross is the

solution to the system of equations.

just graph the points on part b on a cartesian coordinate

system and extend the two lines.   the

solution is, as stated, the point where the two lines cross on the graph.

step-by-step explanation:

mark brainliest!