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.
Learn more about TRIANGLE here brainly.com/question/2217700
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Answer:
Step-by-step explanation:
m=x^3+y^2-6(x-y)-2021
distribute: 6x-6y
now you have m=x^3 + y^2 - 6x - 6y - 2021
They are basically like groups. Take -3x+2x , Just do -3 +2 = -1 and then just add the x. So -1x. Next you have -5y+10y . Just do -5 + 10 = 5 and remember to add the x on so your final answer should be -1x + 5y
The correct answer is 523.33 mm^3
//p.s I already took the test and this was the correct answer//
hope I helped :)
