Answer:
Well 5 times 1 is 2.5 which can't be and integer
so I looked it up an this is what I got How do you write 2.5 as an integer?
If this is the question, then Integer-Part[2.5] = 2, and the Fractional-Part[2.5] = . 5.
I would put 3 it because 2.5 as a hole number is 3
Step-by-step explanation:
Answer:
D: {9, 21, 33}
Step-by-step explanation:
Ken worked 2, 8 and 14 hours on 3 separate days.
For working 2 hours, his earnings were f(2) = 2(2) + 5, or 9;
For working 8 hours, his earnings were f(28) = 2(8) + 5, or 21; and
For working 14 hours, his earnings were f(14) = 2(14) + 5, or 33
Thus, the range of this function for the days given is {9, 21, 33} (Answer D)
Answer:
Cylinder. The explanation is not very exact, more intuition and whatnot, so let me know if there was anything you did not understand.
Step-by-step explanation:
This is assuming the rectangle has its sides parallel with the x and y axes.
Let's try to eliminate them one by one. A cone does not seem likely since it has slanted sides, and spinning the two straight sides will keep the sides straight. This can be used for the pyramid as well. Let me know if this doesn't make sense though.
Now for me I would imagine that rectangle being a sheet of paper, or metal. if you set that sheet atop some dirt or something and then rotated it, what shape would be drawn into the material? It would be a circle. So we know the 3d shapewill have a base of a circle, which leaves us with a cylinder. Again, let me know if this method did not make sense.
The answer is m = 250 - 25w.
This is because this answer is in y=mx+b format (slope-intercept form) and has the correct values.
In slope-intercept form, m is the slope and b is y-intercept.
In m = 250 - 25w, the y-intercept is 250 and the slope is -25.
This matches the graph.
Y-intercept is where a line crosses the y-axis, which is 250 in this case.
The slope is negative since the values are decreasing over time (as x-values increase, y-values decrease).
Answer:
The domain of the inverse of a relation is the same as the range of the original relation. In other words, the y-values of the relation are the x-values of the inverse.
Thus, domain of f(x): x∈R = range of f¯¹(x)
and range of f(x): x∈R =domain of f¯¹(x)