Question about Super Tutor Trigonometry (ESDTRIG) for PC

Rounding off some numbers to keep it simple would give a starting amount of 4500

Loringh

Posted on Nov 07, 2008

Hi,

a 6ya expert can help you resolve that issue over the phone in a minute or two.

best thing about this new service is that you are never placed on hold and get to talk to real repairmen in the US.

the service is completely free and covers almost anything you can think of (from cars to computers, handyman, and even drones).

click here to download the app (for users in the US for now) and get all the help you need.

goodluck!

Posted on Jan 02, 2017

The formula for exponential growth and exponential decay is

A =I (1 + r)^n, where A is the amount, I is the initial amount, r is the rate of exponential growth or decay and n is the number of periods.

For example, if we start with 100 units and the growth is 5% per period, we get the equation A = 100 (1+0.5) ^n. When n is 0, anything to the exponent 0 is 1, so we start with 100 units. When n is 1, we get 105 units.

Similarly, if a car costing $50,000 depreciates in value 30% per year, we have A= 50000 (1-0.3)^n, Again, when n = 0, anything to the power of 0 is 1, so we start at $50,000. When n=1, A = 35,000. When n = 2, A = $24,500.

Good luck,

Paul

A =I (1 + r)^n, where A is the amount, I is the initial amount, r is the rate of exponential growth or decay and n is the number of periods.

For example, if we start with 100 units and the growth is 5% per period, we get the equation A = 100 (1+0.5) ^n. When n is 0, anything to the exponent 0 is 1, so we start with 100 units. When n is 1, we get 105 units.

Similarly, if a car costing $50,000 depreciates in value 30% per year, we have A= 50000 (1-0.3)^n, Again, when n = 0, anything to the power of 0 is 1, so we start at $50,000. When n=1, A = 35,000. When n = 2, A = $24,500.

Good luck,

Paul

Nov 05, 2015 | Texas Instruments Ti-30Xs Multiview...

There are 2 values for interest, say R and r, and R = 0.05, r = 0.04

There are 2 values for the amount earned, from the 2 banks, say I and i, and we know I + i = 100

There are 2 values for principal invested, say P and p, and we know P + p = 2125

There is only one value for period, say t = 1

So then I = P * 0.05 * 1 and

i = p * 0.04 * 1 and

I + i = 100 and

P + p = 2125

then

0.05P + 0.04p = 100

P + p = 2125

P = 2125 - p

0.05 ( 2125 - p) +0.04p = 100

106.25 - 0.05p + 0.04p = 100

106.25 - 0.01p = 100

0.01p = 6.25

p = 625

so

P = 2125 - 625 = 1500

so the amounts were $1500 and $625

There are 2 values for the amount earned, from the 2 banks, say I and i, and we know I + i = 100

There are 2 values for principal invested, say P and p, and we know P + p = 2125

There is only one value for period, say t = 1

So then I = P * 0.05 * 1 and

i = p * 0.04 * 1 and

I + i = 100 and

P + p = 2125

then

0.05P + 0.04p = 100

P + p = 2125

P = 2125 - p

0.05 ( 2125 - p) +0.04p = 100

106.25 - 0.05p + 0.04p = 100

106.25 - 0.01p = 100

0.01p = 6.25

p = 625

so

P = 2125 - 625 = 1500

so the amounts were $1500 and $625

Oct 20, 2014 | Mathsoft StudyWorks! Middle School Deluxe...

$15,000 at 3% and $6,000 at 7%.

If this is homework, be sure to show your work.

If this is homework, be sure to show your work.

Sep 06, 2014 | Office Equipment & Supplies

Invest R10000 in a bank investing at 14% compounded twice a year.

A = P(1+i)^n, where A is the amount, P is the principal or initial investment, i is the interest rate per period, and n is the number of periods.

If the annual rate is 14%, the semi-annual rate is 7%. One year is now composed of 2 6-month periods.

