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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Answer:
m∠CEB is 55°
Step-by-step explanation:
Since ∠ADE = 55°, and ∠ADE is half of ∠ADC because ED bisects ∠ADC. Bisect means to cut in half.
∠ADC = 110° because it is double of ∠ADE.
Since AB║CD and AD║BC, the two sets of parallel lines means this shape is a parallelogram. In parallelograms, <u>opposite angles have equal measures</u>.
∠ADC = ∠CBE = 110°
All quadrilaterals have a sum of angles 360°. Since ∠DCB = ∠BAD and we know two of these other angles are each 110°:
360° - 2(110°) = 2(∠DCB)
∠DCB = 140°/2
∠DCB = ∠BAD = 70°
∠DCB was bisected by EC, which makes each divided part half.
∠DCE = ∠BCE = (1/2)(∠DCB)
∠DCE = ∠BCE = (1/2)(70°)
∠DCE = ∠BCE = 35°
All triangles' angles sum to 180°.
In ΔBCE, ∠BCE = 35° and ∠CBE = 110°.
∠CEB = 180° - (∠BCE + ∠CBE)
∠CEB = 180° - (35° + 110°)
∠CEB = 55°
Therefore m∠CEB is 55°.
