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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You have distribute the 5 into the 2 which is 10w +15 +30w.
Answer:
C)2 : 1
Step-by-step explanation:
AEFG ≈ ALMN
FG / MN = 4 / 2 = 2/1
So the ratio of AEFG to ALMN = 2 / 1 or 2 : 1
Because he has 4 pounds of apples left, you are to subtract that from the total pounds picked, which was a total of 55 pounds. So 55 with 4 taken away, or 55 - 4, equals 51. Since he had equally put 51 pounds into six bags, you are to then divide . 51 / 6 = 8.5. He had put 8.5 pounds worth of apples into each bag.
