Evaluate Expressions, Prime Factorization, And Simplification In Mathematics

Let's dive into evaluating the expression $\frac{0.05 \times 0.48}{0.062 \times 0.10}$ and expressing the result in standard form. This problem requires careful handling of decimal numbers and the application of scientific notation. To begin, we must first calculate the numerator and the denominator separately. The numerator is the product of 0.05 and 0.48, while the denominator is the product of 0.062 and 0.10. Accurate multiplication is crucial here to avoid any errors in the final result. After finding these products, we will divide the numerator by the denominator. The final step involves converting the result into standard form, which is a way of writing numbers as a product of a number between 1 and 10 and a power of 10. This form is particularly useful for expressing very large or very small numbers concisely. The concept of standard form, also known as scientific notation, is a fundamental tool in various scientific and mathematical contexts, enabling efficient handling and comparison of numbers across vastly different scales. Understanding and applying standard form not only simplifies calculations but also enhances the clarity of numerical expressions, making it an essential skill for students and professionals alike. We will carefully review each step to ensure the final answer is accurate and presented in the correct standard form. The process involves basic arithmetic operations, such as multiplication and division, as well as an understanding of decimal places and powers of ten. With a systematic approach, we can confidently solve this problem and gain a better understanding of how to work with decimal numbers and scientific notation.

Firstly, multiply the numbers in the numerator: 0.05 \times 0.48 = 0.024. Then, multiply the numbers in the denominator: 0.062 \times 0.10 = 0.0062. Now, divide the numerator by the denominator: $\frac{0.024}{0.0062}$. To simplify the division, we can multiply both the numerator and the denominator by 10,000 to remove the decimals: $\frac{0.024 \times 10000}{0.0062 \times 10000} = \frac{240}{62}$. Next, simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 2: $\frac{240}{62} = \frac{120}{31}$. Now, divide 120 by 31 to get the decimal value: $\frac{120}{31} \approx 3.871$. Finally, express the result in standard form: $3.871 = 3.871 \times 10^0$. This final step ensures that the answer is presented in the correct scientific notation format, adhering to the rules of standard form. The exponent of 10 indicates the magnitude of the number, and in this case, it is 0, meaning the number is already in the required range of 1 to 10. This comprehensive approach not only provides the solution but also reinforces the understanding of decimal operations and standard form representation.

Next, let's tackle the problem of expressing 2517 as a product of prime factors. Prime factorization is a fundamental concept in number theory, and it involves breaking down a composite number into its prime number constituents. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. The prime factorization of a number is unique, meaning that every composite number can be expressed as a product of primes in only one way, disregarding the order of the factors. This uniqueness is a cornerstone of the fundamental theorem of arithmetic. To find the prime factors of 2517, we will use the method of successive division by prime numbers. We start by dividing 2517 by the smallest prime number, 2. If it is not divisible by 2, we move to the next prime number, 3, and so on, until we have completely factored the number into its prime components. The process of prime factorization is essential in various mathematical applications, such as simplifying fractions, finding the greatest common divisor (GCD), and the least common multiple (LCM) of two or more numbers. It also plays a crucial role in cryptography and other advanced mathematical fields. Understanding how to efficiently perform prime factorization is a valuable skill for any student of mathematics. Prime factorization not only helps in simplifying mathematical problems but also provides insights into the structure and properties of numbers. The systematic approach we will use ensures that we identify all the prime factors accurately and express 2517 in its unique prime factorization form. This meticulous process demonstrates the application of number theory principles in practical problem-solving.

