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3 years ago

9

Solve the inequality.

2 answers:

kaheart [24]3 years ago

7 0

Answer:

B: $x>\frac{3}{2}$

Step-by-step explanation:

$12\left(\frac{1}{2}x-\frac{1}{3}\right)>8-2x$

Expand:

$6x-4>8-2x$

Add 4 to both sides:

$6x-4+4>8-2x+4$

$6x>-2x+12$

Add 2x to both sides:

$6x+2x>-2x+12+2x$

$8x>12$

Divide both sides by 8:

$\frac{8x}{8}>\frac{12}{8}$

Hence, $x>\frac{3}{2}$ (Option B)

Vinvika [58]3 years ago

3 0

Answer:

B. x> 3 /2

Step-by-step explanation:

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2 years ago

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If X²⁰¹³ + 1/X²⁰¹³ = 2, then find the value of X²⁰²² + 1/X²⁰²² = ?

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Step-by-step explanation:

$\bf➤ \underline{Given-} \\$

$\sf{x^{2013} + \frac{1}{x^{2013}} = 2}\\$

$\bf➤ \underline{To\: find-} \\$

$\sf {the\: value \: of \: x^{2022} + \frac{1}{x^{2022}}= ?}\\$

$\bf ➤\underline{Solution-} \\$

<u>Let us assume that:</u>

$\rm: \longmapsto u = {x}^{2013}$

<u>Therefore, the equation becomes:</u>

$\rm: \longmapsto u + \dfrac{1}{u} = 2$

$\rm: \longmapsto \dfrac{ {u}^{2} + 1}{u} = 2$

$\rm: \longmapsto{u}^{2} + 1 = 2u$

$\rm: \longmapsto{u}^{2} - 2u + 1 =0$

$\rm: \longmapsto {(u - 1)}^{2} =0$

$\rm: \longmapsto u = 1$

<u>Now substitute the value of u. We get:</u>

$\rm: \longmapsto {x}^{2013} = 1$

$\rm: \longmapsto x = 1$

<u>Therefore:</u>

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$\rm: \longmapsto {x}^{2022} + \dfrac{1}{ {x}^{2022} } = 2$

★ <u>Which is our required answer.</u>

$\textsf{\large{\underline{More To Know}:}}$

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

a² - b² = (a + b)(a - b)

(a + b)³ = a³ + 3ab(a + b) + b³

(a - b)³ = a³ - 3ab(a - b) - b³

a³ + b³ = (a + b)(a² - ab + b²)

a³ - b³ = (a - b)(a² + ab + b²)

(x + a)(x + b) = x² + (a + b)x + ab

(x + a)(x - b) = x² + (a - b)x - ab

(x - a)(x + b) = x² - (a - b)x - ab

(x - a)(x - b) = x² - (a + b)x + ab

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3 years ago

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Answer:

See below in bold.

Step-by-step explanation:

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