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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the awnser to your question is 48 because you are just subtracting
Answer:
11. C. 315< 6561
12. D. 729>243
13. C. 1
14. B. 1/16
15. B. 1.61*10^8
16. A. 8.5*10^-9
17. A. 49t^8
18. C. -7x^2+8x
19. A. -12k^4-16k^3+20k^2
20. D. 2k^2-7k-4
I didn't guess. It would just take a long time to show all the work I did
Answer:
1) 18
2) 144
3) 51
4) 99.6
Step-by-step explanation:
1) Multiply 1.5 by 12, as 1.5 divided by 1 is 1
2) Multiply 12 by 12, as 12 divided by 1 is 1
3) Multiply 4.25 by 12, as 4.25 divided by 1 is 1
4) Multiply 8.3 by 12, as 8.3 divided by 1 is 1
9514 1404 393
Answer:
39 units
Step-by-step explanation:
The angle markings tell you this is an equilateral triangle, so all sides are the same length.
10x -11 = 2x +29
8x = 40
x = 5
Then the side length is ...
2x +29 = 2·5 +29 = 39
The length of side KL is 39 units.
