1. A relation is plotted as a linear function on a coordinate plane starting at point C at (3, –2) and ending at point D at (–2,
3). What is the rate of change for the linear function and what is its initial value?
The rate of change is ______ and the initial value is ______.
A. 1 and -1
B. -1 and 1
C. 5 and -2
D. -2 and 5
2. Ava drove her car at a constant rate to the train station. At the train station, she waited for the train to arrive. After she boarded the train, she traveled at a constant rate, faster than she drove her car. She entered the taxi and traveled at a constant speed. This speed was equal to the speed at which she had driven her car earlier. After some time, she arrived at her destination.
Which graph represents Ava’s travel plans? (First 3 graphs are the options to this question.)
The rate of change is ______ and the initial value is ______.
A. 1 and -1
B. -1 and 1
C. 5 and -2
D. -2 and 5
2. Ava drove her car at a constant rate to the train station. At the train station, she waited for the train to arrive. After she boarded the train, she traveled at a constant rate, faster than she drove her car. She entered the taxi and traveled at a constant speed. This speed was equal to the speed at which she had driven her car earlier. After some time, she arrived at her destination.
Which graph represents Ava’s travel plans? (First 3 graphs are the options to this question.)
1 answer:
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From the given information, we get the value of ABC = 120°.
<h3>How to estimate the value of ABC?</h3>
Given: In the figure, O exists the center of the circle and OABC exists as a parallelogram.
Now, the radius of the circle exists
OA = OB = OC
Opposite sides of a parallelogram are equal
AB = OC and OA = BC
In ∆OAB,
OA = OB = AB and,
In ∆OCB,
OC = OB = BC
Therefore, ∆OAB and ∆OCB exist in equilateral triangles.
All angles of an equilateral triangle are equivalent to 60°.
Hence, ∠ABC = ∠OBA + ∠OBC
∠ABC = 60° + 60°
∠ABC = 120°
Therefore, the value of ∠ABC = 120°.
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