Example Question - common factors
Here are examples of questions we've helped users solve.
Finding a Common Factor and Multiple
<p>Given that the highest common factor (HCF) of 20 and p is 10, we have:</p> <p>Factors of 20: 1, 2, 4, 5, 10, 20. The common factor with p must include 10.</p> <p>To satisfy the HCF condition, p must include the factor 10. Therefore, p could be 10, 20, or another multiple of 10.</p> <p>Given that the lowest common multiple (LCM) of 20 and p is 60, we calculate:</p> <p>LCM(20, p) = (20 * p) / HCF(20, p)</p> <p>Setting the equation for LCM:</p> <p>(20 * p) / 10 = 60</p> <p>20p = 600</p> <p>p = 30</p> <p>The smallest possible value of p is therefore:</p> <p>p = 30</p>
Simplifying Rational Expression with Exponents
The expression given in the image needs to be simplified. It's a rational expression involving terms with exponents. To simplify, I'll first rewrite the expression to help clarify the steps: \[ \frac{10x^{5n+4} + 1.25x^{5n+2}}{3x^{5n+3} - 20x^{5n+1}} \] Now, let's simplify step by step: First, find the common factor in the numerator and denominator to factor out. We can see that \( x^{5n+1} \) is the greatest common factor in all terms, which means each term can be expressed as a multiple of \( x^{5n+1} \). So, we factor \( x^{5n+1} \) out of each term: \[ = \frac{x^{5n+1}(10x^3 + 1.25x)}{x^{5n+1}(3x^2 - 20)} \] Since \( x^{5n+1} \) is in both the numerator and the denominator, we can cancel it out: \[ = \frac{10x^3 + 1.25x}{3x^2 - 20} \] The expression is now simplified. However, if you need to further simplify by reducing the coefficients, you can factor constants out or simplify fractions if possible. In this case, the coefficients of the terms in the numerator and denominator don't seem to share any common factors other than 1, so this is the simplest form of the expression: \[ \frac{10x^3 + 1.25x}{3x^2 - 20} \]
Simplifying Fraction with Common Factors
To simplify the expression given in the image, look for common factors in the numerator and denominator. The expression is: \[ \frac{10 \cdot 5^{n+4} + 125 \cdot 5^{n+2}}{3 \cdot 5^{n+3} - 20 \cdot 5^{n+1}} \] We can factor out the common powers of 5 from each term and rewrite the expression as follows: \[ \frac{5^{n+2}(10 \cdot 5^2 + 125)}{5^{n+1}(3 \cdot 5^2 - 20)} \] Now simplify inside the parentheses: \[ \frac{5^{n+2}(10 \cdot 25 + 125)}{5^{n+1}(3 \cdot 25 - 20)} \] This simplifies further to: \[ \frac{5^{n+2}(250 + 125)}{5^{n+1}(75 - 20)} \] Combine the numbers inside the parentheses: \[ \frac{5^{n+2}(375)}{5^{n+1}(55)} \] Now that the fractions have a common base of 5, we can cancel out the $5^{n+1}$, leaving $5^1$ in the numerator: \[ \frac{5 \cdot 375}{55} \] Now divide both numerator and denominator by their greatest common divisor, which is 5: \[ \frac{5 \cdot 75}{11} \] This gives us the final simplified result: \[ \frac{375}{11} \] The expression can't be simplified any further, so this is the simplest form of the original expression.
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