Example Question - variable
Here are examples of questions we've helped users solve.
Finding the Value of a Variable in a Triangle
<p>Para encontrar el valor de \( x \), utilizamos la propiedad de que la suma de los ángulos en un triángulo es igual a 180 grados.</p> <p>Los ángulos dados son \( 37^\circ \) y \( x \). Si el tercer ángulo se denota como \( 120^\circ \), entonces podemos plantear la ecuación:</p> <p> \( 37^\circ + x + 120^\circ = 180^\circ \)</p> <p>Resolviendo, tenemos:</p> <p> \( x = 180^\circ - 37^\circ - 120^\circ \)</p> <p> \( x = 23^\circ \)</p> <p>En conclusión, el valor de \( x \) es \( 23^\circ \).</p>
Finding the Value of a Variable
<p>To solve for \( q \), start with the equation:</p> <p> \( 16 \times 5 \times \frac{q}{10} = \text{{value}} \)</p> <p>First, simplify the left side:</p> <p> \( 16 \times 5 = 80 \), hence:</p> <p> \( 80 \times \frac{q}{10} = \text{{value}} \)</p> <p>Next, multiply \( 80 \) by \( \frac{1}{10} \):</p> <p> \( 8q = \text{{value}} \)</p> <p>Finally, solve for \( q \):</p> <p> \( q = \frac{\text{{value}}}{8} \)</p>
Finding the Value of a Variable
<p>To solve for \( q \), start with the equation:</p> <p>\( \frac{3}{8} \times 4 \times q = \text{blank} \)</p> <p>First, calculate \( \frac{3}{8} \times 4 \):</p> <p>\( \frac{3 \times 4}{8} = \frac{12}{8} = \frac{3}{2} \)</p> <p>Now substitute this into the equation:</p> <p>\( \frac{3}{2} \times q = \text{blank} \)</p> <p>Solving for \( q \):</p> <p>\( q = \frac{\text{blank}}{\frac{3}{2}} = \text{blank} \times \frac{2}{3} \)</p>
Finding the Value of n
<p>Given \( (x^n)^3 = \frac{x^{18}}{x^{-6}} \), we can start by simplifying the right side:</p> <p>First, rewrite \( x^{-6} \) as \( \frac{1}{x^6} \), so we have:</p> <p>\( \frac{x^{18}}{x^{-6}} = x^{18} \cdot x^{6} = x^{18 + 6} = x^{24} \)</p> <p>Now we have:</p> <p> \( (x^n)^3 = x^{24} \)</p> <p>Using the property of exponents, we get:</p> <p> \( x^{3n} = x^{24} \)</p> <p>Since the bases are the same, set the exponents equal:</p> <p> \( 3n = 24 \)</p> <p>Now, solving for \( n \):</p> <p> \( n = \frac{24}{3} = 8 \)</p> <p>Thus, the value of \( n \) is \( 8 \).</p>
Finding the Value of a Variable in an Equation
<p>Given the equation \( x^{2m} = \frac{(x^3)^8}{x^6} \).</p> <p>First, simplify the right side:</p> <p>\( \frac{(x^3)^8}{x^6} = \frac{x^{24}}{x^6} = x^{24-6} = x^{18} \).</p> <p>Now, equate the exponents:</p> <p>So, \( 2m = 18 \).</p> <p>To find \( m \), divide both sides by 2:</p> <p>Therefore, \( m = \frac{18}{2} = 9 \).</p>
Determine the value of a variable
<p> Multiplique ambos lados de la ecuación por 6 para eliminar los denominadores: </p> <p> 6 \left( \frac{x - 1}{2} + \frac{x - 2}{3} \right) = x </p> <p> 3(x - 1) + 2(x - 2) = x </p> <p> 3x - 3 + 2x - 4 = x </p> <p> 5x - 7 = x </p> <p> 5x - x = 7 </p> <p> 4x = 7 </p> <p> x = \frac{7}{4} </p>
Finding the Value of a Variable
<p>Empezamos con la ecuación: </p> <p>\(\frac{x}{5} + \frac{x}{3} + \frac{x}{15} = 9\)</p> <p>El común denominador de los denominadores 5, 3 y 15 es 15. Multiplicamos toda la ecuación por 15:</p> <p> \(15 \left(\frac{x}{5}\right) + 15 \left(\frac{x}{3}\right) + 15 \left(\frac{x}{15}\right) = 15 \cdot 9\)</p> <p>Esto nos da:</p> <p> \(3x + 5x + x = 135\)</p> <p>Sumamos los términos similares:</p> <p> \(9x = 135\)</p> <p>Ahora dividimos ambos lados por 9:</p> <p> \(x = \frac{135}{9}\)</p> <p>Por lo tanto, \(x = 15\).</p>
Solving a Basic Algebra Problem
Vamos resolver a questão dada na imagem. A questão diz que "Henry adiciona 5 a um número. Ele divide a soma por 2 e o quociente é 6". Seja \( x \) o número desconhecido. Então, a expressão para o problema é: \[ \frac{x + 5}{2} = 6 \] Para encontrar o valor de \( x \), siga os passos: <p>\( x + 5 = 6 \cdot 2 \)</p> <p>\( x + 5 = 12 \)</p> <p>\( x = 12 - 5 \)</p> <p>\( x = 7 \)</p> Portanto, o número desconhecido é 7.
Solving for a Variable with Square Roots
<p>\(\sqrt{a} = -\sqrt{a}\)</p> <p>1) Поднимите обе стороны уравнения в квадрат, чтобы избавиться от квадратных корней:</p> <p>\((\sqrt{a})^2 = (-\sqrt{a})^2\)</p> <p>2) Применяя свойства степеней, уберите квадратные корни:</p> <p>\(a = a \cdot (-1)^2\)</p> <p>3) Упростите правую сторону уравнения:</p> <p>\(a = a\)</p> <p>Полученное уравнение верно для всех \(a\), кроме тех случаев, когда \(a\) меньше 0, так как квадратный корень не может быть отрицательным числом. Однако, так как обе части уравнения одинаковы, это уравнение не имеет решения, потому что изначально невозможно, чтобы \(\sqrt{a}\) был равен \(-\sqrt{a}\), если \(a \geq 0\).</p>
Solving a Linear Equation with a Variable
<p>Given equation: \(-3(-8 - 7b) = -27 + 4b\)</p> <p>Distribute the \(-3\) inside the parentheses: \(24 + 21b = -27 + 4b\)</p> <p>Subtract \(4b\) from both sides: \(24 + 21b - 4b = -27 + 4b - 4b\)</p> <p>Simplify: \(24 + 17b = -27\)</p> <p>Subtract \(24\) from both sides: \(17b = -27 - 24\)</p> <p>Simplify: \(17b = -51\)</p> <p>Divide both sides by \(17\): \(b = \frac{-51}{17}\)</p> <p>Solution: \(b = -3\)</p>
Solve the Equation with Fractions and a Variable
<p>La ecuación dada es: $\frac{1}{t-2} = \frac{t+3}{t+5}$.</p> <p>Multiplica ambos lados por $(t-2)(t+5)$ para eliminar los denominadores:</p> <p>$(t-2)(t+5) \cdot \frac{1}{t-2} = (t-2)(t+5) \cdot \frac{t+3}{t+5}$,</p> <p>$t+5 = (t-2)(t+3)$.</p> <p>Expande el lado derecho de la ecuación:</p> <p>$t+5 = t^2 + 3t - 2t - 6$,</p> <p>$t+5 = t^2 + t - 6$.</p> <p>Reorganiza los términos para formar una ecuación cuadrática:</p> <p>$t^2 + t - t - 6 - 5 = 0$,</p> <p>$t^2 - 11 = 0$.</p> <p>Finalmente, resuelve para la variable $t$:</p> <p>$t^2 = 11$,</p> <p>$t = \pm\sqrt{11}$.</p> <p>Así que las soluciones para la ecuación son $t = \sqrt{11}$ y $t = -\sqrt{11}$.</p>
Comparing Integrals with Different Powers of Variable in the Denominator
<p>\( \int \frac{3}{x} dx \neq 3 \int \frac{1}{x} dx \)</p> <p>Let's solve the integral on the left side:</p> <p>\( \int \frac{3}{x} dx = 3 \int \frac{1}{x} dx \)</p> <p>Since \( \int \frac{1}{x} dx = \ln|x| + C \), where C is the constant of integration, we have:</p> <p>\( 3 \int \frac{1}{x} dx = 3(\ln|x| + C) = 3\ln|x| + C' \)</p> <p>So, the equality given in the question is incorrect, because:</p> <p>\( \int \frac{3}{x} dx = 3\ln|x| + C' \)</p> <p>And not:</p> <p>\( 3 \int \frac{1}{x} dx = 9\ln|x| + C' \)</p> <p>There seems to be a misunderstanding or typo in the provided equality.</p>
Indefinite Integration of a Constant Divided by a Variable
<p>\int \frac{3}{x} \, dx = 3 \int \frac{1}{x} \, dx</p> <p>3 \int \frac{1}{x} \, dx = 3 \ln|x| + C</p> <p>\therefore \int \frac{3}{x} \, dx = 3 \ln|x| + C</p>
Solving a Linear Equation Step by Step
Claro, vamos a resolver la ecuación paso por paso. Tenemos la ecuación \( 11u = 16 + 3u \). Primero, vamos a mover los términos con la variable \( u \) al mismo lado restando \( 3u \) en ambos lados de la ecuación. Esto nos deja con: \( 11u - 3u = 16 \) Resolvemos en el lado izquierdo: \( 8u = 16 \) Ahora, para despejar \( u \), dividimos ambos lados de la ecuación por 8: \( u = \frac{16}{8} \) Simplificamos la fracción: \( u = 2 \) Por lo tanto, la solución es \( u = 2 \). La respuesta está simplificada tanto como es posible.
Solving Matrix Equation with Variable
The image you’ve provided shows a matrix equation that needs to be solved. The quality of the image is quite blurry and the text is not very clear, but I can decipher that the equation is of the form: \[ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 1 \\ x \end{bmatrix} = \begin{bmatrix} e & f & g \\ h & i & j \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ -x \end{bmatrix} \] Where \( a, b, c, d, e, f, g, h, i, j \) are numbers and \( x \) is the variable we need to solve for. However, I'm unable to precisely read the numbers due to the lack of clarity. To solve the matrix equation, one would typically multiply out both sides to obtain two systems of linear equations, and then solve for \( x \). Since I cannot provide an accurate answer without the specific values, I can only guide you with the general approach: 1. Multiply the left-hand side matrix by the column matrix with variables \( 1 \) and \( x \). 2. Multiply the right-hand side matrix by the column matrix with variables \( 1 \), \( 0 \), and \( -x \). 3. Equate the corresponding elements of the resulting matrices from both sides. 4. Solve the system of equations to obtain the value of \( x \). If you can provide a clearer image or the specific values of the numbers, I could proceed with the calculation and give you a more precise answer.
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