Answer:
12.5 x 10^12
Step-by-step explanation:
12,500,000,000,000
12 zeroes following 12.5
Ok. First, Kym went to bed at 8:30. There are only 60 minutes in an hour. 10 minutes to make her bed. 8:30 - 10 = 8:20. She had a 25 minute bath "jeez that's long" 8:20 - 25 = 7:55. Kym spent 15 minutes folding clothes: 7:55 - 15 = 7:40. She began to start folding clothes at 7:40.
Answer:<span> B. 7:40 p.m.
Or you could do it the easy way.
15 + 25 + 10 = 50
50 - 8:30 = 7:40
</span>
Moving from a to b, the x component of the desired vector is 7-(-9) = 16, and
the y component is 3-9 = -6.
So the vector from a to b is <16,-6>, and the magnitude is
sqrt(16^2 + (-6)^2 ), applying the Pythagorean Theorem.
Answer:
Debra Goforth's savings account shows a balance of $904.51 on March 1. The same day, she ... of $904.51 on March 1. The same day, she made a deposit of $375.00 to the account. She also made deposits of $500.00 on April 1 and May 1. The bank pays interest at a rate of 5.5 percent compounded daily.
Step-by-step explanation:
Hope this helps!
The correct option is (B) yes because all the elements of set R are in set A.
<h3>
What is an element?</h3>
- In mathematics, an element (or member) of a set is any of the distinct things that belong to that set.
Given sets:
- U = {x | x is a real number}
- A = {x | x is an odd integer}
- R = {x | x = 3, 7, 11, 27}
So,
- A = 1, 3, 5, 7, 9, 11... are the elements of set A.
- R ⊂ A can be understood as R being a subset of A, i.e. all of R's elements can be found in A.
- Because all of the elements of R are odd integers and can be found in A, R ⊂ A is TRUE.
Therefore, the correct option is (B) yes because all the elements of set R are in set A.
Know more about sets here:
brainly.com/question/2166579
#SPJ4
The complete question is given below:
Consider the sets below. U = {x | x is a real number} A = {x | x is an odd integer} R = {x | x = 3, 7, 11, 27} Is R ⊂ A?
(A) yes, because all the elements of set A are in set R
(B) yes, because all the elements of set R are in set A
(C) no because each element in set A is not represented in set R
(D) no, because each element in set R is not represented in set A
