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Anna [14]
4 years ago
6

1. 8y^2-4y+1 divided by 2y-1

Mathematics
1 answer:
____ [38]4 years ago
5 0
Q1. The answer is 4y+ \frac{1}{2y-1}

\frac{8y^{2}-4y+1 }{2y-1} = \frac{4y*2y-4y+1}{2y-1} = \frac{4y(2y-1)+1}{2y-1} = \frac{4y(2y-1)}{2y-1}+ \frac{1}{2y-1} =4y+ \frac{1}{2y-1}


Q2. The answer is 2a+3+ \frac{6}{a-1}

\frac{2 a^{2}+a+3 }{a-1} = \frac{a*2a-2a+3a+3}{a-1} = \frac{2a(a-1)+3a+3}{a-1}=  \frac{2a(a-1)}{a-1}+ \frac{3a+3}{a-1} = \\  \\ =2a+ \frac{3a+3}{a-1}=2a+ \frac{3a-3+3+3}{a-1}=2a+ \frac{3(a-1)+6}{a-1} =2a+ \frac{3(a-1)}{a-1} + \frac{6}{a-1}= \\  \\ =2a+3+ \frac{6}{a-1}
<span>

Q3. The answer is </span>2 x^{2} +5x+2<span>

</span>\frac{6 x^{3} +11 x^{2} -4x-4}{3x-2} = \frac{3x*2 x^{2}-2 x^{2} *2+15 x^{2} -4x-4  }{3x-2} = \frac{2 x^{2} (3x-2)+15 x^{2} -4x-4}{3x-2}= \\  \\ =  \frac{2 x^{2} (3x-2)}{3x-2} + \frac{15 x^{2} -4x-4}{3x-2} =2 x^{2} +\frac{15 x^{2} -4x-4}{3x-2}=2 x^{2} + \frac{15 x^{2} -10x+6x-4}{3x-2}= \\  \\ =2 x^{2} + \frac{5x*3x-5x*2+6x-4}{3x-2} =2 x^{2} + \frac{5x(3x-2)+3x*2-2*2}{3x-2} = \\  \\ =2 x^{2} + \frac{5x(3x-2)}{3x-2}  + \frac{3x*2-2*2}{3x-2} =2 x^{2} +5x+ \frac{2(3x-2)}{3x-2} =2 x^{2} +5x+2
<span>

Q4. The answer is 2x + 7

</span>\frac{6 x^{2} +11x-35}{3x-5} = \frac{6 x^{2} -10x+21x-35 }{3x-5} =  \frac{3 x *2x-5*2x+7*3x-7*5 }{3x-5} = \\  \\ = \frac{2x(3x-5)+7(3x-5)}{3x-5}= = \frac{(3x-5)(2x+7)}{3x-5} =2x+7
<span>

Q5. The answer is </span>x+1- \frac{3}{x-1}<span>
      
</span>\frac{ x^{2} -4}{x-1} = \frac{ x^{2} -x+x-1-3 }{x-1} = \frac{x*x-x+x-1-3}{x-1} = \frac{x(x-1)+(x-1)-3}{x-1} =  \\  \\ \frac{(x+1)(x-1)-3}{x-1} =  \frac{(x+1)(x-1)}{x-1}  -\frac{3}{x-1} =x+1- \frac{3}{x-1}


Q6. The answer is y^{2} -2y+3

\frac{ y^{3}-4 y^{2}+7y-6  }{y-2} = \frac{y* y^{2} -2y^{2}-2 y^{2} +7y-6 }{y-2} = \frac{y^{2}(y-2)-2 y^{2} +7y-6}{y-2}= \\  \\ = \frac{y^{2}(y-2)}{y-2}+   \frac{-2 y^{2} +7y-6}{y-2} = y^{2} + \frac{-2 y^{2} +4y + 3y-6}{y-2} =  \\  \\ =y^{2} + \frac{-2y*y-2y(-2)+3y-3*2}{y-2} = y^{2} + \frac{(-2y)(y-2)+3(y-2)}{y-2} = \\  \\ = y^{2} + \frac{(-2y+3)(y-2)}{y-2} = y^{2} +(-2y+3) =y^{2} -2y+3
<span>

Q7. The answer is </span>x^{2} +xy+ y^{2}}{x-y}<span>

</span>\frac{ x^{3} - \frac{x}{y}  y^{3} }{x-y} =  \frac{(x-y)( x^{2} +xy+ y^{2}) }{x-y} = \frac{ x^{2} +xy+ y^{2}}{x-y}
<span>

Q8. The answer is </span>(a^{2} +2ab+2b^{2})<span>

</span>\frac{a^{4} +4b^{4} }{a^{2}-2ab+2 b^{2} } = \frac{ (a^{2})^{2} +(2b)^{2}}{a^{2}-2ab+2 b^{2}} = \frac{(a^{2} -2ab+2b^{2})(a^{2} +2ab+2b^{2}) }{(a^{2} -2ab+2b^{2})} =(a^{2} +2ab+2b^{2})<span>


