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Answer:
Yes, △ABC ∼ △FED by AA postulate.
Step-by-step explanation:
Given:
Two triangles ABC and FED.
m∠A = m∠B
m∠C = m∠A + 30°
m∠E = m∠F =
m∠D = °.
Now, let m∠A = m∠B =
So, m∠C = m∠A + 30° =
Now, sum of all interior angles of a triangle is 180°. Therefore,
m∠A + m∠B + m∠C = 180
Therefore, m∠A = 50°, m∠B = 50° and m∠C = m∠A + 30° = 50 + 30 = 80°.
Now, consider triangle FED,
m∠D+ m∠E + m∠F = 180
Therefore, m∠F = 50°
m∠E = 50° and
m∠D =
So, both the triangles have congruent corresponding angle measures.
m∠A = m∠F = 50°
m∠B = m∠E = 50°
m∠C = m∠D = 80°
Therefore, the two triangles are similar by AA postulate.
Check the picture below.
the piecewise function, has two subfunctions or behaviours, one if x < 1, meaning less, not equals, but less than 1, so it doesn't include one, so it has a "hole" on that endpoint.
and the second behaviour of the piecewise is if x ⩾ 1, larger or equals than 1, so it includes 1, so it has a solid ball on that endpoint.
the piecewise function, has two subfunctions or behaviours, one if x < 1, meaning less, not equals, but less than 1, so it doesn't include one, so it has a "hole" on that endpoint.
and the second behaviour of the piecewise is if x ⩾ 1, larger or equals than 1, so it includes 1, so it has a solid ball on that endpoint.