Hi there!
We can begin by making each fraction have a common denominator:
If we make each have a common denominator of ab, we get:
Simplify:
Multiply both sides by ab:
Move b to the opposite side:
Factor out b:
Divide by (ac - 1):
Teresa graphs the following 33 equations: y=2xy=2x, y=x2+2y=x2+2, and y=2x2y=2x2. She says that the graph of y=2xy=2x will event
ually surpass both of the other graphs. Is Teresa correct? Why or why not?
1 answer:
The correct question is
<span>Teresa graphs the following 3 equations: y=2x, y=x2+2, and y=2x2. She says that the graph of y=2x will eventually surpass both of the other graphs. Is Teresa correct? Why or why not?
we have that
y=2x
y=x</span>²+2
y=2x²
using a graph tool
see the attached figure
<span>We can affirm the following
</span>the three graphs present the same domain-----> the interval (-∞,∞)
The range of the graph y=2x is the interval (-∞,∞)
The range of the graphs y=x²+2 and y=2x² is the interval [0,∞)
therefore
<span>Teresa is not correct because the graph of y = 2x will not surpass the other two graphs since in the interval of [0, infinite) the three graphs present the same range</span>
<span>Teresa graphs the following 3 equations: y=2x, y=x2+2, and y=2x2. She says that the graph of y=2x will eventually surpass both of the other graphs. Is Teresa correct? Why or why not?
we have that
y=2x
y=x</span>²+2
y=2x²
using a graph tool
see the attached figure
<span>We can affirm the following
</span>the three graphs present the same domain-----> the interval (-∞,∞)
The range of the graph y=2x is the interval (-∞,∞)
The range of the graphs y=x²+2 and y=2x² is the interval [0,∞)
therefore
<span>Teresa is not correct because the graph of y = 2x will not surpass the other two graphs since in the interval of [0, infinite) the three graphs present the same range</span>
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