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Dmitry_Shevchenko [17]
4 years ago
12

If a triangle has side lengths 14,48, and 50 units what type of triangle is it an acute an obtuse a right or no solution.

Mathematics
1 answer:
Snezhnost [94]4 years ago
4 0

Answer:

No solution

Step-by-step explanation:

we know that

The <u>Triangle Inequality Theorem</u> states that the sum of any 2 sides of a triangle must be greater than the measure of the third side

so

In this problem

48+50>14

98>14 -----> is not true

therefore

You can't build a triangle with those dimensions

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Factor f(x)=x4-x3-7x2+13x-6 completely. Then Sketch the graph
Dennis_Churaev [7]

ANSWER TO QUESTION 1


f(x)=(x-2)(x+3)(x-1)^2.



EXPLANATION


The function given to us is,


f(x)=x^4-x^3-7x^2+13x-6


According to rational roots theorem,


\pm1,\pm2,\pm3,\pm6 are possible rational zeros of  


f(x)=x^4-x^3-7x^2+13x-6.


We find out that,


f(-3)=(-3)^4-(-3)^3-7(-3)^2+13(-3)-6


f(-3)=81+27-63-39-6



f(-3)=6-6



f(-3)=0


Also




f(2)=(2)^4-(2)^3-7(2)^2+13(2)-6



f(2)=16-8-28+26-6



f(2)=6-6



f(2)=0


This implies that  


x-2 and x+3 are factors of



f(x)=x^4-x^3-7x^2+13x-6 and hence (x-2)(x+3)=x^2+x-6 is also a factor.



We perform the long division as shown in the diagram.



Hence,


f(x)=(x-2)(x+3)(x-1)^2.


ANSWER TO QUESTION 2

Sketching the graph

We can see from the factorization that the roots  

x=2 and x=-3 have a multiplicity of 1, which is odd. This means that the graph crosses the x-axis at this intercepts.


Also the root x=1 has a multiplicity of 2, which is even. This means the graph does not cross the x-axis at this intercept.



Now we determine the position of the graph on the following intervals,


x\le -3


f(-4)=(-4)^4-(-4)^3-7(-4)^2+13(-4)-6


f(-4)=150\:>0



-3\le x \le 1


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1\le x\le 2


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x \ge 2



f(3)=24\:>0



We can now use these information to sketch the function as shown in diagram



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