Google: Gemma 4 26B A4B

600Watts

26

The capital of Australia is **Canberra**.

Walk

To find the total profit, we need to calculate the total cost Sam spent and the total revenue he earned from selling the packs. **Step 1: Calculate the total cost** Sam bought 12 boxes at $10 each. $$\text{Total Cost} = 12 \text{ boxes} \times \$10/\text{box} = \$120$$ **Step 2: Calculate the total number of pens** Each of the 12 boxes contains 30 pens. $$\text{Total Pens} = 12 \text{ boxes} \times 30 \text{ pens/box} = 360 \text{ pens}$$ **Step 3: Calculate the number of packs created** He rearranged the 360 pens into packs of 6. $$\text{Total Packs} = 360 \text{ pens} \div 6 \text{ pens/pack} = 60 \text{ packs}$$ **Step 4: Calculate the total revenue** He sold the 60 packs for $5 each. $$\text{Total Revenue} = 60 \text{ packs} \times \$5/\text{pack} = \$300$$ **Step 5: Calculate the profit** Profit is the Total Revenue minus the Total Cost. $$\text{Profit} = \$300 - \$120 = \$180$$ **Total Profit: $180**

The primary purpose of a dictionary is to serve as a **reference tool for the meaning, usage, and properties of words.** While people often use them simply to find definitions, a dictionary actually serves several integrated functions: ### 1. Semantics (Meaning) The most fundamental role is to provide the **definition** of a word. This helps users understand what a word signifies, distinguishing between its literal meaning and its figurative or metaphorical meanings. ### 2. Orthography (Spelling) Dictionaries act as the authoritative guide for the **correct spelling** of words. They help users navigate irregular spellings, silent letters, and common mistakes. ### 3. Phonology (Pronunciation) Most dictionaries provide a guide on **how to say a word.** This is typically done through phonetic symbols (like the International Phonetic Alphabet - IPA) or pronunciation respellings, helping users master the sounds, syllable breaks, and stress patterns of new vocabulary. ### 4. Morphology (Structure and Grammar) Dictionaries provide essential information about how a word functions within a sentence, including: * **Part of speech:** Identifying if a word is a noun, verb, adjective, etc. * **Inflection:** Showing how a word changes (e.g., past tense of a verb or plural forms of a noun). * **Etymology:** Explaining the history and origin of a word (e.g., whether it comes from Latin, Greek, or Old English). ### 5. Lexicography (Usage and Context) Modern dictionaries often provide guidance on **connotation and usage**. They tell the user if a word is formal, informal, slang, archaic, or offensive, and often provide example sentences to show how the word is naturally used in context. ### Summary In short, the primary purpose is **to facilitate communication** by ensuring that the sender and the receiver of a message have a shared understanding of what words mean, how they are spelled, and how they are pronounced.

111

The time complexity of binary search is **$O(\log n)$**, where $n$ is the number of elements in the sorted array. Here is a detailed breakdown of why it is logarithmic, along with its other complexity measures. --- ### 1. Why is it $O(\log n)$? (The Intuition) In each step of a binary search, the algorithm compares the target value to the middle element of the current range. Based on that comparison, it eliminates **half** of the remaining elements. Imagine you have $n$ elements: * **Step 0:** $n$ elements remaining. * **Step 1:** $n/2$ elements remaining. * **Step 2:** $n/4$ elements remaining. * **Step 3:** $n/8$ elements remaining. * **Step $k$:** $n/2^k$ elements remaining. The search terminates when we are left with only 1 element (or zero). To find how many steps ($k$) it takes to reach 1, we solve for $k$: $$\frac{n}{2^k} = 1$$ $$n = 2^k$$ $$\log_2(n) = k$$ Therefore, the maximum number of iterations required is proportional to the logarithm of $n$. --- ### 2. Summary Table | Case | Time Complexity | Description | | :--- | :--- | :--- | | **Best Case** | $O(1)$ | The target element is exactly at the middle index on the very first try. | | **Average Case** | $O(\log n)$ | The target is found after several divisions of the array. | | **Worst Case** | $O(\log n)$ | The target is at the far ends of the array or is not present at all. | --- ### 3. Space Complexity The space complexity depends on whether the algorithm is implemented **iteratively** or **recursively**: * **Iterative Implementation: $O(1)$** Since we only use a few variables (like `low`, `high`, and `mid`) to track indices, the space used does not grow with the size of the input. * **Recursive Implementation: $O(\log n)$** Each recursive call adds a new layer (frame) to the **call stack**. Since the maximum depth of the recursion is the number of times we can divide $n$ by 2, the space complexity is $O(\log n)$. --- ### 4. Essential Requirement It is important to remember that binary search **only works on sorted arrays** (or similar data structures that allow direct access to middle elements). If the data is unsorted, you must either sort it first—which takes $O(n \log n)$—or use Linear Search, which takes $O(n)$.

