How do you find g(h(t)) given $g (t) = \frac{t + 1}{2}$ $g(t) = (t+1)/2$ , $h (t) = \frac{t - 3}{4}$ $h(t) = (t-3)/4$ ?

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How do you find g(h(t)) given $g (t) = \frac{t + 1}{2}$ $g(t) = (t+1)/2$ , $h (t) = \frac{t - 3}{4}$ $h(t) = (t-3)/4$ ?

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Answer 1 · 2016-03-02T14:00:14

$g (h (t)) = \frac{t + 1}{8}$ $g(h(t))=(t+1)/8$

Explanation:

Given: $g (t) = \frac{t + 1}{2}$ $g(t)=(t+1)/2$ and $h (t) = \frac{t - 3}{4}$ $h(t)=(t-3)/4$

$g (h (t)) = \frac{h (t) + 1}{2} = \frac{\frac{t - 3}{4} + 1}{2} = \frac{t - 3 + 4}{8} = \frac{t + 1}{8}$ $g(h(t))=(h(t)+1)/2=((t-3)/4+1)/2=(t-3+4)/8=(t+1)/8$

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How do you find g(h(t)) given $g (t) = \frac{t + 1}{2}$ $g(t) = (t+1)/2$ , $h (t) = \frac{t - 3}{4}$ $h(t) = (t-3)/4$ ?

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How do you find g(h(t)) given $g (t) = \frac{t + 1}{2}$ $g(t) = (t+1)/2$ , $h (t) = \frac{t - 3}{4}$ $h(t) = (t-3)/4$ ?