At Westonci.ca, we provide clear, reliable answers to all your questions. Join our vibrant community and get the solutions you need. Discover in-depth solutions to your questions from a wide range of experts on our user-friendly Q&A platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
Answer:
y=n^2+3n+4
Step-by-step explanation:
Sequence,S: 8,14,22,32 ,44, 58, 74...
First differences of S: 6,8,10,12,14,16,...
Second differences of S: 2,2,2,2,2,2,....
It is indeed a quadratic since the 1st differences that are the same are the second differences.
So the 1st term is 8. If the 0th term was listes it would be 4 since 8-4=4 and 6-4=2.
This means our quadratic it is in the form
y=an^2+bn+c where c=4.
So we have y=an^2+bn+4.
Now we can find a and b using two points from our sequence to form a system of equations to solve.
(1,8) gives us 8=a(1)^2+b(1)+4
Simplifying gives 8=a+b+4
Subtracting 4 on both sides gives: 4=a+b
(2,14) gives us 14=a(2)^2+b(2)+4
Simplifying gives 14=4a+2b+4
Subtracting 4 on both sides gives 10=4a+2b
So we have the following system to solve:
4=a+b
10=4a+2b
I'm going to solve using elimination/linear combination style.
Dividing second equation by 2 gives the system as:
4=a+b
5=2a+b
Subtract the 2nd equation from the first gives:
-1=-1a
This implies a=1 since -1=-1(1) is true.
If a+b=4 and a=1, then b=3. Because 3+1=4.
So the equation is y=1n^2+3n+4 or y=n^2+3n+4.
We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.
