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masya89 [10]
3 years ago
14

Which statement best describes how to determine whether f(x) = x3 + 5x + 1 is an even function?

Mathematics
2 answers:
balandron [24]3 years ago
7 0

<u>The right answer is:</u> Determine whether (-x)^3+5(-x)+1 is equivalent to x^3+5x+1


A function is said to be even if its graph is symmetric with respect to the y-axis. That is:

A \ function \ y=f(x) \ is \ \mathbf{even} \ if, \ for \ each \ x \ in \ the \ domain \ of \  f:\\ \\ f(-x)=f(x)


According to this definition, the statement that best describes if the function:

f(x)=x^3+5x+1

is even, is:


Determine whether (-x)^3+5(-x)+1 is equivalent to x^3+5x+1


By doing this, we have:

f(-x)=(-x)^3+5(-x)+1 \\ \\ \therefore \ f(-x)=-x^3-5x+1


As you can see:

f(x) \neq f(-x)


Conclusion:<em> </em><em>The function is not even.</em>

777dan777 [17]3 years ago
4 0
Its B -Determine whether (–x)3<span> + 5(–</span>x<span>) + 1 is equivalent to </span>x3<span> + 5</span>x<span> + 1.</span>

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Answer:

Lisa's team have played 24 home games and 25 away games.

Step-by-step explanation:

Given:

Let x represents games played at home.

also y represent games played away.

Total number of games played = 49

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Now according to given data:

Number of home games won = \frac{2}{3}

Number of away game won = \frac{2}{5}

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\frac{2}{3}x+\frac{2}{5}y=26\\\\\frac{2\times 5}{3\times 5}x+\frac{2\times 3}{5\times 5}y=26\\\\\frac{10}{15}x+\frac{6}{15}y=26\\\\\frac{10x+6y}{15}=26\\\\10x+6y= 26\times 15\\10x+6y = 390\\2(5x+3y)=390\\5x+3y=\frac{390}{5}\\\\5x+3y=195 \ \ \ \ equation \ 2\\

Now multiplying equation 1 by 3 we get,

3(x+y)=3\times49\\3x+3y=147 \ \ \ \ \ equation \ 3\\

Now Subtracting equation 2 by equation 3 we get,

(5x+3y=195)-(3x+3y=147)\\2x=48\\x= \frac{48}{2}=24

Substituting value of x in equation 1 we get.

x+y=49\\24+y=49\\y=49-24=25

Hence Lisa's team have played 24 home games and 25 away games.

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Step-by-step explanation:

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