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nadya68 [22]
3 years ago
5

3√6-4√6 Please simplify

Mathematics
1 answer:
Eva8 [605]3 years ago
3 0

Answer:

= -√6

Step-by-step explanation:

1) collect like terms by subtracting coefficients:

3√6 - 4√6

(3-4)√6

2) calculate (3-4):

(3-4)√6

-1√6

3) the coefficient of -1 does not have to be written, but the negative sign remains:

-1√6

-√6

so therefore:

-√6 is your answer

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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>

<em />cos\theta = \frac{a\sqrt{a^2 + b^2}}{a^2 + b^2}

Step-by-step explanation:

Given

P\ (a,b)

r = \± \sqrt{(a)^2 + (b)^2}

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

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}

So, from the list of given options;

<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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2 years ago
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Answer:

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trasher [3.6K]

Answer:

1/4 is ur answer :)

Step-by-step explanation:

Hoped It Helped

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ANTONII [103]

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5. 15 hours

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3 0
3 years ago
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dedylja [7]

1. \: MC  \: is \:  tangent \:  to \:  (B)\Rightarrow \widehat{BCM}=90° \\ 2. \: MC  \: and \: MZ \: are \:  tangent  \: to  \: (B)\Rightarrow MC = MZ\Leftrightarrow 5x - 9 = x + 7\Leftrightarrow x = 4 \\ 3. \: BM =  \sqrt{ {BC}^{2} + {MC}^{2}  }  =  \sqrt{ {5}^{2} +  {12}^{2}  }  = 13 \\ \Rightarrow EM = BM - BE = BM - BC = 13 - 5 = 8 \\ 4. \:  \tan(60°) = \frac{MC}{BC}  =  \frac{13 \sqrt{3}  }{ BC}\Leftrightarrow BC =  \frac{13 \sqrt{3} }{\tan(60°)} =   \frac{13 \sqrt{3} }{ \sqrt{3} }  = 13 \\ 5. \: MC =  \sqrt{ {BM}^{2}  -  {BC}^{2} }  =  \sqrt{ {20}^{2}  -  {12}^{2} }  = 16 \\ 6. \:BZ =  \sqrt{ {BM}^{2}  -  {MZ}^{2} } =  \sqrt{ {25}^{2}  -  {20}^{2} }  = 15 \\ \Rightarrow XE=BX+BE=BZ+BZ=15+15=30

7 0
3 years ago
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