The average of the five consecutive numbers ending with b in discuss when expressed in terms of a is; Choice D; a+3.
<h3>What is the average of five consecutive integers ending with b?</h3>
First, since it was given in the task content that the average of six positive consecutive odd integers starting with a is equal to b, it therefore follows that;
(a+a+2+a+4+a+6+a+8+a+10)/6 = b
6b=6a+30
b=a+5
Also, let the average of the consecutive intergers ending with b be denoted by; x.
(b+b-1+b-2+b-3+b-4)/5 = x
=(5b-10)/5
=b–2
The average, x=b – 2 (where b = a-5)
Ultimately, the value of the required average is; = a+5-2 = a+3.
Read more on average of integers;
brainly.com/question/24369824
#SPJ1
Answer:
The Answer is 0.20833333333
Step-by-step explanation:
Let us evaluate each pair of expressions one at a time.
8. (n²+4) - n² and 4n.
(n²+4) - n² = n²+ 4 - n² = 4 which is not equal to 4n.
NOT EQUIVALENT
9. 3x + 5 and 2(x + 3)
2(x + 3) = 2x + 6 which is not equal to 3x + 5.
NOT EQUIVALENT
10. 15 - 6x and 15(1 - 6x)
15(1 - 6x) = 15 - 90x which is not equal to 15 - 6x
NOT EQUVALENT
11. (y + y + 2 + y) + 3y and 6y + 2
(y +y +2 + y) + 3y = y+y+y +2 + 3y = 3y + 2 + 3y = 6y + 2.
It matches the other expression.
EQUIVALENT
12. 8y - 3 + 10y and 3(6 - 1)
8y - 3 + 10y = 8y + 10y - 3 = 18y - 3
3(6 - 1) = 3(5) = 15
The two expressions do not match.
NOT EQUIVALENT.
Answer: Only the expressions in number 11 are equivalent.
Answer:
addition because it's in parentheses and following PEMDAS parentheses comes first so you do addition first.
I hope you find this helpful :D
Answer:
Check below, please
Step-by-step explanation:
Hello!
1) In the Newton Method, we'll stop our approximations till the value gets repeated. Like this
2) Looking at the graph, let's pick -1.2 and 3.2 as our approximations since it is a quadratic function. Passing through theses points -1.2 and 3.2 there are tangent lines that can be traced, which are the starting point to get to the roots.
We can rewrite it as: 
As for
3) Rewriting and calculating its derivative. Remember to do it, in radians.
For the second root, let's try -1.5
For x=-3.9, last root.
5) In this case, let's make a little adjustment on the Newton formula to find critical numbers. Remember their relation with 1st and 2nd derivatives.
For -1.2
For x=0.4
and for x=-0.4
These roots (in bold) are the critical numbers
