Write the subtraction expression as an equivalent addition expression and then evaluate it.
1 answer:
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Answer:
<h2>y = 2</h2>Step-by-step explanation:
To find the value of y when x = 8 we must first find the relationship between the two variables.
The statement
y varies directly with variable x is written as
y = kx
where k is the constant of proportionality
when
x = 12
y = 3
Substitute the values into the above formula and solve for k
That's
3 = 12k
Divide both sides by 12
So the formula for the variation is
when
x = 8
we have the final answer as
<h3>y = 2</h3>Hope this helps you
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>