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Law Incorporation [45]
3 years ago
10

The sum of $3,000 is deposited into an account paying 10% annually. If $1,206 is withdrawn at the end of years 1 and 2, how much

then remains in the account?"
Mathematics
1 answer:
Yakvenalex [24]3 years ago
7 0
What we have so far: 
INITIAL CASH AMOUNT IN THE BANK: USD3,000
ANNUAL INCREASE OF THE CASH AMOUNT IN THE BANK: 10%
YEARS THE CASH STAYED IN THE BANK: 2 years.
AMOUNT WITHDRAWN AT THE END OF YEAR 1: USD1,206
AMOUNT WITHDRAWN AT THE END OF YEAR 2: USD1,206

First, we need to solve for YEAR 1:
FOR YEAR 1: 
Initial Deposit * Annual Increase Rate = Annual Increase
3,000 * 0.10 = Year 1's Annual Increase
Year 1's  Annual Increase = USD300
∴The YEAR 1'S ANNUAL INCREASE IS USD300.
∴The NEW AMOUNT is now USD3,300.

BUT NOT SO FAST! After the year, you took out USD1,206.
New Amount - USD1,206 = Year 1 Amount
3,300 - 1,206 = Year 1 Amount
Year 1 Amount = USD2094
∴The YEAR 1 AMOUNT which will carry over to YEAR 2 is USD2094.

Now, let us solve for the REMAINING BALANCE.
FOR YEAR 2's Annual Increase: 
YEAR 1 AMOUNT * Annual Increase = Year 2's Annual Increase
2094*0.10 = Year 2's Annual Increase
Year 2's Annual Increase = USD209.4
∴The YEAR 1'S ANNUAL INCREASE IS USD209.4.
∴The NEW AMOUNT is now USD2,303.4.

But you took out USD1,206
USD2,303.4 - USD1,206 = Remaining Balance
Remaining Balance = USD1097.4

∴The Answer is: USD1097.4

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Consider this equation: –3x – 5 2x = 6which is an equivalent equation after combining like terms?
Alla [95]
-3x-5+2x=6
2x-3x-5=6
-1x-5=6
-x-5=6
-x=11
x=11
an equivilent equaiton is x=11
7 0
3 years ago
Use Newton's Method to approximate the zero(s) of the function. Continue the iterations until two successive approximations diff
liubo4ka [24]

Answer:

There is only one real zero and it is located at x = 1.359

Step-by-step explanation:

After the 4th iteration the solution was repeating the first 3 decimal places.  The formula for Newton's Method is

x_{n}-\frac{f(x_{n}) }{f'(x_{n}) }

If our function is

f(x)=x^5+x-6

then the first derivative is

f'(x)=5x^4+1

I graphed this on my calculator to see where the zero(s) looked like they might be, and saw there was only one real one, somewhere between 1 and 2.  I started with my first guess being x = 1.

When I plugged in a 1 for x, I got a zero of 5/3.  

Plugging in 5/3 and completing the process again gave me 997/687

Plugging in 997/687 and completing the process again gave me 1.36976

Plugging in 1.36976 and completing the process again gave me 1.359454

Plugging in 1.359454 and completing the process again gave me 1.359304

Since we are looking for accuracy to 3 decimal places, there was no need to go further.

Checking the zeros on the calculator graphing program gave me a zero of 1.3593041 which is exactly the same as my 5th iteration!

Newton's Method is absolutely amazing!!!

5 0
3 years ago
Find a fraction or mixed number that is between the two numbers -3/4 -1/3
Mandarinka [93]

Hey there!


-3/4 - 1/3

= -3/4 + (-1/3)

≈ -9/12 - 4/12

= -9 - 4 / 12 - 0

≈ -9 - 4 / 12

= -13 / 12

≈ -1 1/12


Therefore, your answer is/

-13/12 or -1 1/12

EITHER OF THOSE SHOULD BE CORRECT BECAUSE THEY ARE EQUIVALENT OR THE SAME THING.


Good luck on your assignment & enjoy your day!



~Amphitrite1040:)

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2 years ago
Please help mee like pretty please
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answer

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Step-by-step explanation:

6 0
3 years ago
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Helppppp please i tont get these answers will get brainly i swear
Marrrta [24]

Answer:15-A

Step-by-step explanation:

5 0
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