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:
Rotation then reflection
Step-by-step explanation:
The triangle starts where the light green one is. Then it is rotated 90 degrees clockwise around the point where the hypotenuse and longer leg meet. From the sark green triangle it is reflectd to where the purple triangle is.
Answer: Your answer is the third one, a function assigns to each input exactly one output.
If you have a true function, there will only be one output for each input, if you had more than one, it isn't a function.
Answer:
Step-by-step explanation:
Eliminating a negative and changing our operation
Rewriting our equation with parts separated
Solving the whole number parts
Solving the fraction parts
Find the LCD of 5/6 and 1/4 and rewrite to solve with the equivalent fractions.
LCD = 12
Combining the whole and fraction parts
[RevyBreeze]
Answer:
y = -2.8x +69.4
Step-by-step explanation:
The 2-point form of the equation of a line can be used to find the equation of the line through points (3, 61) and (13, 33). The general form of it is ...
y = (y2-y1)/(x2-x1)·(x -x1) +y1
For the given points, this is ...
y = (33 -61)/(13 -3)·(x -3) +61
y = -28/10(x -3) +61
y = -2.8x +69.4 . . . . . the equation of the line through the given points
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<em>Comment on the problem</em>
A "line of best fit" is one that minimizes some measure of deviation from the line. Usually, what is minimized is the square of the deviations. Choosing two points to draw the line through may be convenient, but does not necessarily result in a line of best fit.