Mr. Clarence E. Harris
Unit 5 Math 133 Discussion Board

Grace is a grandmother who is looking for a long-range investment for her grandchild Elliott’s education. I searched the internet to find a Certificate of Deposit, which is a CD plan that earns compounded interest to invest her $25,000.

I used the advertised rates, the number of compounding periods per year and the time the funds will be invested, from the web site I researched to calculate the future value of her investment. Here is a step by step walk through of how I calculated Grace’s future investment.

Tip: You’re going to need a special calculator to apply your values and get the correct result. The calculator I used was at the following website:

Retrieved from https://www.bankoncit.com/calculators-savings.htm

Step #1. Was to apply the principle investment $25,000.
The principal is $25,000 Principle =P P=$25,000

Step #2. Was to apply the advertised annual interest rate for her investment. Remember to convert your percent to a decimal dividing it by 100

Use advertised annual interest rate from the website

Step #3. Was to apply the time in years for the investment.
State the time in years for the investment.

Time in years = t T=10

In this scenario we were not given the number of compounding periods a year. That means you’ll have to make it up.

Some options you may choose would be: Annual, Quarterly, Monthly or Daily. I chose annual because it made the math simple for me making the value for (n) #1.

Step #4. Was to apply the number of compounding periods per year. State Compounding Periods Per Year which will =(n)

Compounded Periods = 1 (Annual) N=1

Now I’ll put the assigned values in a simple column for easy reading going into the next step.

P=$25,000
R=0.0115
T=10
N=1

Step #5. Now we can apply these values to the formula F(t) = P(1 + r/n) nt...

...1. Solve
a. e^.05t = 1600
0.05t = ln(1600)
0.05t = 7.378
t = 7.378/.05
t = 147.56
b. ln(4x)=3
4x = e^3
x = e^3/4
x = 5.02
c. log2(8 – 6x) = 5
8-6x = 2^5
8-6x = 32
6x = 8-32
x = -24/6
x = -4
d. 4 + 5e-x = 0
5e^(-x) = -4
e^(-x) = -4/5
no solution, e cannot have a negative answer
2. Describe the transformations on the following graph of f (x) log( x) . State the
placement of the vertical asymptote and x-intercept after the transformation. For
example, vertical shift up 2 or reflected about the x-axis are descriptions.
a. g(x) = log( x + 5)
horizontal left shift 5
Vertical asymptote x = -5
x-intercept: (-4, 0)
b. g(x)=log(-x)
over the x-axis
vertical asymptote x=0
no x-intercept
3. Students in an English class took a final exam. They took equivalent forms of the exam at monthly intervals thereafter. The average score S(t), in percent, after t months was found to be given by S(t) = 68 - 20 log (t + 1), t ≥ 0.
a. What was the average score when they initially took the test, t = 0? Round your answer to a whole percent, if necessary.
S(0)=68-20xlog(0+1) =
68-20x0
= 68%
b. What was the average score after 4 months? after 24 months?
Round your answers to two decimal places.
-S(4) = 68-20xlog(4+1)
68-20x0.699
68-13.98
=54.02
-S(24) = 68-20xlog(24+1) = 40.04
68-20x1.398
68-27.96
=40.04
c. After what time t was the average...

...Answer Form Unit5 IP_2
April 22, 2011
(1.)
a.
e^(0.05t)=1600
Answer: t=147.556
Show your work:
ln(e^(0.05t))=ln(1600)
0.05t=ln(1600)
(0.05t)/(0.05)=(ln(1600))/(0.05)
t=(ln(1600))/(0.05)
t=20ln(1600)
t=147.556
b.
ln(4x)=3
Answer: x= 5.0214
Show your work:
4x = e^3/4
c.
log2(8-6x) = 5
Answer: x= -4
Show work:
8 – 6x = 2^5
6x = 8 – 32
X = - 24/6
d.
4 + 5e – x = 0
Answer: no solution found
Show work:
e ^ - x = - 4/5
(2.)
a.
g(x) = log (x + 5)
Description of transformation: shifted left 5units
Equation for vertical asymptote: x = -5
X-intercept in (x, y) form: (- 4, 0)
b.
g(x) = log ( - x)
Description of transformation: no units were moved, stayed in same place
Equation for vertical asymptote: x = 0
X-intercept in (x, y) form: (-1, 0)
(3.)
a.
Answer: 68 %
Show work:
S (0) = 68 – 20 * log (0 + 1)
S (0) = 68 – 20 * 0 = 68
b.
Answer:
After 4 months:
54.02%
After 24 months:
40.04%
Show work:
S (4) = 68 – 20 * log (4 + 1) = 54.02
S (24) = 68 – 20 * log (24 + 1) = 40.04
c.
Answer: t = 6.94
Show work:
50 = 68 – 20 log ( t + 1)
t + 1 = 10 ^ (18 / 20) = 7.9433
t= 7.9433 – 1 = 6.94
(4.)
a.
Answer: A= $2,938.66
Show work:
A= 2000 * (1+ 0.08/1) ^ (1*5) = 2938.66
b.
Answer: A=...

