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Elza [17]
3 years ago
14

Someone help please !!!!!

Mathematics
1 answer:
k0ka [10]3 years ago
4 0

Answer:

<h2><em><u>no, it's 18 dear</u></em></h2>

Step-by-step explanation:

please mark as brain and like please please follow me and I'm waiting excitedly for the reaction of you and your like

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Combine like terms
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What is the genral form of (x-6)2+(y-4)2=56
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X+y=38 that’s your answer
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labwork [276]

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Step-by-step explanation:

3 0
3 years ago
Consider the following division of polynomials.
Bond [772]

x^4=x^2\cdot x^2. Multiplying the denominator by x^2 gives

x^2(x^2+2x+8)=x^4+2x^3+8x^2

Subtracting this from the numerator gives a remainder of

(x^4+x^3+7x^2-6x+8)-(x^4+2x^3+8x^2)=-x^3-x^2-6x+8

-x^3=-x\cdot x^2. Multiplying the denominator by -x gives

-x(x^2+2x+8)=-x^3-2x^2-8x

and subtracting this from the previous remainder gives a new remainder of

(-x^3-x^2-6x+8)-(-x^3-2x^2-8x)=x^2+2x+8

This last remainder is exactly the same as the denominator, so x^2+2x+8 divides through it exactly and leaves us with 1.

What we showed here is that

\dfrac{x^4+x^3+7x^2-6x+8}{x^2+2x+8}=x^2-\dfrac{x^3+x^2+6x-8}{x^2+2x+8}

=x^2-x+\dfrac{x^2+2x+8}{x^2+2x+8}

=x^2-x+1

and this last expression is the quotient.

To verify this solution, we can simply multiply this by the original denominator:

(x^2+2x+8)(x^2-x+1)=x^2(x^2-x+1)+2x(x^2-x+1)+8(x^2-x+1)

=(x^4-x^3+x^2)+(2x^3-2x^2+2x)+(8x^2-8x+8)

=x^4+x^3+7x^2-6x+8

which matches the original numerator.

3 0
3 years ago
Read 2 more answers
What’s the name of each quadrilateral and the answer
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h) a square has angles that must be 90degree big. A rhombus doesn’t.
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