Q9. The answer is a^{n-8} - a^{-14}

</span>\frac{ (a^{2}) ^{n} - a^{n-6}  }{ a^{n+8} }= \frac{(a^{2}) ^{n} }{a^{n+8}}- \frac{ a^{n-6} }{a^{n+8}}    \\  \\   (x^{y}) ^{z}= x^{y*z}   \\  \\  \frac{ x^{y} }{ x^{z} } = x^{y-z}  \\  \\ &#10;\frac{(a^{2}) ^{n} }{a^{n+8}}- \frac{ a^{n-6} }{a^{n+8}}  =\frac{a^{2n}  }{a^{n+8}}- \frac{ a^{n-6} }{a^{n+8}}  =  a^{2n-(n+8)} - a^{n-6-(n+8)} = \\  \\ =a^{2n-n-8} - a^{n-6-n-8} = a^{n-8} - a^{-14}
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Arte-miy333 [17]
5 dozen= 60 doughnuts
3;43-3;27 = 16
So it took him 16 minutes for 48 doughnuts
16÷48=0,3 so one doughnut takes him 30 seconds
60×0,3=19,99=20 so it takes him about 20 minutes for 5 dozens of doughnuts
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Answer:

As a decimal its 6.34, If you round up then its 6.35

As a fraction its 400/63

As a mixed number its  6 22/63

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How many units away is 1 from -6 on a number line? <br> a. -7 <br> b.-5 <br> c. 5<br> d. 7
Kobotan [32]

\huge\text{Hey there!}

\huge\text{How many units away is 1 from -6 on a}\\\huge\text{ number line?}

\huge\text{We say \bf{ units away}}\huge\text{ means substraction}\\\huge\text{in this.}

\huge\text{1 - (-6) = the answer}\\\huge\text{negative \& negative = positive}\\\huge\text{New equation: 1 + 6}\\\huge\text{1 + 6 = 7}\\\boxed{\text{Answer: D) 7}}\checkmark

\text{Good luck on your assignment and enjoy your day!}\\\frak{LoveYourselfFirst:)}

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3 years ago
Find the imaginary part of\[(\cos12^\circ+i\sin12^\circ+\cos48^\circ+i\sin48^\circ)^6.\]
iren [92.7K]

Answer:

The imaginary part is 0

Step-by-step explanation:

The number given is:

x=(\cos(12)+i\sin(12)+ \cos(48)+ i\sin(48))^6

First, we can expand this power using the binomial theorem:

(a+b)^k=\sum_{j=0}^{k}\binom{k}{j}a^{k-j}b^{j}

After that, we can apply De Moivre's theorem to expand each summand:(\cos(a)+i\sin(a))^k=\cos(ka)+i\sin(ka)

The final step is to find the common factor of i in the last expansion. Now:

x^6=((\cos(12)+i\sin(12))+(\cos(48)+ i\sin(48)))^6

=\binom{6}{0}(\cos(12)+i\sin(12))^6(\cos(48)+ i\sin(48))^0+\binom{6}{1}(\cos(12)+i\sin(12))^5(\cos(48)+ i\sin(48))^1+\binom{6}{2}(\cos(12)+i\sin(12))^4(\cos(48)+ i\sin(48))^2+\binom{6}{3}(\cos(12)+i\sin(12))^3(\cos(48)+ i\sin(48))^3+\binom{6}{4}(\cos(12)+i\sin(12))^2(\cos(48)+ i\sin(48))^4+\binom{6}{5}(\cos(12)+i\sin(12))^1(\cos(48)+ i\sin(48))^5+\binom{6}{6}(\cos(12)+i\sin(12))^0(\cos(48)+ i\sin(48))^6

=(\cos(72)+i\sin(72))+6(\cos(60)+i\sin(60))(\cos(48)+ i\sin(48))+15(\cos(48)+i\sin(48))(\cos(96)+ i\sin(96))+20(\cos(36)+i\sin(36))(\cos(144)+ i\sin(144))+15(\cos(24)+i\sin(24))(\cos(192)+ i\sin(192))+6(\cos(12)+i\sin(12))(\cos(240)+ i\sin(240))+(\cos(288)+ i\sin(288))

The last part is to multiply these factors and extract the imaginary part. This computation gives:

Re x^6=\cos 72+6cos 60\cos 48-6\sin 60\sin 48+15\cos 96\cos 48-15\sin 96\sin 48+20\cos 36\cos 144-20\sin 36\sin 144+15\cos 24\cos 192-15\sin 24\sin 192+6\cos 12\cos 240-6\sin 12\sin 240+\cos 288

Im x^6=\sin 72+6cos 60\sin 48+6\sin 60\cos 48+15\cos 96\sin 48+15\sin 96\cos 48+20\cos 36\sin 144+20\sin 36\cos 144+15\cos 24\sin 192+15\sin 24\cos 192+6\cos 12\sin 240+6\sin 12\cos 240+\sin 288

(It is not necessary to do a lengthy computation: the summands of the imaginary part are the products sin(a)cos(b) and cos(a)sin(b) as they involve exactly one i factor)

A calculator simplifies the imaginary part Im(x⁶) to 0

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3 years ago
8. Write y-x-y-x in simplest form.<br><br><br><br>10 points ​
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Answer:

Easy -2x

Step-by-step explanation:Because both a positive y and a negative y cancle out eachother so all thats left is both of the negative x's

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