What number must you add to -6 to get a sum of zero? Explain

1 answer:

6 0

You would have to add a positive 6 to the negative 6 to get zero. Lets say you have -2 in order to get it to zero or just any positive number, you have to add a positive of the same value or higher to be able to get it there. I hope you understand that.

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Solve for x. <br><br> x = ___°

irina [24]

x = 109

this is pentagon

the sum of the interior angles of a polygon is

sum = 180°( n - 2) where n is the number of sides.

sum = 180° × 3 = 540° thus

x + 121+ 108 + 102 + 100 = 540

x + 431 = 540

subtract 431 from both sides

x = 540 - 431 = 109

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If x+1/x= 3, then prove that m^5+1/m^5= 123

alina1380 [7]

9514 1404 393

Explanation:

We can start with the relations ...

$\displaystyle\left(x+\frac{1}{x}\right)^3=\left(x^3+\frac{1}{x^3}\right)+3\left(x+\frac{1}{x}\right)\\\\\left(x+\frac{1}{x}\right)^5=\left(x^5+\frac{1}{x^5}\right)+5\left(x^3+\frac{1}{x^3}\right)+10\left(x+\frac{1}{x}\right)\\\\\textsf{From these, we can derive ...}\\\\x^5+\frac{1}{x^5}=\left(x+\frac{1}{x}\right)^5-5\left(\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)\right)-10\left(x+\frac{1}{x}\right)$

$\displaystyle x^5+\frac{1}{x^5}=\left(x+\frac{1}{x}\right)^5-5\left(x+\frac{1}{x}\right)^3+5\left(x+\frac{1}{x}\right)\right)\\\\x^5+\frac{1}{x^5}=3^5 -5(3^3)+5(3)\\\\=((3^2-5)3^2+5)\cdot3=(4\cdot9+5)\cdot3=(41)(3)\\\\=\boxed{123}$