To begin, we test if 2517 is divisible by the smallest prime number, 2. Since 2517 is an odd number, it is not divisible by 2. Next, we try dividing by 3: $2517 \div 3 = 839$. So, 2517 = 3 \times 839. Now, we need to factor 839. We continue testing prime numbers. It's not divisible by 5 (since it doesn't end in 0 or 5) or 7 ($839 \div 7 \approx 119.86$, not an integer). Let's try 11: $839 \div 11 \approx 76.27$, not an integer. Trying 13: $839 \div 13 \approx 64.54$, not an integer. Trying 17: $839 \div 17 \approx 49.35$, not an integer. Trying 19: $839 \div 19 = 44.16$, not an integer. Trying 23: $839 \div 23 = 36.48$, not an integer. Finally, let's try 29: $839 \div 29 = 28.93$, not an integer. It's crucial to methodically test prime numbers to ensure we do not miss any factors. Further testing is needed since the number 839 did not divide evenly by the primes tested so far. Continuing to test higher prime numbers, we find that 839 is divisible by 37: $839 \div 37 = 22.68$, not an integer. Then try dividing by 41: $839 \div 41$ which also is not an integer. Now, consider 43: $839 = 43 \times 19.51$ not an integer. Continuing with this process, we find that 839 is equal to 13 \times 64.5. Let's try dividing 839 by the prime number 31: $839 \div 31=27.06$, it does not divide evenly. However, checking 839 shows that $839 = 13 \times 64 + 7$. If we divide 839 by 17, we get $839 = 17 \times 49 + 6$. Now check to see if we divide 839 by 19, $839 \div 19 = 44.15$, not an integer. Next, check dividing by 23: $839 \div 23 = 36.47$ and so on. When we try to divide 839 by 37, we find $839 \div 37 \approx 22.67$. After testing prime numbers up to the square root of 839 (which is approximately 28.9), we find that 839 is equal to 3 \times 279 + 2. We continue checking other primes. Eventually, we discover that $839 = 13 \times 64 + 7$. It turns out that 839 is a product of 13 and 64. When continuing dividing 64, we find that 839 is not divisible by 13. Thus, we'll continue our efforts to verify 839 is a prime number. Thus, we should test until we reach the $\sqrt{839}$ or about 29. After testing prime numbers, we find 839 is not divisible by any prime number. This means that 839 itself is a prime number. Thus, 2517 expressed as a product of prime factors is 3 \times 839.

Now, let's simplify the algebraic expression $\frac{x+2}{5} \cdot \frac{6-(8+x)}{6}$. Simplifying algebraic expressions is a critical skill in algebra, enabling us to work with more complex equations and functions effectively. This particular expression involves the multiplication of two fractions, each containing a linear term involving the variable x. Simplifying expressions like these often involves several steps, including expanding parentheses, combining like terms, and canceling common factors. The goal is to reduce the expression to its simplest form, making it easier to analyze and use in further calculations. In this case, we will first address the subtraction within the second fraction, then multiply the two fractions together, and finally, simplify the resulting expression. This process demonstrates the application of algebraic rules and techniques to manipulate and reduce expressions. Mastering the art of simplification is crucial for solving equations, graphing functions, and understanding more advanced algebraic concepts. By carefully applying the order of operations and algebraic principles, we can systematically simplify the given expression and arrive at its most concise form. This step-by-step approach not only provides the solution but also reinforces the understanding of fundamental algebraic concepts and techniques.

To simplify the expression, we start by simplifying the second fraction's numerator: $6 - (8 + x) = 6 - 8 - x = -2 - x$. So, the expression becomes: $\frac{x+2}{5} \cdot \frac{-2-x}{6}$. We can rewrite $-2-x$ as $-(x+2)$, so we have: $\frac{x+2}{5} \cdot \frac{-(x+2)}{6}$. Now, multiply the two fractions: $\frac{(x+2) \cdot -(x+2)}{5 \cdot 6} = \frac{-(x+2)^2}{30}$. Expand the numerator: $-(x+2)^2 = -(x^2 + 4x + 4) = -x^2 - 4x - 4$. Thus, the simplified expression is: $\frac{-x^2 - 4x - 4}{30}$. This final form represents the simplified version of the original expression, achieved through careful application of algebraic principles and simplification techniques. The process involved distributing negative signs, multiplying polynomials, and combining like terms to arrive at the most concise representation of the expression. This comprehensive approach not only yields the solution but also reinforces the understanding of algebraic manipulation and simplification strategies. The simplified form is crucial for further analysis or use in more complex equations or problems.

In this exploration, we tackled three distinct mathematical problems, each requiring a different set of skills and concepts. We started by evaluating a numerical expression involving decimal numbers and converting the result into standard form, emphasizing the importance of accurate arithmetic operations and the understanding of scientific notation. We then moved on to expressing a number as a product of prime factors, which highlighted the significance of prime factorization in number theory and its applications in various mathematical contexts. Finally, we simplified an algebraic expression, demonstrating the application of algebraic rules and techniques to manipulate and reduce expressions to their simplest form. These exercises collectively reinforce the importance of a strong foundation in fundamental mathematical concepts and the ability to apply these concepts to solve a variety of problems. The skills acquired through these exercises are not only valuable for academic pursuits but also for real-world applications where mathematical thinking and problem-solving are essential. By mastering these fundamental concepts, students can build a solid mathematical foundation and confidently tackle more complex challenges in the future. The combination of numerical, prime factorization, and algebraic simplification problems offers a comprehensive review of essential mathematical skills. These skills are crucial for success in higher-level mathematics and related fields. Each problem requires a methodical approach and a clear understanding of the underlying principles, reinforcing the importance of precision and attention to detail in mathematical problem-solving. The ability to apply these techniques effectively demonstrates a strong grasp of fundamental mathematical concepts.