So after one year, we have A = 10 000 (1.07)^2 or 11,449.

Good luck,

Paul

A = P(1+i)^n, where A is the amount, P is the principal or initial investment, i is the interest rate per period, and n is the number of periods.

If the annual rate is 14%, the semi-annual rate is 7%. One year is now composed of 2 6-month periods.

So after one year, we have A = 10 000 (1.07)^2 or 11,449.

Good luck,

Paul

Nov 19, 2013 | Sharp EL-738 Scientific Calculator

hi.

for a full definition, try wikipedia.

an independent variable can be freely changed.

a dependent variable will BE changed as the independent variable is changed

basically there are 2 main examples. mathematical, and experimental.

with experiments,(science, chemistry etc.) one example would be changing the amount of fertilizer fed to a plant to see how it affects growth. there are 2 variables here. amount of fertilizer used and growth of the plant.

the amount of the fertilizer can be freely changed. (independent variable)

the amount of growth DEPENDS on the amount of fertilizer (dependent variable)

with maths, you can use something like 2x+5=y.

x can freely changed. it is the independent variable. y will change depending on the value of x.

hope this helps

for a full definition, try wikipedia.

an independent variable can be freely changed.

a dependent variable will BE changed as the independent variable is changed

basically there are 2 main examples. mathematical, and experimental.

with experiments,(science, chemistry etc.) one example would be changing the amount of fertilizer fed to a plant to see how it affects growth. there are 2 variables here. amount of fertilizer used and growth of the plant.

the amount of the fertilizer can be freely changed. (independent variable)

the amount of growth DEPENDS on the amount of fertilizer (dependent variable)

with maths, you can use something like 2x+5=y.

x can freely changed. it is the independent variable. y will change depending on the value of x.

hope this helps

Jul 04, 2011 | Cycling

4 5 0 0 0 +/- PV (investment amount, negative because you're paying it out)

2 5 0 0 0 0 FV (desired amount, positive because you're receiving it)

2 0 SHIFT xP/YR (20 years)

I/YR (calculate annual interest rate)

2 5 0 0 0 0 FV (desired amount, positive because you're receiving it)

2 0 SHIFT xP/YR (20 years)

I/YR (calculate annual interest rate)

Jan 23, 2011 | HP 10bII Calculator

The error comes from the vanishing values at the tail of the function. The exponential function does not vanish except for -infinity. Truncate the lists by removing the pairs (5,00 and (6,0) or fudge up the data (5, 1E-5) (6,1E-6) and you will get your equation of best fit.

I am including a link to a post of mine where I show how to perform an exponential regression. JUST IN CASE you need additional information on the subject.

I am including a link to a post of mine where I show how to perform an exponential regression. JUST IN CASE you need additional information on the subject.

Apr 14, 2010 | Texas Instruments TI-84 Plus Calculator

x=7

y=-1

y=-1

Jan 30, 2009 | Bagatrix Algebra Solved! 2005 (105101) for...

Following formula we can use.

S=P(1+RT)

WHERE S =INVEST AMOUNT+PROFIT

P=INVEST AMOUNT

R= RATE OF RETURN

T=TIME IN YEARS

EX : (300+x) = x(1+(5.4/100)*1)

(300+x)*100 = 105.4x

x=30000/5.4

x=5555. 4

S=P(1+RT)

WHERE S =INVEST AMOUNT+PROFIT

P=INVEST AMOUNT

R= RATE OF RETURN

T=TIME IN YEARS

EX : (300+x) = x(1+(5.4/100)*1)

(300+x)*100 = 105.4x

x=30000/5.4

x=5555. 4

Jan 05, 2008 | The Learning Company Achieve! Math &...

Try the FV function
**Syntax**

**FV**(**rate**,**nper**,**pmt**,pv,type)

Nov 03, 2007 | Computers & Internet

95 people viewed this question

Usually answered in minutes!

×