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AleksandrR [38]
3 years ago
10

Aziza has a triangle with two sides measuring 11 in. and 15 in. She claims that the third side can be any length as long as it i

s greater than 4 in. Which statement about Aziza’s claim is correct?
Mathematics
1 answer:
irina [24]3 years ago
4 0

Answer:

Aziza’s claim is incomplete. The third side must be between 4 in. and 26 in.

Step-by-step explanation:

With the Triangle Inequality Theorem, saying that the sum of lengths of any two sides of a triangle is greater than the length of the third side. With this we can develop two inequalities:

11 + 15 > x

26 > x

rewrite this as x < 26

11 + x > 15

x > 15 - 11   Subtract 11 from both sides

x > 4

Therefore, the third side can be anywhere greater than 4 inches and less than and less than 26 inches.

4 <  x < 26

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Ans(a):

Given function is f(x)=\frac{3x-1}{x+4}

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so let's solve

x+4≠0 for x

x≠0-4

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Hence at x=4, function can't have solution.


Ans(b):

We know that vertical shift occurs when we add something on the right side of function so vertical shift by 4 units means add 4 to f(x)

so we get:

g(x)=f(x)+4

g(x)=\frac{3x-1}{x+4}+4

We may simplify this equation but that is not compulsory.

Comparision:  

Graph of g(x) will be just 4 unit upward than graph of f(x).


Ans(c):

To find value of x when g(x)=8, just plug g(x)=8 in previous equation

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8-4=\frac{3x-1}{x+4}


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