Mathematics: Long division 3x+1/6x^6+5x^5+2x^4-9x^3+7x^2-10x+2

Long division 3x+1/6x^6+5x^5+2x^4-9x^3+7x^2-10x+2 - id3166552493721021363978, mazolethrin6876

Ответ:

Drreeaa

14.03.2022

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

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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Ответ:

billydeck30

17.01.2023

Mathematics

The probability of choosing of the blue tile and flip a coin are

1. P(BT) = 1/7 = 4

2. P(H or T) = 1/2 = 1.5

Step-by-step explanation:

1. P(BT) = 1/7

P(BT) = 1/7 + 2/7+ 3/7 + 4/7 + 5/7 + 6/7 + 7/7 = 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28/7

P(BT) = 4

2. P(H or T) = 1/2 + 2/2 = 1 + 2/2 = 3/2 = 1.5

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

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