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fiasKO [112]
3 years ago
8

Write your answer in order of decreasing powers of x. (18x 5 + 6x 4 - 12x 3) ÷ 6x 2

Mathematics
2 answers:
miv72 [106K]3 years ago
8 0
\frac{18x^5 +6x^4 - 12x^3}{6x^2} = \frac{(6x^2)(3x^3 +x^2 - 2x)}{6x^2} = 3x^3 +x^2 - 2x
Dvinal [7]3 years ago
5 0
The answer is:  " 3x³ + x² − 2x " .
________________________________________________________
Explanation:
_________________________________________________________

Take the first expression; or the "numerator" ; 

and  "factor out"  a:  " 6x² " :
___________________________________________________________
 
 \frac{18x^5+6x^4-12x^3}{6 x^{2} } = \frac{6x^2(3x^3+x^2-2x)}{6x^2}  ;

→  Both  " 6x² " values "cancel out to 1" ;  

     →  since:  " { 6x² / 6x² = 1 } " ; 

And we have:   " 3x³ + x² <span>− 2x " . 
</span>________________________________________________________
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A student is given that point P(a, b) lies on the terminal ray of angle Theta, which is between StartFraction 3 pi Over 2 EndFra
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Answer:

<em>A.</em>

<em>The student made an error in step 3 because a is positive in Quadrant IV; therefore, </em>

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Step-by-step explanation:

Given

P\ (a,b)

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Required

Where and which error did the student make

Given that the angle is in the 4th quadrant;

The value of r is positive, a is positive but b is negative;

Hence;

r = \sqrt{(a)^2 + (b)^2}

Since a belongs to the x axis and b belongs to the y axis;

cos\theta is calculated as thus

cos\theta = \frac{a}{r}

Substitute r = \sqrt{(a)^2 + (b)^2}

cos\theta = \frac{a}{\sqrt{(a)^2 + (b)^2}}

cos\theta = \frac{a}{\sqrt{a^2 + b^2}}

Rationalize the denominator

cos\theta = \frac{a}{\sqrt{a^2 + b^2}} * \frac{\sqrt{a^2 + b^2}}{\sqrt{a^2 + b^2}}

cos\theta = \frac{a\sqrt{a^2 + b^2}}{a^2 + b^2}

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<em>The student's mistake is that a is positive in quadrant iv and his error is in step 3</em>

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