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UNO [17]
4 years ago
8

If f(x) and 1(x) are inverse functions of each other and f(x) = 2x+5, what is ri(8)?

Mathematics
1 answer:
Yakvenalex [24]4 years ago
3 0

Answer:

3/2

Step-by-step explanation:

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Rhombus ABCD with vertices A(2, 6), B(6, 7),<br> C(5, 3), and D(1, 2): 180°
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Step-by-step explanation:

Rule for the rotation of a point 180° about the origin is,

(x, y) → (-x, -y)

Therefore, images of the points after the rotation of 180° will be,

A(2, 6) → A'(-2, -6)

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The table shows the snack preferences of 50 shoppers at the mall. What is the probability that one shopper, selected at random f
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What are the roots of the polynomial equation x^3-5x+5=2x^2-5? Use a graphing calculator and a system of equations. Round nonint
leonid [27]

The right answer is c. –2.24, 2, 2.24


This question needs to be solved in two ways. First, using a graphing calculator. Next, using a system of equations.


1. Using a graphing calculator.


We have the following polynomial equation:

x^3-5x+5=2x^2-5


By ordering this equation we have:

x^3-2x^2-5x+10=0


So, we can say that this equation comes from a function given by:

f(x)=x^3-2x^2-5x+10


Thus, by plotting this function, we have that the graph of this function is indicated in Figure 1. By zooming, we can see, in Figure 2, that the roots of the polynomial equation are the x-intercepts of f(x) which are:


x_{1}=-2.236 \\ \\ x_{2}=2 \\ \\ x_{3}=2.236


Finally, rounding noninteger roots to the nearest hundredth we have:


\boxed{Root_{1}=-2.24} \\ \\ \boxed{Root_{2}=2} \\ \\ \boxed{Root_{3}=2.24}


2. Using a system of equations.


The ordered equation is:

x^3-2x^2-5x+10=0


By arranging to factor out we have:

x^3-5x-2x^2+10=0


Then, by factoring:

x(x^2-5)-2(x^2-5)=0


Term (x^2-5) is a common factor, thus:


(x-2)(x^2-5)=0 \\ \\ (x-2)(x-\sqrt{5})(x+\sqrt{5})=0 \\ \\ Finally: \\ \\ \boxed{Root_{1}=-\sqrt{5}=-2.24} \\ \\ \boxed{Root_{2}=2} \\ \\ \boxed{Root_{3}=\sqrt{5}=2.24}

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3 years ago
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