Substitute , so that
Then the resulting ODE in is separable, with
On the left, we can split into partial fractions:
Integrating both sides gives
Now solve for :
3)A triangle has all integer side lengths and two of those sides have lengths 9 and 16. Consider the altitudes to the three side
s. What is the largest possible value of the ratio of any two of those altitudes
2 answers:
Answer:
2 is the largest possible value.
Step-by-step explanation:
Given Data:
Length of one side = 9
Length of second side = 16
Let AB and BC sides of triangle be of length 9 and 16 respectively as shown in figure attached.
Now, let sides AD and CF be the respective altitudes.
Also, ∠ABC = ∅ (as shown in figure)
If AD and CF are the respective altitudes then,
we have
AD = 9Sin∅ ;
CF = 16Sin∅;
By dividing both sides, we get
AD/CF = 9/16
This equation shows that is independant of angle ∅.
Now, let ∠BAC = α
Now we have, BE = 9 sinα
and FC = AC sinα
By dividing both sides, we get
BE/FC=9/AC
Similarly we also have,
BE/AD = 16/AC
ABC is a triangle as long as length of AC is ithin range of 8 to 24 i.e, 8≤AC≤24 (because sum of any two sides of triangle should be greater than length of third side)
Using these values we get ranges of:
9/24 ≤ BF/FC ≤ 9/8 ; 2/3 ≤ BE/AD ≤ 2
So,
2 is the largest possible value of the ratio of any two of these altitudes.
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Answer: OPTION D:
Step-by-step explanation:
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Where "a" is called numerator and "b" is the denominator. The numerator and the denominator are Integers.
In order to solve this exercise you need to remember de Division of fractions:
In this case, according to the information given in the exercise, you know that of rubies are divided into 4 lots.
Therefore, in order to calculate the weight in pounds of each lot, you need to divide by 4.
Then, you get the following result:
As you can notice, this result matches with the Option D.