For a), this is clearly a given as it is literally to the right of where it says “Given:”
For b), since ON bisects ∠JOH, this means that it splits it into two separate angles - JON and HON, which are similar due to that bisects mean that it splits it equally into two halves
For c), since NO is the same thing as NO, it is equal to itself
For d), since AAS (angle-angle-side) congruence states that if there are two angles that are congruent (proved in a) and b) ) as well as that a side is congruent (proved in c) ), two triangles are congruent
For e), since two triangles are congruent, every side must have one side that it matches up to in the other triangle. As the opposite side of angle H is JO and the opposite side of angle J is OH, and ∠J=∠H, those two are congruent. As JN and HN are the two sides left, they must be congruent.
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The longest possible altitude of the third altitude (if it is a positive integer) is 83.
According to statement
Let h is the length of third altitude
Let a, b, and c be the sides corresponding to the altitudes of length 12, 14, and h.
From Area of triangle
A = 1/2*B*H
Substitute the values in it
A = 1/2*a*12
a = 2A / 12 -(1)
Then
A = 1/2*b*14
b = 2A / 14 -(2)
Then
A = 1/2*c*h
c = 2A / h -(3)
Now, we will use the triangle inequalities:
2A/12 < 2A/14 + 2A/h
Solve it and get
h<84
2A/14 < 2A/12 + 2A/h
Solve it and get
h > -84
2A/h < 2A/12 + 2A/14
Solve it and get
h > 6.46
From all the three inequalities we get:
6.46<h<84
So, the longest possible altitude of the third altitude (if it is a positive integer) is 83.
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Well on a table of data, if the y coordinates increase a lot over a short x distance then it will be steep.
In an equation, you would put it into y = mx + b form and whatever your m is will be your slope. If m is large, it will be steep.
To complete the square (x+2/5)^2 it would be x^2+(4/5)x+4/25
