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:
634
Step-by-step explanation:
659-25=634
Poor little Francis. :(
Your answer is A.
why:
because if the y value changes then your vertical will change. if your x changes than your horizontal will change. if it’s compressed, your x value is less than 1.
Answer:
A. (7n-1) (8n-1)
Step-by-step explanation:
56n^2 -8n-7n+1
56n^2 -8n -7n+1
Factor out 8n from the first group and -1 from the second group
8n (7n-1) -1(7n-1)
Factor out the (7n-1)
(7n-1) (8n-1)
V=(1/3)hpir^2
lets say this is original, undoubled volume
so o=v=(1/3)hpir^2
so for new volume, or n, that is r to 2r, doubled radius
n=(1/3)hpi(2r)^2
n=(1/3)hpi4r^2
n=4((1/3)hpir^2)
remmember that o=(1/3)hpir^2
n=4(o)
it is 4 times the old one
