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Vsevolod [243]
3 years ago
9

10-3•2/1+1 what is the answer???

Mathematics
2 answers:
bearhunter [10]3 years ago
6 0

Answer:

5

Step-by-step explanation:

PEMDAS ( or for whatever one you do... )

You read it from left to right. So, whatever comes first you do then you continue.

Multiplication came BEFORE division in this case.

3 x 2 = 6

Then Division.

You had gotten 6 FROM 3 x 2. Divide that TOTAL with 1.

6/1 = 6

[ Then subtraction is infront of addition. It may seem confusing but regardless how the letters are formatted in "PEMDAS" you ALWAYS read from left to right. "PEMDAS" has Addition infront of Subtraction but in this case, subtraction goes first because it's infront of the addition. ]

Then you subtract.

You had gotten 6 FROM 3x2/1

Subtract that TOTAL with 10. 10 - 6

What is 10 - 6? 4 because 10 is LARGER than 6

Then you add.

You got the TOTAL of 4 FROM 10-3x2/1.

Add that TOTAL with 1.

4 + 1 = 5.

The answer should be 5 but correct me If I'm wrong.

Ne4ueva [31]3 years ago
3 0
14/2 = 7
The answer can be either depending on it you have to reduce
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Answer:

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Step-by-step explanation:

If a polynomial is expressed with real coefficients (which must be the case if it is a function f(x) in the Real coordinate system), then if it has a complex root "a+bi", it must also have for root the conjugate of that complex root.

This is because in order to render a polynomial with Real coefficients, the binomial factor  (x - (a+bi)) originated using the complex root would be able to eliminate the imaginary unit, only when multiplied by the binomial factor generated by its conjugate: (x - (a-bi)). This is shown below:

(x-(a+bi))*(x-(a-bi))=\\(x-a-bi)*(x-a+bi)=\\([x-a]-bi)*([x-a]+bi)=\\(x-a)^2-(bi)^2=\\(x-a)^2-b^2(-1)=\\(x-a)^2+b^2

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So in our case, a+bi is -6i (real part a=0, and imaginary part b=-6)

Then, the conjugate of this root would be: +6i, giving us the other complex root that also may be present in the real polynomial we are dealing with.

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