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2 years ago

Simplify (5n^4)^-3. Assume n = 0.

Mathematics

2 answers:

denpristay [2]2 years ago

8 0

Answer:

Step-by-step explanation:

5*0=0

0^4=0

0^-3=0

kkurt [141]2 years ago

7 0

Answer:

Undefined

Step-by-step explanation:

$(5n^4)^{-3}$

$(5(0)^4)^{-3}$

$(5(0))^{-3}$

$0^{-3}$

$\frac{1}{0^3}$

Since dividing by 0 is undefined, then the answer is undefined when n=0.

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raketka [301]

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Long division<br> 3x+1/6x^6+5x^5+2x^4-9x^3+7x^2-10x+2

inna [77]

$(6 x^{6}+5x^{5}+2x^{4}-9x^{3}+7x^{2}-10x+2) / (3x+1)$

We divide first number from first parenthesis with first number from second parenthesis. Then the resulting number we multiply by all numbers in second parenthesis and substract from first parenthesis.

$6 x^{6}/3x = 2 x^{5} \\ \\ 6 x^{6}+5x^{5}+2x^{4}-9x^{3}+7x^{2}-10x+2 - 6 x^{6} -2 x^{5} = 3x^{5}+2x^{4}-9x^{3}+7x^{2}-10x+2\\$

We repeat previous steps until we run out of numbers:
$3x^{5}/3x=x^{4} \\ \\ 3x^{5}+2x^{4}-9x^{3}+7x^{2}-10x+2-3x^{5}-x^{4}= \\ \\ x^{4}-9x^{3}+7x^{2}-10x+2 \\ \\ \\ x^{4}/3x= \frac{1}{3} x^{3} \\ \\ x^{4}-9x^{3}+7x^{2}-10x+2-x^{4}- \frac{1}{3} x^{3}= \\ \\ - \frac{28}{3} x^{3}+7x^{2}-10x+2 \\ \\ \\ - \frac{28}{3} x^{3}/3x= - \frac{28}{9} x^{2} \\ \\ - \frac{28}{3} x^{3}+7x^{2}-10x+2+ \frac{28}{3} x^{3}+ \frac{28}{9} x^{2} = \\ \\ \frac{91}{9} x^{2}-10x+2$
$\frac{91}{9} x^{2}/3x=\frac{91}{27} x \\ \\ \frac{91}{9} x^{2}-10x+2-\frac{91}{9} x^{2}-\frac{91}{27} x= \\ \\ -\frac{361}{27} x+2 \\ \\ \\ -\frac{361}{27} x/3x=-\frac{361}{81} \\ \\ -\frac{361}{27} x+2+\frac{361}{27}x+\frac{361}{27}= \\ \\ \frac{415}{27}$

We are left with a number that has no x inside. This is remainder.
The final solution is sum of all these solutions and remainder:
$(2 x^{5}+x^{4}+\frac{1}{3} x^{3} - \frac{28}{9} x^{2} +\frac{91}{27} x)+(-\frac{361}{81} )$

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