Triangle ABC has been translated to create triangle A'B'C'. Angles C and C' are both 32 degrees, angles B and B' are both 72 deg
rees, and sides BC and B'C' are both 5 units long. Which postulate or theorem below would prove the two triangles are congruent?
2 answers:
Answer:
ΔABC ≅ ΔA'B'C' are congruent by ASA postulate
Step-by-step explanation:
Let's draw ΔABC and ΔA'B'C'
<u>Statement Reason</u>
i) ∠C = ∠C' = 32° Given
ii) ∠B = ∠B' = 72° Given
iii) BC = B'C' = 5 Given
IV) ΔABC ≅ ΔA'B'C' If two angles and the included
side of one triangle equal to the
corresponding angles and
included side of another triangle
then the two triangles are
congruent. By ASA postulate.
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For the first circle, let's use the pythagorean theorem
now, it just so happen that the hypotenuse on that triangle, is actually 17, but we used the pythagorean theorem to find it, and the pythagorean theorem only works for right-triangles.
so if the hypotenuse is actually 17, that means that triangle there is actually a right-triangle, meaning that the radius there, and the outside line there, are both meeting at a right-angle.
when an outside line touches the radius line, and they form a right-angle, the outside line is indeed a tangent line, since the point of tangency is always a right-angle with the radius.
now, let's check for second circle
well, low and behold, we didn't get our hypotenuse as 16 after all, meaning, that triangle is NOT a right-triangle, and that outside line is not touching the radius at a right-angle, therefore is NOT a tangent line.
let's check the third circle
this time, we did get our hypotenuse to 65, the triangle is a right-triangle, so the outside line is indeed a tangent line.
now, it just so happen that the hypotenuse on that triangle, is actually 17, but we used the pythagorean theorem to find it, and the pythagorean theorem only works for right-triangles.
so if the hypotenuse is actually 17, that means that triangle there is actually a right-triangle, meaning that the radius there, and the outside line there, are both meeting at a right-angle.
when an outside line touches the radius line, and they form a right-angle, the outside line is indeed a tangent line, since the point of tangency is always a right-angle with the radius.
now, let's check for second circle
well, low and behold, we didn't get our hypotenuse as 16 after all, meaning, that triangle is NOT a right-triangle, and that outside line is not touching the radius at a right-angle, therefore is NOT a tangent line.
let's check the third circle
this time, we did get our hypotenuse to 65, the triangle is a right-triangle, so the outside line is indeed a tangent line.
Answer:
Step-by-step explanation:
Given
See attachment for complete question
Required
Match equivalent expressions
Solving (a):
The expression can be written as:
--- 0
---- 1
--- 2
---- 3
---- 4
For the nth term, the expression is:
---- n
So, the summation is:
Solving (b):
The expression can be written as:
--- 0
---- 1
--- 2
---- 3
---- 4
For the nth term, the expression is:
---- n
So, the summation is:
Solving (c):
The expression can be written as:
--- 0
---- 1
--- 2
---- 3
---- 4
For the nth term, the expression is:
---- n
So, the summation is:
Solving (d):
The expression can be written as:
--- 0
---- 1
--- 2
---- 3
---- 4
For the nth term, the expression is:
---- n
So, the summation is:
