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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Answer:
4 significant figures are in the number 0.3368
Answer:
B
Step-by-step explanation:
In the attached file
The least common denominator for the question you have supplied me with is 10. The least common multiple of the denominators, 5 and 10, is 10, making the LCD 10.
Answer:
C. (1, -3)
Step-by-step explanation:
Plug in the x to get the y in the equation. The point must fit the formula.
Plug in C. (1, -3), in which x = 1, y = -3:
(-3) = 2(1) - 5
Simplify:
(-3) = 2(1) - 5
-3 = 2 - 5
-3 = -3 (True).
C is your answer.
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