Answer:
Step-by-step explanation:
Hello, we want to prove that a proposition depending on n, that we can note P(n), is true for any n positive integer greater than 1. We need to follow several steps.
Step 1 - prove P(1)
For n = 1, n(2n+1)=1*3 =3 so we have
3 = 3, which is obviously true.
First step done!
Step 2 - for we assume P(k) and we need to prove P(k+1)
We assume that 3+7+11+...+(4k-1)=k(2k+1)
so we can write that
3+7+11+...+(4k-1)+(4(k+1)-1)=k(2k+1)+(4k+4-1)=k(2k+1)+4k+3
and
(k+1)(2(k+1)+1)=(k+1)(2k+3)
These two expressions are the same so it means that P(k+1) is true, meaning that
3+7+11+...+(4k-1)+(4(k+1)-1)=(k+1)(2(k+1)+1)
Step 3 - The conclusion
Finally, we have just proved that
3+7+11+...+(4n-1)=n(2n+1) for any n positive integer > 0
Thank you
Answer:
2x^2 +4x +4
Step-by-step explanation:
5x ^2 - 3x + 8 - (3x ^2 -7x + 4)
Distribute the minus sign
5x ^2 - 3x + 8 - 3x ^2 +7x - 4
I like to put them vertically
5x ^2 - 3x + 8
- 3x ^2 +7x - 4
-----------------------
2x^2 +4x +4
Answer:
Down here ↓↓↓↓↓
Step-by-step explanation:
(a) Does the Parabola open upward or downward
Upward - the open "space" is above the vertex, so it opens upward
(b) X and y intercepts
x - intercepts : where the graph touches the x - axis
x - intercepts = (2, 0) , (4, 0)
Warning, don't get it mixed up! X - intercepts have y - values of 0, and y - intercepts have x - values of 0
y - intercepts : where the graph touches the y - axis
y - intercepts = (0, 3)
(c) Vertex of the parabola
vertex: the point of symmetry
vertex : (4, -1)
(d) Axis of symmetry
axis of symmetry : x - value of vertex
axis of symmetry : x = 4
-Chetan K
Answer:
B
Step-by-step explanation:
You cannot have two or more domains.
'
3 appears twice and is located in the x axis, which is basically the domain.
Hope this helps