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A function h(t)h(t) increases by a factor of 66 over every unit interval in tt and h(0)=1h(0) = 1.\newlineWhich could be a function rule for h(t)h(t)?\newlineChoices:\newline(A)h(t)=6th(t) = 6^t\newline(B)h(t)=0.94th(t) = 0.94^t\newline(C)h(t)=16th(t) = \frac{1}{6^t}\newline(D)h(t)=6th(t) = 6t

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Q. A function h(t)h(t) increases by a factor of 66 over every unit interval in tt and h(0)=1h(0) = 1.\newlineWhich could be a function rule for h(t)h(t)?\newlineChoices:\newline(A)h(t)=6th(t) = 6^t\newline(B)h(t)=0.94th(t) = 0.94^t\newline(C)h(t)=16th(t) = \frac{1}{6^t}\newline(D)h(t)=6th(t) = 6t
  1. Find Initial Value: h(t)=a×6th(t) = a \times 6^t, where aa is the initial value.\newlineSince h(0)=1h(0) = 1, let's find aa.\newlineh(0)=a×60=a×1=1h(0) = a \times 6^0 = a \times 1 = 1, so a=1a = 1.
  2. Calculate Final Expression: Now we know a=1a = 1, so h(t)=1×6th(t) = 1 \times 6^t.\newlineh(t)=6th(t) = 6^t.

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