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A function h(t)h(t) increases by 55 over every unit interval in tt and h(0)=0h(0) = 0.\newlineWhich could be a function rule for h(t)h(t)?\newlineChoices:\newline(A) h(t)=t5h(t) = \frac{t}{5}\newline(B) h(t)=t+5h(t) = t + 5\newline(C) h(t)=15th(t) = \frac{1}{5^t}\newline(D) h(t)=5th(t) = 5t

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Q. A function h(t)h(t) increases by 55 over every unit interval in tt and h(0)=0h(0) = 0.\newlineWhich could be a function rule for h(t)h(t)?\newlineChoices:\newline(A) h(t)=t5h(t) = \frac{t}{5}\newline(B) h(t)=t+5h(t) = t + 5\newline(C) h(t)=15th(t) = \frac{1}{5^t}\newline(D) h(t)=5th(t) = 5t
  1. Identify Linear Function: h(t)h(t) increases by 55 for each unit interval, so the function should be linear and have a slope of 55.
  2. Check Answer Choices: Check each choice to see if it fits the description:\newline(A) h(t)=t5h(t) = \frac{t}{5} increases by 15\frac{1}{5} for each unit interval, not 55.
  3. Evaluate Choice (A): (B) h(t)=t+5h(t) = t + 5 starts at 55 when t=0t = 0, not at 00.
  4. Evaluate Choice (B): (C) h(t)=15th(t) = \frac{1}{5^t} is an exponential decay function, not linear.
  5. Evaluate Choice (C): (D) h(t)=5th(t) = 5t increases by 55 for each unit interval, and h(0)=0h(0) = 0.
  6. Evaluate Choice (D): So, the correct function rule is (D) h(t)=5th(t) = 5t.

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