...April 25, 2013
Math133
DB Unit5
I have put a side 62,000 for my grandsons college tuition and I have found that the best most safe and effective way to earn money for the next 12 yrs until he goes to college is by investing it into a CD with an annual rate of 0 .90% (bankrate, 2013) and it is compounded monthly.
I have researched to see if my investment would help with tuition in 12 years. I used an inflation calculator and with the current trend of 45% increases in college tuition per year in 12 years college will cost 99,900.49 in the first year and 8% inflation increase each year after. This was done for a 4yrs public college (College Cost Projecter, 2013).
My Principal starting fund in 2013 grandson age 6 is $62,000.
a) Annual Interest rate r= 0.90%
b) Investment time t= 18yrs
c) Interest compounded monthly n=12
d) Model of my investment earns as a function is f(t)=P(1+r/n)^nt f(12)=62,000(1+.009/12)^12t
f(12)= 62,000(1+.009/12)^12*12
f(12)=62000(1+.00075)^12*12
f(12)=62000(1.00075)^12*12
f(12)=62000(1.00075)^144
f(12)=62000*144.108
f(12)=89,346.96 Future value of my investment
With the average cost of college today (National Center for Education Statistics, 2013) , I have found that I would have enough money for the first two years of college, if he were going today. In the future I would have enough for a portion of the first year. If available he...

...NAME : MATH133 Unit5 Individual Project – A
Describe the transformations on the following graph of f ( x) log( x) . State the placement of the vertical asymptote and x-intercept after the transformation. For example, vertical shift up 2 or reflected about the x-axis are descriptions. 1)
10 9 8 7 6 5 4 3 2 1 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 -1 -2 -3 -4 -5 -6 -7 -8 -9 -10
Y
X 1 2 3 4 5 6 7 8 9 10
a)
g(x) = log(x - 5)
Description of transformation: Equation(s) for the Vertical Asymptote(s): x-intercept in (x, y) form: b)
g ( x) log( x) 2
Description of transformation: Equation(s) for the Vertical Asymptote(s): x-intercept in (x, y) form:
2) Students in an English class took a final exam. They took equivalent forms of the exam at monthly intervals thereafter. The average score S(t), in percent, after t months was found to be given by S(t) = 68 − 20 log (t + 1), t ≥ 0. a) What was the average score when they initially took the test, t = 0?
Answer: Show your work in this space: b) What was the average score after 14 months?
Answer:
Show your work in this space:
c) After what time t was the average score 40%? Answer: Show your work in this space:
3) The formula for calculating the amount of money returned for an initial deposit into a bank account or CD (certificate of deposit) is given by
r A P1 n A is the amount of...

...=(8700)
(0.0325)(3)
I = ₱848.25
http://www.investopedia.com/terms/s/simple_interest.asp
Math Of Investment (Calculator-Based) by Zorilla, Partible, Esller, Mendoza, Bansa, Apuyan
B. Discount Interest Method
The interest rate charged to commercial banks and other depository institutions for loans received from the Federal Reserve Bank’s discount window.
Pr = Proceeds
M = Maturity Value
B = Bank Discount
B = Bank Discount
D = Discount Rate
T = Time (in years)
M = Maturity Value
Pr = Proceeds
B = Bank Discount
D = Discount Rate
T = Time
M = ₱32000
D= 10%
T= 1.5 years
B=MDT
B = (32000)(0.10)(1.5)
=₱4800
http://www.investopedia.com/terms/d/discountrate.asp
Math Of Investment (Calculator-Based) by Zorilla, Partible, Esller, Mendoza, Bansa, Apuyan
C. Add-on Interest Method
The interest rate charged to commercial banks and other depository institutions for loans received from the Federal Reserve Bank’s discount window.
P =Principal
I =Interest
R =Rate
T =Time(in years)
I =Interest Amount
P =Principal
R =Rate
T =Time
M =Maturity Value
P =Principal
I =Interest
R =Rate
T =Time
₱5000 at 12% add-on interest rate for 36 months
I = (5000)(0.12)(36/12)
=₱1800+5000
=₱6800/36
=₱188.8889(monthly payment)
http://www.investopedia.com/terms/a/add-on_interest.asp
Math Of Investment...

