The given statement "The exterior angles of a triangle are always obtuse." is not always true.
The given statement is "The exterior angles of a triangle are always obtuse."
We need to provide a counterexample to the given statement.
The angle is formed by a polygonalside and its extended neighbouring side. When a transversal cuts one of two lines, it creates an angle that is outside of the line.
Take an obtuse-angled triangle as a counter-example:
An obtuse-angled triangle or obtuse triangle is a type of triangle whose one of the vertex angles is bigger than 90°. An obtuse-angled triangle has one of its vertex angles as obtuse and other angles as acute angles i.e. if one of the angles measure more than 90°, then the sum of the other two angles is less than 90°.
In the figure given below, we can see in ΔABC, ∠A=110°.
Exterior angle to ∠A measures 70°.
Therefore, the given statement "The exterior angles of a triangle are always obtuse." is not always true.
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Any answer choices? Im sorry
Answer:
y = (-3/7)x + 3
Step-by-step explanation:
As we go from (0,3) to (7,0), x increases by 7 from and y decreases by 3 to 0.
Thus, the slope of this line is m = rise / run = -3/7.
Subst. the appropriate values from this info into the slope-intercept form of the equation of a straight line:
y = (-3/7)x + 3 Note that the y-intercept is given: (0,3).