The interval where the function is nonlinear and decreasing is 0 < x < 4
<h3>How to determine the interval where the function is nonlinear and decreasing?</h3>
The straight lines on the graph are the intervals where the graph is linear
This means that the straight lines on the graph will not be considered
Considering the curve, the graph decrease from x = 0 to x = 4
This can be rewritten as:
0 < x < 4
Hence, the interval where the function is nonlinear and decreasing is 0 < x < 4
Read more about function intervals at:
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Isosceles triangle I think
-2x = -1
x = 1/2
Any equation that returns x=1/2 is a solution to this problem. For example, x= (3•4)/6 can be simplified so that x=1/2
Ordered pairs that work for this direct variation are (4, 3), (8, 6) and (12, 9).
In order to find these, we must first find the value of the direct variation coefficient. We can do that using the base equation y = kx and then by plugging in to find k.
y = kx
12 = k(16)
3/4 = k
Now that we have k, we can model the equation as y = 3/4x. We can also find any number of ordered pairs by using the x value and finding the y value. All of the above answers work.
2/5 of 50 is 20. 50-20=30. Making FG=30
