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neonofarm [45]
3 years ago
5

Eight less than twice a number is the same as six times the number. Find the number.

Mathematics
1 answer:
xz_007 [3.2K]3 years ago
6 0

Answer:

Step-by-step explanation:

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

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

Learn more about composite functions here: brainly.com/question/20379727

#SPJ1

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