Answer:
Step-by-step explanation:
Let the integer be 6 for even and 7 for odd (say)
For 6, we divide by 2, now get 3. Now we multiply by 3 and add 1 to get 10. Now since 10 is even divide by 5, now multiply by 3 and add 1 to get 16. Now divide by 2 again by 2 again by 2 again by 2 till we get rid of even numbers.
The result is 1, so multiply by 3 and add 1 we get 4 now divide 2 times by 2 to get 1, thus this result now again repeats after 2 times.
Say if we select off number 3, multiply by 3 and add 1 to get 10 now divide by 5, now repeat the same process as above for 5 until we get 1 and it gets repeated every third time.
Thus whether odd or even after some processes, we get 1 and the process again and again returns to 1.
Answer:
7 800 pour 1/2 heure
Step-by-step explanation:
15 600/2=7 800
Sammy calculated the total amount that he deposited by <em>adding up the total amount deposited in each account.</em>
- The Checking account is used to pay for fixed and variable essential expenses. In this account, Sammy deposited $1,712.
- The Cash account is used for non-essential expenses, like snacks, etc. A total of $260 was deposited into this account.
- The Emergency Savings account is used for both other predictable and unpredictable expenses. A total of $340 was deposited into the account.
- For the Vacation Savings account, it is used to carter for other long-term expenses. A total of $50 was deposited into the account.
- When these accounts are added up, the total deposit made by Sammy can be calculated to be $2,362 ($1,712 + $260 + $340 + $50).
Data and Calculations:
Accounts Checking Cash Account Emergency Vacations
Savings Savings
Expenses:
Essential (fixed) $1,292
Essential (variable) 420
Non-essential $260
Other predictable $130
Other (unpredictable) 210
Other (long-term) $50
Total $1,712 $260 $340 $50
Thus, Sammy calculated the total amount that he deposited by adding up the total deposit in each account.
Link for related question:
brainly.com/question/22429629
Figure a- 9×3 = 27 ft^2
figure b- 12 × 5 = 60 ft^2
figure c- 1/2(4×3) = 6 ft^2
figure d- 1/2(5×2) = 5 ft^2
We are given:
138 km south
83 km west of the bearing is the drop-off point.
First, we need to illustrate the problem to clearly see the pattern.
A triangle is formed, 138 km down, then 83 km left.
c^2 = 138^2 + 83^2
c = 161.04 km
The angles are determined using Pythagorean Theorem:
a) 217 degrees
b) 31 degrees
