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-Dominant- [34]
2 years ago
7

What is the degree of the polynomial? 4x3+3x2−9x+7

Mathematics
1 answer:
PolarNik [594]2 years ago
7 0

Answer:

3

Step-by-step explanation:

The highest degree amount or exponet amount is the power of a equation.

3 is the highest exponet so it is 3.

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One factor of the polynomial is . which expression represents the other factor, or factors, of the polynomial?
stich3 [128]

Your question was incomplete. Please refer the content below:

One factor of a polynomial is (x+1) . which expression represents the other factor, or factors, of the polynomial 2 x² + 3 x + 1 ?

We get that if one factor of the polynomial 2 x² + 3 x + 1 is (x+1), then the other factor is (2x+1).

Factor means that we have to split an expression into multiple values and make the power of the variable linear.

For a quadratic equation, there will be 2 factors.

We have the polynomial

2 x² + 3 x + 1

Using middle term splitting, we get that:

= 2 x² + 2 x + x + 1

Taking common factor:

= 2 x ( x + 1) + 1 (x + 1)

= (2 x + 1)( x + 1)

Therefore, we get that if one factor of the polynomial 2 x² + 3 x + 1 is (x+1), then the other factor is (2x+1).

Learn more about polynomials here:

brainly.com/question/4142886

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7 0
1 year ago
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Moira does one-half of a homework assignment on Monday. On Tuesday, Wednesday, and Thursday, she does one-half of the homework s
gladu [14]
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Tuesday 1/4
Wednesday 1/8
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8 0
3 years ago
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Prove: An even plus an even is<br> an even<br> 2n + 2m = [? ](n + m)<br> = even
KonstantinChe [14]

Answer:

2(n + m)

Step-by-step explanation:

3 0
2 years ago
Find x<br><br><br><br><br> Thanks for the help in advance!
Marina86 [1]

I'm not sure if this is the easiest way of doing this, but it surely work.

Let the base of the triangle be AB, and let CH be the height. Just for reference, we have

AH=2,\quad HB=6,\quad AC=x

Moreover, let CH=y and BC=z

Now, AHC, CHB and ABC are all right triangles. If we write the pythagorean theorem for each of them, we have the following system

\begin{cases}4+y^2=x^2\\36+y^2=z^2\\x^2+z^2=64\end{cases}

If we solve the first two equations for y squared, we have

y^2=x^2-4\\y^2=z^2-36

And we can deduce

z^2 = x^2+32

So that the third equation becomes

x^2+x^2+32=64 \iff 2x^2 = 32 \iff x^2=16 \iff x=4

(we can't accept the negative root because negative lengths make no sense)

7 0
2 years ago
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