2m²n² + 6mn² + 16n²
2n²(m²) + 2n²(m) + 2n²(8)
2n²(m² + m + 8)
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: 36 - 100%
43 - x %
Step-by-step explanation 36 - 100%
43 - x %
we multiply by cross -> 36 * x = 100 / 43
we should remove '100%' from new value, because we find only increase.. no new value in percent.
x + 100 = (43 * 100)/36
x = 4300/36 - 100
x = 119+ 16/36 -100
x = 19 +4/9
Answer:
2/8 and 4/16
Step-by-step explanation:
5^3=125; 2.5^2=6.25. So the answer to your first question is 125, and the second one is 6.25
