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:
60
Step-by-step explanation:
First we divide 5 from 20 then we get 4 so 4 x 15 would be 60.
Hey there!!
The given set is a function
( x , y ) - this is the form in which we write co-ordinates
In order to be a function , at least x values shall repeat.
Noted down all the x values
2 , -1, 4 , -2
None of the values is repeating.
Hence, the given data is a function
Hope my answer helps!
Answer:
24, 32, 40, or 48
Step-by-step explanation:
Count by 8.
8
16
24
32
40
48
56
Answer:
25.5 mph
Step-by-step explanation:
So Bradley's speed can be modeled by the equation y=2x+40 where y=speed, x=time in hours after noon, and b=initial speed
So 12:15 is 15 minutes after noon, which is also 0.25 or 1/4 of an hour after noon. This is the x-value. Plug this into the equation to get his speed at 12:15
y=2(0.25)+40
y=0.5+40
y=40.5
So his speed was 40.5 at the time and since he was going 15 miles over the speed limit, the speed limit is 15 less than his speed
40.5 - 15 = 25.5
