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
#SPJ4
Answer:
4.19
Step-by-step explanation:
just use the formula 2pir^2
Answer:
13.2 miles
Step-by-step explanation:
To solve this, we will need to first solve for the base of the triangle and then use the information we find to solve for the shortest route.
(5.5 + 3.5)² + b² = 15²
9² + b² = 15²
81 + b² = 225
b² = 144
b = 12
Now that we know that the base is 12 miles, we can use that and the 5.5 miles in between Adamsburg and Chenoa to find the shortest route (hypotenuse).
5.5² + 12² = c²
30.25 + 144 = c²
174.25 = c²
13.2 ≈ c
Therefore, the shortest route from Chenoa to Robertsville is about 13.2 miles.
Answer:
The probability that the average age of a randomly selected sample of 100 students will be less than 21.8 years is 0.159
Step-by-step explanation:
According to the given data we have the following:
mean = μ= 22
standard deviation = σ = 2
n = 100
μx = 22
σx=σ/√n=2/√100=0.2
Therefore, P( x < 21.8)=P(x-μx)/σx<(21.8-22)/0.2
=P(z<-1)
= 0.159
The probability that the average age of a randomly selected sample of 100 students will be less than 21.8 years is 0.159
