How does $$ h(x)=5 \cdot 0.75^{x} $$ change over the interval from $$ x=-8 $$ to $$ x=-7 $$ ? $$ h(x) $$ decreases by $$ 25 \% $$ $$ h(x) $$ increases by $$ 175 \% $$ $$ h(x) $$ decreases by $$ 75 \% $$ $$ h(x) $$ decreases by 0.75

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To determine how the function $$ h(x) = 5 \cdot 0.75^{x} $$ changes as $$ x $$ increases from $$-8$$ to $$-7$$, let's analyze the behavior step by step.

1. **Understanding the Function Behavior:**
   
   The function $$ h(x) = 5 \cdot 0.75^{x} $$ is an exponential function where the base $$ 0.75 $$ is less than 1. Generally, for exponential functions of the form $$ y = A \cdot b^x $$ where $$ 0 < b < 1 $$, the function decreases as $$ x $$ increases.

2. **Calculating the Change from $$ x = -8 $$ to $$ x = -7 $$:**
   
   - **At $$ x = -8 $$:**
     $$
     h(-8) = 5 \cdot 0.75^{-8} = 5 \cdot \left(\frac{1}{0.75}\right)^8 = 5 \cdot \left(\frac{4}{3}\right)^8
     $$
     (The exact value isn't necessary for understanding the percentage change.)
   
   - **At $$ x = -7 $$:**
     $$
     h(-7) = 5 \cdot 0.75^{-7} = 5 \cdot \left(\frac{1}{0.75}\right)^7 = 5 \cdot \left(\frac{4}{3}\right)^7
     $$
   
   - **Comparing $$ h(-7) $$ to $$ h(-8) $$:**
     $$
     h(-7) = 0.75 \cdot h(-8)
     $$
     This means $$ h(-7) $$ is $$ 75\% $$ of $$ h(-8) $$.

3. **Interpreting the Percentage Change:**

Since $$ h(-7) $$ is $$ 75\% $$ of $$ h(-8) $$, the function has **decreased by $$ 25\% $$** over the interval from $$ x = -8 $$ to $$ x = -7 $$.

**Answer:**  
$$ h(x) $$ decreases by $$ 25 \% $$.
$$ h(x) $$ decreases by $$ 25\% $$.

How does change over the interval from to ?
decreases by
increases by
decreases by
decreases by 0.75

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