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zavuch27 [327]
3 years ago
5

Given the flag ABCDE, determine the single rule (x,y) that creates the

Mathematics
1 answer:
cricket20 [7]3 years ago
6 0

Answer:

(y, -x)

Step-by-step explanation:

Rotation of 90°

(Clockwise)

(x, y)  ---> (y, -x)

Rotation of 90°

(CounterClockwise)

(x, y) --> (-y, x)

Rotation of 180°

(Both Clockwise and Counterclockwise)

(x, y) ---> (-x, -y)

Rotation of 270°

(Clockwise)

(x, y) ---> (-y, x)

Rotation of 270°

(CounterClockwise)

(x, y) ---> (y, -x)

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Factor this polynomial completely x^2-10x+25
san4es73 [151]

<u>x^2-10x+25</u>

=x^2-5x-5x+25

x(x-5)-5(x-5)

<u>=</u><u>(</u><u>x-5)</u><u>(</u><u>x-5)</u>

<u>x=</u><u>5</u><u> </u><u>,</u><u> </u><u>x=</u><u>5</u><u> </u>

hope it's helpful to you

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2 years ago
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an airline is trying to promote its new Boston to Atlanta flight. the usual price for this flight is $315.00. however the airlin
Veronika [31]

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2 years ago
Evaluate the function for the given values to determine if the value is a root. p(−2) = p(2) = The value is a root of p(x).
bija089 [108]

<em>Note: Since you missed to mention the the expression of the function </em>p(x)<em> . After a little research, I was able to find the complete question. So, I am assuming the expression as </em>p(x)=x^4-9x^2-4x+12<em> and will solve the question based on this assumption expression of  </em>p(x)<em>, which anyways would solve your query.</em>

Answer:

As

p\left(-2\right)=0

Therefore, x=-2 is a root of the polynomial <em> </em>p(x)=x^4-9x^2-4x+12

As

p\left(2\right)=-16

Therefore, x=2 is not a root of the polynomial <em> </em>p(x)=x^4-9x^2-4x+12

Step-by-step explanation:

As we know that for any polynomial let say<em> </em>p(x)<em>, </em>c is the root of the polynomial if p(c)=0.

In order to find which of the given values will be a root of the polynomial, p(x)=x^4-9x^2-4x+12<em>, </em>we must have to evaluate <em> </em>p(x)<em> </em>for each of these values to determine if the output of the function gets zero.

So,

Solving for p\left(-2\right)

<em> </em>p(x)=x^4-9x^2-4x+12

p\left(-2\right)=\left(-2\right)^4-9\left(-2\right)^2-4\left(-2\right)+12

\mathrm{Simplify\:}\left(-2\right)^4-9\left(-2\right)^2-4\left(-2\right)+12:\quad 0

\left(-2\right)^4-9\left(-2\right)^2-4\left(-2\right)+12

\mathrm{Apply\:rule}\:-\left(-a\right)=a

=\left(-2\right)^4-9\left(-2\right)^2+4\cdot \:2+12

\mathrm{Apply\:exponent\:rule}:\quad \left(-a\right)^n=a^n,\:\mathrm{if\:}n\mathrm{\:is\:even}

=2^4-2^2\cdot \:9+8+12

=2^4+20-2^2\cdot \:9

=16+20-36

=0

Thus,

p\left(-2\right)=0

Therefore, x=-2 is a root of the polynomial <em> </em>p(x)=x^4-9x^2-4x+12<em>.</em>

Now, solving for p\left(2\right)

<em> </em>p(x)=x^4-9x^2-4x+12

p\left(2\right)=\left(2\right)^4-9\left(2\right)^2-4\left(2\right)+12

\mathrm{Remove\:parentheses}:\quad \left(a\right)=a

p\left(2\right)=2^4-9\cdot \:2^2-4\cdot \:2+12

p\left(2\right)=2^4-2^2\cdot \:9-8+12

p\left(2\right)=2^4+4-2^2\cdot \:9

p\left(2\right)=16+4-36

p\left(2\right)=-16

Thus,

p\left(2\right)=-16

Therefore, x=2 is not a root of the polynomial <em> </em>p(x)=x^4-9x^2-4x+12<em>.</em>

Keywords: polynomial, root

Learn more about polynomial and root from brainly.com/question/8777476

#learnwithBrainly

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Answer:

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Step-by-step explanation:

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Simplify the expression .
GaryK [48]
The answer is B
Hope this helps;)
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