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schepotkina [342]

4 years ago

13

A triangle has a side length of 11, and a side length of 17. The third side length is also a positive integer. How many differen

t possible values are there for the third side length? (Assume that the triangle is non-degenerate.)

2 answers:

damaskus [11]4 years ago

4 0

Answer:

21 different positive integers could be the length of the third side.

Step-by-step explanation:

By triangle inequality, the sum of two sides cannot be greater than the third.

This means that the third side is between (17-11)=6 and (17+11)=28.

How many integers are between there? You could write them out and count them!

Or!

The highest integer the third side can be is 27 and the lowest integer it can be is 7.

So 27-7+1=20+1=21.

Murljashka [212]4 years ago

3 0

It would be in between 28-6

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From here we can solve for $a,b$ :

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$f(x)=(x-2)(x-3)(cx-d)+3x-2$

$f(x)=(cx^3-(5c+d)x^2+(6c+5d)x-6d)+3x-2$

$f(x)=cx^3-(5c+d)x^2+(6c+5d+3)x-(6d+2)$

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$\implies2d=2\implies d=1$

so that

$Q(x)=x-1$

4 0

3 years ago