We find the first differences between terms:
7-4=3; 12-7=5; 19-12=7; 28-19=9.
Since these are different, this is not linear.
We now find the second differences:
5-3=2; 7-5=2; 9-7=2. Then:
Since these are the same, this sequence is quadratic.
We use (1/2a)n², where a is the second difference:
(1/2*2)n²=1n².
We now use the term number of each term for n:
4 is the 1st term; 1*1²=1.
7 is the 2nd term; 1*2²=4.
12 is the 3rd term; 1*3²=9.
19 is the 4th term; 1*4²=16.
28 is the 5th term: 1*5²=25.
Now we find the difference between the actual terms of the sequence and the numbers we just found:
4-1=3; 7-4=3; 12-9=3; 19-16=3; 28-25=3.
Since this is constant, the sequence is in the form (1/2a)n²+d;
in our case, 1n²+d, and since d=3, 1n²+3.
The correct answer is n²+3
Answer:
1. the range of f^-1(x) is {10, 20, 30}.
2. the graph of f^-1(x) will include the point (0, 3)
3. n = 8
Step-by-step explanation:
1. The domain of a function is the range of its inverse, and vice versa. The range of f^-1(x) is {10, 20, 30}.
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2. See above. The domain and range are swapped between a function and its inverse. That means function point (3, 0) will correspond to inverse function point (0, 3).
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3. The n-th term of an arithmetic sequence is given by ...
an = a1 +d(n -1)
You are given a1 = 2, a12 = 211, so ...
211 = 2 + d(12 -1)
209/11 = d = 19 . . . . . solve the above equation for the common difference
Now, we can use the same equation to find n for an = 135.
135 = 2 + 19(n -1)
133/19 = n -1 . . . . . . . subtract 2, divide by 19
7 +1 = n = 8 . . . . . . . . add 1
135 is the 8th term of the sequence.
Answer:
12 if x_y
Step-by-step explanation:
12= 3x4 (I'm sorry for this, I need points)
Answer:
B. 36
Step-by-step explanation:
Answer:
Yes
Step-by-step explanation:
But if you stuck a pencil through the center of the picture the line that it makes would not be coplanar.
