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4 years ago

11

Solve the equation x^16-2x^15-x^14+4x^13-x^12-2x^11+x^10=0 in the real number system

1 answer:

Svetach [21]4 years ago

3 0

While I'm not too sure what you mean the real number system, I can help you simplify this equation. Through the process of factoring, this equation simplifys down to:

x^10(x^6−2x^5−x^4+4x^3−x^2−2x+1)

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Brand on has 25 buttons.Mason has 6 times as many buttons.How many buttons does Mason have?

Lady_Fox [76]

Answer: Mason has 150 buttons.

Step-by-step explanation: Brandon has 25 buttons. If Mason has 6 times as many, all you have to do is multiply 25 times 6. You end up with 150 buttons.

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3 years ago

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Thx for helping if u do

Fofino [41]

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3 years ago

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Find the exact value of cos(theta) for an angle (theta) with tan (theta)= -2/3 and with its terminal side in Quadrant II.

Naya [18.7K]

Answer:

$-\frac{3\sqrt{13}}{13}$ .

Step-by-step explanation:

Since we are in quadrant two, cosine value is negative while sine value is positive.

We are going to use the Pythagorean Identity: $1+\tan^2(\theta)=\sec^2(\theta)$ .

$1+(\frac{-2}{3})^2=\sec^2(\theta)$

$1+\frac{4}{9}=\sec^2(\theta)$

$\frac{9+4}{9}=\sec^2(\theta)$

$\frac{13}{9}=\sec^2(\theta)$

$\pm \sqrt{\frac{13}{9}}=\sec(\theta)$

$\pm \frac{\sqrt{13}}{3}=\sec(\theta)$

Since cosine and secant are reciprocals then they will have the same sign as along as they both exist.

$\sec(\theta)=-\frac{\sqrt{13}}{3}$

$\cos(\theta)=-\frac{3}{\sqrt{13}}$ .

I don't see this answer as I'm going to rationalize the denominator.

$\cos(\theta)=-\frac{3}{\sqrt{13}} \cdot \frac{\sqrt{13}}{\sqrt{13}}$ .

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4 years ago

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4m-3n=8, for m wats yeah

stealth61 [152]

The answer would be (8+3n)/4

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3 years ago