Answer:
Step-by-step explanation:
The first differences of the sequence are ...
- 5-2 = 3
- 10-5 = 5
- 17-10 = 7
- 26-17 = 9
- 37-26 = 11
Second differences are ...
- 5 -3 = 2
- 7 -5 = 2
- 9 -7 = 2
- 11 -9 = 2
The second differences are constant, so the sequence can be described by a second-degree polynomial.
We can write and solve three equations for the coefficients of the polynomial. Let's define the polynomial for the sequence as ...
f(n) = an^2 + bn + c
Then the first three terms of the sequence are ...
- f(1) = 2 = a·1^2 + b·1 + c
- f(2) = 5 = a·2^2 +b·2 + c
- f(3) = 10 = a·3^2 +b·3 +c
Subtracting the first equation from the other two gives ...
3a +b = 3
8a +2b = 8
Subtracting the first of these from half the second gives ...
(4a +b) -(3a +b) = (4) -(3)
a = 1 . . . . . simplify
Substituting into the first of the 2-term equations, we get ...
3·1 +b = 3
b = 0
And substituting the values for a and b into the equation for f(1), we have ...
1·1 + 0 + c = 2
c = 1
So, the formula for the sequence is ...
f(n) = n^2 + 1
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The 20th term is f(20):
f(20) = 20^2 +1 = 401
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<em>Comment on the solution</em>
It looks like this matches the solution of the "worked example" on your problem page.
(0.5,2)
0.5 is x value
2 is y value
l
l
---------------
l
l
x value is on dashed line (0.5)
y value is on more separated line going up and down. (2)
To plot 0.5, you can put it as an equation. 0.5= 1/2. So to plot it, 0.5 is half of 1. So plot it as in between the 1 and 0.
To plot 2 go up to the two on the y axis and then go over 0.5 and plot a point.
That is you're plotted point for (0.5,2)
Hope it helps!
Answer:
The first column is transformation
The second one I have forgotten
Need other help
ANSWER:
6_34/99
STEP:
So yes. When a decimal is repeating, you can take the repeating number (most likely a decimal) and put 99 under it. Since 99 cannot be solved, you put 99. So, 34/99. Though we are not finished. There is still the whole 6 number left. So, you do 6_34/99.
Proof:
10x=6.6...
-x=-0.6...
9x=6
x=6/9=1/3.
