1. 3/6 = 1/2
2. 5/6
3. 3/6 : 3/6 = 3 : 3
4. 3/6 : 3/6 = 3 : 3
5. 5/6 : 1/6 = 5 : 1
7. 3/6 × 1/2 = 1/4
Formula to convert F to C:
(F -32) x 5/9
F = 63 + 10 = 73
73-32 = 41
41 x 5/9 = 22.77777
Rounded to the nearest degree = 23 C
Answer:

Step-by-step explanation:

Given integral:



Using <u>integration by parts</u>:


Therefore:





Therefore:



Divide both sides by 2:

Rewrite in the same format as the given integral:

There is one answer, -24.
While yes, absolute value does indicate both a positive and negative will work, for every x value there will be only one y value associated with it; if there were more, then it would not meet the criteria to be a function.
Answer:
<u>The correct answer is C. It is the original amount of money the bank loans the borrower.</u>
Step-by-step explanation:
Let's recall that are five basic elements for calculating the payment of a loan:
1. The principal. How much money you borrow.
2. The interest rate. How much money you will pay in addition to the principal.
3. The period of time. How long will it takes you to pay the loan.
4. The frequency of payment. Will you pay every month?, every quarter?, every year or maybe every two weeks?
5. Additional payments. When you have additional income seasonally and you want to lower either the period of time or the interests to pay.