Abbigail2017
Abbigail2017 Abbigail2017
  • 24-07-2016
  • Mathematics
contestada

Given that f(x) = 3x + 1 and g(x) = the quantity of 4x plus 2 divided by 3, solve for g(f(0))

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jhonyy9
jhonyy9 jhonyy9
  • 24-07-2016
f(0)=1

g(1)=(4*1+2)/3 = 6/3=2

hope helped 
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Panoyin
Panoyin Panoyin
  • 24-07-2016
[tex]f(x) = 3x + 1[/tex]
[tex]g(x) = \frac{4x + 2}{3}[/tex]

[tex]g[f(0)] = \frac{4[3(0) + 1] + 2}{3}[/tex]
[tex]g[f(0)] = \frac{4[0 + 1] + 2}{3}[/tex]
[tex]g[f(0)] = \frac{4(1) + 2}{3}[/tex]
[tex]g[f(0)] = \frac{4 + 2}{3}[/tex]
[tex]g[f(0)] = \frac{6}{3}[/tex]
[tex]g[f(0)] = 2[/tex]
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