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WARRIOR [948]
3 years ago
15

(03.07) Select the equation of the line that passes through the point (5, 7) and is perpendicular to the line x = 4. (2 points)

Select one: a. y = 4 b. x = 5 c. y = 7 d. x = 7
Mathematics
2 answers:
borishaifa [10]3 years ago
6 0

Answer:

It's y=7 Guys!!!

Step-by-step explanation:

Andru [333]3 years ago
3 0

Answer:

y=7

Step-by-step explanation:

It has to be go in the opposite direction and x=4 so the answer will have to be y=something and if the line has the point (5,7) then the line must be y=7

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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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