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AMOUNT WITHDRAWN AT THE END OF YEAR 1: USD1,206
AMOUNT WITHDRAWN AT THE END OF YEAR 2: USD1,206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
Daniel's age is X
Daniel's brother age is Y
X = 3 + ( 2Y )
Y = ( X - 3 ) / 2
In case of Daniel is 17
then his brother's age is = ( 17 - 3 ) / 2 = 7 years
3/10 due to there being 10 marbles (for white) for blue 7/10
Answer:
3x(2z+4z)+(−4y+7y)−6
Step-by-step explanation:
1 Collect like terms.
3x+(2z+4z)+(−4y+7y)−6
2 Simplify.
3x+(2z+4z)+(−4y+7y)−6
3x+6z+3y−6
Answer:
x = -7/3
Step-by-step explanation:
