Example Question - like terms
Here are examples of questions we've helped users solve.
Algebraic Expressions
1. 8p 2. ab 3. b^2 4. 6x + 6
Simplifying Expression with Variables
На картинке представлено выражение: \(8p - 4g + 4p - g\) Чтобы решить это выражение, мы должны объединить подобные слагаемые. Подобные слагаемые — это те слагаемые, которые содержат одинаковые переменные в одинаковой степени. В данном случае, \(8p\) и \(4p\) являются подобными слагаемыми, так же как \(-4g\) и \(-g\). Сложим подобные слагаемые: \(8p + 4p = 12p\), \(-4g - g = -5g\). Теперь объединим их вместе: \(12p - 5g\). Это и есть упрощенный ответ выражения с картинки.
Adding Functions
The question is asking to find the sum of the functions \(f(x)\) and \(g(x)\). To do this, we simply add the corresponding terms from each function. Given: \(f(x) = -3x^2 + 5\) \(g(x) = 3x^2 + 2x + 1\) Adding \(f(x) + g(x)\), we align the like terms and add them: \(-3x^2 + 5\) (From \(f(x)\)) \(+ 3x^2 + 2x + 1\) (From \(g(x)\)) Now, add the like terms: \(-3x^2 + 3x^2\) (These terms cancel each other out) \(+ 2x\) (Only term with \(x\), so it remains as is) \(+ 5 + 1\) (Adding the constant terms) So we are left with: \(2x + 6\) Therefore, the sum of \(f(x)\) and \(g(x)\) or \(f + g\) is \(2x + 6\). The correct answer is: A) \(2x + 6\)
Understanding Equation Solving With Like Terms
To solve the equation \( 403s + 1002s = 1410s \), we can combine like terms on the left side by adding the coefficients of \( s \) together. Here's how you do it step by step: 1. Combine like terms (the terms with \( s \)): \[ 403s + 1002s = (403 + 1002)s = 1405s \] 2. Now that we have \( 1405s \) on the left side and \( 1410s \) on the right side, we set them equal to each other: \[ 1405s = 1410s \] 3. In order to find the solution for \( s \), you generally would isolate \( s \) by dividing both sides of the equation by the coefficient of \( s \). However, since the coefficients of \( s \) are equal and there is no constant term, this would result in an identity where any value of \( s \) satisfies the equation. But since the coefficients are actually different (1405 vs. 1410), there seems to be a mistake in the original problem or a transcription error. If the equation were correctly stated as \( 1405s = 1410s \), the only solution would be \( s = 0 \) because that would be the only way the two sides could be equal. Any non-zero value of \( s \) would result in the two sides not being equal. Since the equation you provided is already balanced with \( 403s + 1002s = 1410s \), which simplifies to \( 1405s = 1410s \), this equation has no solution for \( s \) unless there is a mistake in the terms provided. If the equation is correct, then the solution is that there is no value for \( s \) that would satisfy this equation.
Simplifying Algebraic Fraction with Exponents
To simplify the expression, we start by simplifying both the numerator and the denominator separately. We can simplify by combining like terms and applying the laws of exponents. Given expression: \[ \frac{10x^{n+4} + 125x^{n+2}}{3x^{n+3} - 20x^{n+1}} \] First, let's simplify the numerator and the denominator by factoring out the common x term: Numerator: \[ 10x^{n+4} + 125x^{n+2} = x^{n+2}(10x^2 + 125) \] Denominator: \[ 3x^{n+3} - 20x^{n+1} = x^{n+1}(3x^2 - 20) \] Now, rewrite the expression using the factored terms: \[ \frac{x^{n+2}(10x^2 + 125)}{x^{n+1}(3x^2 - 20)} \] Next, cancel out the common x term from both the numerator and the denominator: since we have \(x^{n+2}\) in the numerator and \(x^{n+1}\) in the denominator, we can divide each by \(x^{n+1}\) so that: \[ x^{n+2} / x^{n+1} = x \] The simplified expression will be: \[ \frac{x(10x^2 + 125)}{(3x^2 - 20)} \] Now, we can't simplify further without knowing the specific value of \(n\), as the remaining terms in the parentheses do not have a common x term with which to factor out or cancel. The simplified expression is: \[ x \cdot \frac{10x^2 + 125}{3x^2 - 20} \]
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