The time complexity of binary search is **$O(\log n)$**, where $n$ is the number of elements in the sorted array. Here is a detailed breakdown of why it is logarithmic, along with its other complexity measures. --- ### 1. Why is it $O(\log n)$? (The Intuition) In each step of a binary search, the algorithm compares the target value to the middle element of the current range. Based on that comparison, it eliminates **half** of the remaining elements. Imagine you have $n$ elements: * **Step 0:** $n$ elements remaining. * **Step 1:** $n/2$ elements remaining. * **Step 2:** $n/4$ elements remaining. * **Step 3:** $n/8$ elements remaining. * **Step $k$:** $n/2^k$ elements remaining. The search terminates when we are left with only 1 element (or zero). To find how many steps ($k$) it takes to reach 1, we solve for $k$: $$\frac{n}{2^k} = 1$$ $$n = 2^k$$ $$\log_2(n) = k$$ Therefore, the maximum number of iterations required is proportional to the logarithm of $n$. --- ### 2. Summary Table | Case | Time Complexity | Description | | :--- | :--- | :--- | | **Best Case** | $O(1)$ | The target element is exactly at the middle index on the very first try. | | **Average Case** | $O(\log n)$ | The target is found after several divisions of the array. | | **Worst Case** | $O(\log n)$ | The target is at the far ends of the array or is not present at all. | --- ### 3. Space Complexity The space complexity depends on whether the algorithm is implemented **iteratively** or **recursively**: * **Iterative Implementation: $O(1)$** Since we only use a few variables (like `low`, `high`, and `mid`) to track indices, the space used does not grow with the size of the input. * **Recursive Implementation: $O(\log n)$** Each recursive call adds a new layer (frame) to the **call stack**. Since the maximum depth of the recursion is the number of times we can divide $n$ by 2, the space complexity is $O(\log n)$. --- ### 4. Essential Requirement It is important to remember that binary search **only works on sorted arrays** (or similar data structures that allow direct access to middle elements). If the data is unsorted, you must either sort it first—which takes $O(n \log n)$—or use Linear Search, which takes $O(n)$.

The time complexity of binary search is **$O(\log n)$**, where $n$ is the number of elements in the sorted array. Here is a detailed breakdown of why it is logarithmic, along with its other complexity measures. --- ### 1. Why is it $O(\log n)$? (The Intuition) In each step of a binary search, the algorithm compares the target value to the middle element of the current range. Based on that comparison, it eliminates **half** of the remaining elements. Imagine you have $n$ elements: * **Step 0:** $n$ elements remaining. * **Step 1:** $n/2$ elements remaining. * **Step 2:** $n/4$ elements remaining. * **Step 3:** $n/8$ elements remaining. * **Step $k$:** $n/2^k$ elements remaining. The search terminates when we are left with only 1 element (or zero). To find how many steps ($k$) it takes to reach 1, we solve for $k$: $$\frac{n}{2^k} = 1$$ $$n = 2^k$$ $$\log_2(n) = k$$ Therefore, the maximum number of iterations required is proportional to the logarithm of $n$. --- ### 2. Summary Table | Case | Time Complexity | Description | | :--- | :--- | :--- | | **Best Case** | $O(1)$ | The target element is exactly at the middle index on the very first try. | | **Average Case** | $O(\log n)$ | The target is found after several divisions of the array. | | **Worst Case** | $O(\log n)$ | The target is at the far ends of the array or is not present at all. | --- ### 3. Space Complexity The space complexity depends on whether the algorithm is implemented **iteratively** or **recursively**: * **Iterative Implementation: $O(1)$** Since we only use a few variables (like `low`, `high`, and `mid`) to track indices, the space used does not grow with the size of the input. * **Recursive Implementation: $O(\log n)$** Each recursive call adds a new layer (frame) to the **call stack**. Since the maximum depth of the recursion is the number of times we can divide $n$ by 2, the space complexity is $O(\log n)$. --- ### 4. Essential Requirement It is important to remember that binary search **only works on sorted arrays** (or similar data structures that allow direct access to middle elements). If the data is unsorted, you must either sort it first—which takes $O(n \log n)$—or use Linear Search, which takes $O(n)$.