Solve the inequality

Mathematics

1 answer:

Nady [450]3 years ago

5 0

Answer:

x >= -6

Step-by-step explanation:

Multiply:

(1/3)x + 1 >= (1/2)x + 1/2 - (-1/4)x

Combine like terms:

(1/3)x + 1 >= (1/4)x + 1/2

Move the x's on one side and the integers on another:

(1/3)x - (1/4)x >= 1/2 - 1

(1/12)x >= -1/2

x >= -6

I hope this helped! :D

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Given that f.x 3x-2 over x+1 g[x] x +5 evaluate f[-4] and gf [-2]

Jobisdone [24]

The value of f[ -4 ] and g°f[-2] are $\frac{14}{3}$ and 13 respectively.

<h3>What is the value of f[-4] and g°f[-2]?</h3>

Given the function;

$f(x) = \frac{3x-2}{x+1}$
$g(x)=x+5$
f[ -4 ] = ?
g°f[ -2 ] = ?

For f[ -4 ], we substitute -4 for every variable x in the function.

$f(x) = \frac{3x-2}{x+1}\\\\f(-4) = \frac{3(-4)-2}{(-4)+1}\\\\f(-4) = \frac{-12-2}{-4+1}\\\\f(-4) = \frac{-14}{-3}\\\\f(-4) = \frac{14}{3}$

For g°f[-2]

g°f[-2] is expressed as g(f(-2))

$g(\frac{3x-2}{x+1}) = (\frac{3x-2}{x+1}) + 5\\\\g(\frac{3x-2}{x+1}) = \frac{3x-2}{x+1} + \frac{5(x+1)}{x+1}\\\\g(\frac{3x-2}{x+1}) = \frac{3x-2+5(x+1)}{x+1}\\\\g(\frac{3x-2}{x+1}) = \frac{8x+3}{x+1}\\\\We\ substitute \ in \ [-2] \\\\g(\frac{3x-2}{x+1}) = \frac{8(-2)+3}{(-2)+1}\\\\g(\frac{3x-2}{x+1}) = \frac{-16+3}{-2+1}\\\\g(\frac{3x-2}{x+1}) = \frac{-13}{-1}\\\\g(\frac{3x-2}{x+1}) = 13$