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Whitepunk [10]
4 years ago
13

.2 to the 5th power

Mathematics
1 answer:
Aleonysh [2.5K]4 years ago
7 0

Answer:   The correct answer is:

____________________________________________

        →    " 0.00032 " ; or, write as:  " \frac{1}{3125}  " .

____________________________________________

Step-by-step explanation:

_______________________________________

Method 1)

____________________________________________

(0.2)⁵ = ? ;

Note:  " 0.2 " = 2/10  = (2÷2) / (10÷2) = 1/5 .

→ (0.2)⁵ = (1/5)⁵ = (1⁵) / (5)⁵ = ? ;

Note:  1⁵ = 1 × 1 × 1 × 1× 1 = 1 ;  

Note:  5⁵ = 5 × 5 × 5 × 5 × 5 ;

               = (5 × 5) × 5 × 5 × 5 ;

               =  (25) × 5 × 5 × 5 ;

               =  (25 × 5) × 5 × 5 ;

               =  125 × 5 × 5 ;

               =  (125 × 5)  * 5 ;

               =  (625 ) × 5 ;

               =   3125 .

___________________________________________

 

→  (0.2)⁵  =  (1/5)⁵ = (1⁵ / 5⁵) =   \frac{1}{3125} .

___________________________________________

Method 2)

____________________________________________

(0.2)⁵ = 0.2 ×  0.2 × 0.2 × 0.2 × 0.2 ;

          =  (0.2 × 0.2) × 0.2 × 0.2 × 0.2 ;

          = (0.04)  ×  (0.2)  × (0.2) × (0.2) ;

          =  (0.04 * 0.2)  × 0.2 × 0.2 ;

         =  (0.008)  ×  0.2 × 0.2 ;

          =  (0.008)  ×  (0.2 × 0.2) ;

          =  (0.008)  × (0.04) ;

(0.2)⁵ =  0.00032  

____________________________________________

Now, just for curiosity, let us see if the values obtained using

      both "Method 1" and "Method 2" are equal:

 \frac{1}{3125}  =?  0.00032  ?? ;

     →   (1 ÷ 3125)  =?  0.00032 ?? ; Yes!

____________________________________________

Also, check work using calculator:

____________________________________________

 (0.2)⁵  = 0.00032 .  Yes!

____________________________________________

The correct answer is:

____________________________________________

  " 0.00032 " ; or, write as:  " \frac{1}{3125}  " .

____________________________________________

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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:

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=\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:

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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

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

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