...52 weeks a year. If his current wage is $15.00 per hour, how much will he make per year?
Before taxes Joe will make $31,200.
2. How much would Joe’s wages be in an average month?
Joe’s wages would average $2400 per month.
3. Based upon the information in the video, Joe has a car payment of $249 per month. The average utilities are: Electricity, $79; Gas, $49; Water, Sewer, and Trash, $27. In addition, Joe is saving $100 per month. How much of his monthly salary would be committed to utilities and other expenses?
$504.00 would be committed to utilities and other expenses.
4. If Joe’s net income after taxes each month is equal to 74% of his gross wages, how much is his net income?
$1776 a month would be Joe’s net income.
5. Joe’s total monthly payment to the bank will consist of the mortgage interest plus principal amount plus monthly payments for insurance and taxes. What will the total monthly payments be for a 15-year and a 30-year loan including taxes and insurance?
15 year loan - $575.00 mortgage payment, $31.25 taxes, $20.83 insurance. Total monthly payment $575.00+$31.25+20.83 = $627.08 per month
30 year loan - $439.00 mortgage payment, $31.25 taxes, $20.83 insurance. Total monthly payment $439.00+$31.25+$20.83 = $491.08
6. After paying for his car, utilities, and a 15-year mortgage payment how much will he have left to cover other living expenses? If he goes with a 30-year mortgage, how much will he have left over to cover...

...includes all the items listed below with the exception of which of the following:
(D) Purchase of fixed asset for use in the business
3. Avis Ltd sold an old car for $300,000 to May Ltd for cash. The cost $650,000 and had accumulated depreciation amounting to $400,000. Prepare the journal entries related to the disposal of the car.
Date
Details
Debit
$
Credit
$
Disposal A/C
650,000
Motor Car
650,000
Transferring the Cost of Asset
Accumulated Depreciation
400,000
Disposal
600,000
Transfer of Accumulated Appreciation
Cash
300,000
Disposal
300,000
Sales Proceeds
Disposal Account
50,000
Profit and Loss Account
50,000
Profit has been made
4. $700,000 – $100,000/5 = 120,000
3 years = $360,000
Dr
Machine Account
Cr
Date
Particular
Folio
$
Date
Particular
Folio
$
20X10
20X10
Jan 31
Cash
700,000
Jan 31
Disposal Account
700,000
Jan 31
Depreciation
900,000
Balance c/d
900,000
1,600,000
1,600,000
Feb 1
Bal b/d
900,000
Dr
Disposal Account
Cr
Date
Particular
Folio
$
Date
Particular
Folio
$
20X10
20X10
Jan 31
Machine
700,000
Jan 31
Provision
420,000
Jan 31
Cash
750,000
Jan 31
Fixed Asset
900,000
Jan 31
Profit and Loss Account
130,000
1,450,000
1,450,000
Dr
Provision Account
Cr
Date
Particular
Folio
$
Date
Particular
Folio
$
Year 1990
Balance c/d
60,000...

...Student Answer form Unit 2
1.
a. x-4x=-6
A) 1
B) -10
C) -6
X^2-10x-24=0
(X-4) (x+6)
X-4=0 x=4
X+6=0 x+-6
b. x=7+4=5.5 x=7-4=1.5
x=-b±b2-4ac2a
x= (-7) ± (-7)2-4(3) (20)2a
x=7±64-802a
x=7±-16
x=7+4/2=5.5
x=7-4/2=1.5
c. 10x^2+x-3=0
x=-b±b2-4ac2a
x=-1± (1)2-4(10) (-3)2(10)
x=1±1-12020
x=1±10
x=1+320 = 5
x=1-320= 10
2.
a. (-2.3, 0), (0, 6.3)
b. This is a maximum function.
This is a maximum function because it is at the peak of the parabola.
c. (2, 9)
d. x=2
3.
a. s= -16t^2+64t+25
b. 73 Feet
S= -16◦(1)^2+64◦(1)+25
S= -16+64+25
S=73 feet
c. 2 seconds
-b2a= -642(-16)= -64/-32=2 seconds
d. 89 feet
S= -16*(2)^2+64*(2)+25
S= 64+128+25= 89 feet
4.
a. x | y |
-2 | 4 |
-1 | 0 |
0 | -2 |
1 | -2 |
2 | 0 |
3 | 4 |
Y=x^2-x-2
X=-2 x= -1 x=0 x=1
Y=-2^2-(-2)-2 y= -1^2-(-1)-2 y= 0^2-0-2 y=1^2-1-2
Y=4+2-2 y= 1+1 -2 y= -2 y= 1-1-2
Y=6-2 y= 2-2 y= 0-2
Y=4 y=0...

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