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:
2+32
34
Step-by-step explanation:
First we should multiply before adding after multiplying 8×4 is 32 and add 32+2 then it is 34
#1) 5/20
#2) 7/300
#3) 3/50
#4) 5/20
Explanation
For #1:
There are 15 even numbers out of 30. Since it is replaced before drawing the second ball, there will be 15 odd numbers out of 30. This gives us
15/30(15/30) = 225/900 = 5/20.
For #2:
There is 1 7 out of 30; then there are 14 numbers greater than 16 out of 30:
1/30(14/30) = 14/900 = 7/300
For #3:
There are 6 multiples of 5 out of 30; then there are 9 prime numbers out of 30:
6/30(9/30) = 54/900 = 3/50
For #4:
There are 15 even numbers out of 30; then there are still 15 even numbers out of 30:
15/30(15/30) = 225/900 = 5/20
Yes, you can use division to solve a missing factor problem
Answer:
14.625
Step-by-step explanation:
i believe this is the answer all you do is so
78/5.3
