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frosja888 [35]
3 years ago
8

A triangle has side lengths of n, n – 3, and 2(n − 2). If the perimeter of the triangle is at least 37 units, what is the value

of n?
OA.

n> 11

OB

n> 8

O C.

n > 10.5

OD

n > 7.5
Mathematics
1 answer:
Vilka [71]3 years ago
3 0

Answer:

n = 11

Step-by-step explanation:

The side length of a triangle are n,n-3 and 2(n-2).

The perimeter of the triangle, P = 37 units

We need to find the value of n.

Perimeter = sum of all sides

n+n-3 +2(n-2) = 37

2n-3+2n-4 = 37

4n -7 = 37

4n = 37+7

4n = 44

n = 11

So, the value of n is 11.

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If z^2+1/z^2=11 , find the value of z^3-1/z^3,using only the positive value of z-1/z
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Hello,

Let's assume

t=z- \dfrac{1}{z} \\\\

t^2=z^2-2+ \dfrac{1}{z^2}= 11-2=9==\ \textgreater \ t=3\\\\

t^3=z^3-3z^2* \dfrac{1}{z} +3z* \dfrac{1}{z^2} - \dfrac{1}{z^3} \\\\
=z^3 - \dfrac{1}{z^3} -3(z-\dfrac{1}{z})\\\\
z^3 - \dfrac{1}{z^3}=t^3+3t=3^3+9=27+9=36



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