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3 years ago

How do you solve for the quotient of (x^-1) - 1 ÷ x - 1?

1 answer:

Ber [7]3 years ago

3 0

$\bf x^{-1}-1\div x-1\implies \implies \cfrac{1}{x}-1\div x-1\implies \cfrac{\frac{1}{x}-1}{~~x-1~~}\implies \cfrac{~~\frac{1-x}{x}~~}{\frac{x-1}{1}} \\\\\\ \cfrac{1-x}{x}\cdot \cfrac{1}{x-1}\implies \cfrac{-(\begin{matrix} x-1 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix})}{x}\cdot \cfrac{1}{\begin{matrix} x-1 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}}\implies -\cfrac{1}{x}\implies -x^{-1}$

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2 years ago

Four people enter their names into a drawing. The winner receives one of two different prizes that are randomly selected. The tr

galina1969 [7]

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2 years ago

7 is 1/10 of what number

umka21 [38]

We should find the original number, 7 of 1/10. We divide the known number for the numerator, than multiply for denominator.

7 : 1 = 7 * 10 = 70 <- answer

So 7 is 1/10 of 70

We can check our work:

1/10 of 70 = 7?

70 : 10 = 7*1 = 7

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2 years ago

What's 3570 divided by 7

mart [117]

Answer:

510

Step-by-step explanation:

$\frac{3570}{7}$ = 510

7 0

2 years ago

Read 2 more answers

Sin4x.sin5x+sin4x.sin3x-sin2x.sinx=0

andreev551 [17]

Recall the angle sum identity for cosine:

cos(x + y) = cos(x) cos(y) - sin(x) sin(y)

cos(x - y) = cos(x) cos(y) + sin(x) sin(y)

==> sin(x) sin(y) = 1/2 (cos(x - y) - cos(x + y))

Then rewrite the equation as

sin(4x) sin(5x) + sin(4x) sin(3x) - sin(2x) sin(x) = 0

1/2 (cos(-x) - cos(9x)) + 1/2 (cos(x) - cos(7x)) - 1/2 (cos(x) - cos(3x)) = 0

1/2 (cos(9x) - cos(x)) + 1/2 (cos(7x) - cos(3x)) = 0

sin(5x) sin(-4x) + sin(5x) sin(-2x) = 0

-sin(5x) (sin(4x) + sin(2x)) = 0

sin(5x) (sin(4x) + sin(2x)) = 0

Recall the double angle identity for sine:

sin(2x) = 2 sin(x) cos(x)

Rewrite the equation again as

sin(5x) (2 sin(2x) cos(2x) + sin(2x)) = 0

sin(5x) sin(2x) (2 cos(2x) + 1) = 0

sin(5x) = 0 or sin(2x) = 0 or 2 cos(2x) + 1 = 0

sin(5x) = 0 or sin(2x) = 0 or cos(2x) = -1/2

sin(5x) = 0 ==> 5x = arcsin(0) + 2nπ or 5x = arcsin(0) + π + 2nπ

… … … … … ==> 5x = 2nπ or 5x = (2n + 1)π

… … … … … ==> x = 2nπ/5 or x = (2n + 1)π/5

sin(2x) = 0 ==> 2x = arcsin(0) + 2nπ or 2x = arcsin(0) + π + 2nπ

… … … … … ==> 2x = 2nπ or 2x = (2n + 1)π

… … … … … ==> x = nπ or x = (2n + 1)π/2

cos(2x) = -1/2 ==> 2x = arccos(-1/2) + 2nπ or 2x = -arccos(-1/2) + 2nπ

… … … … … … ==> 2x = 2π/3 + 2nπ or 2x = -2π/3 + 2nπ

… … … … … … ==> x = π/3 + nπ or x = -π/3 + nπ

(where n is any integer)

5 0

2 years ago