Example Question - value of x
Here are examples of questions we've helped users solve.
Finding the Value of a Variable
<p>Primero, expandimos cada término en la ecuación:</p> <p>3(x + 2) + 2(x - 3) = 6(x - 5)</p> <p>Esto se convierte en:</p> <p>3x + 6 + 2x - 6 = 6x - 30</p> <p>Combinando términos similares:</p> <p>5x = 6x - 30</p> <p>Ahora, restamos 6x de ambos lados:</p> <p>5x - 6x = -30</p> <p>-x = -30</p> <p>Multiplicamos por -1:</p> <p>x = 30</p>
Determining the Value of x in an Exponential Expression
\[ \begin{align*} \frac{a^3 \cdot b^x \cdot c^4}{a^{1/2} \cdot b^{5/2} \cdot c^5} &= a^2 \cdot c^{-1} \\ a^{3 - \frac{1}{2}} \cdot b^{x - \frac{5}{2}} \cdot c^{4 - 5} &= a^2 \cdot c^{-1} \\ a^{\frac{5}{2}} \cdot b^{x - \frac{5}{2}} \cdot c^{-1} &= a^2 \cdot c^{-1} \\ \end{align*} \] Kita dapat memisahkan persamaan berdasarkan basis variabel a, b, dan c untuk membuatnya lebih mudah diselesaikan. \[ \begin{align*} a^{\frac{5}{2}} &= a^2 \\ b^{x - \frac{5}{2}} &= 1 \\ c^{-1} &= c^{-1} \\ \end{align*} \] Dari persamaan tersebut dapat disimpulkan bahwa: \[ \begin{align*} \frac{5}{2} &= 2 \quad \text{(Ini benar, tidak perlu dipecahkan lebih lanjut)} \\ x - \frac{5}{2} &= 0 \quad \text{(Ini yang akan kita selesaikan)} \\ \end{align*} \] \[ \begin{align*} x - \frac{5}{2} &= 0 \\ x &= \frac{5}{2} \\ \end{align*} \]
Finding the Domain of a Function with a Square Root
Para encontrar el dominio de la función f(x) = 3 / √(x - 3), necesitamos identificar todos los valores de x para los cuales la función está definida. La única restricción aquí es que el denominador no puede ser cero y debe ser un número real, lo que significa que x - 3 debe ser positivo porque la raíz cuadrada de un número negativo no está definida en el conjunto de los números reales. Por lo tanto, configuramos la desigualdad: x - 3 > 0 Resolvemos para x: x > 3 Esto significa que el dominio de f(x) son todos los valores de x mayores que 3. En notación de intervalo, el dominio se escribe como: (3, +∞) En resumen, el dominio de la función f(x) = 3 / √(x - 3) es todos los valores reales de x mayores que 3.
Solving a Linear Equation with One Variable
The image shows two systems of equations. The first one is a two-variable linear system, and the y-terms have been eliminated, resulting in a single linear equation -14x = 7. To find the value of x, we need to solve the equation -14x = 7. We do so by isolating x: -14x = 7 x = 7 / (-14) x = -1/2 Thus, the value of x is -1/2, which corresponds to option A: -1/2.
Finding the Value of x with Parallel Lines
In the given image, there are two intersecting lines, one of which is straight (m) and the other one (n) has a bend. You are given that line m is parallel to line n and you need to find the value of x. The straight angle between the upper part of line n and the line m is given as 140°. Since lines m and n are parallel, and we have a transversal that intersects both lines (the bent segment of line n), the alternate interior angles are equal. Therefore, the angle measured by x will also be 140°. Hence, the value of x is 140°.
Finding the Value of x in Angle Bisector Problem
The task is to find the value of x given that M is a point on the bisector of the angle AOB. Since OM is the bisector, it divides the angle AOB into two equal angles. The expressions for the angles on either side of OM are given as (3x + 12)° and (5x + 6)°. Since these two angles are equal, we can set up the following equation: 3x + 12 = 5x + 6 Now let's solve for x: 3x + 12 − 12 = 5x + 6 − 12 3x = 5x − 6 Now, subtract 5x from both sides: 3x − 5x = −6 −2x = −6 Divide both sides by -2: x = −6 / −2 x = 3 Therefore, the value of x is 3.
Finding the Value of x when y is Given
To find the value of \( x \) when \( y \) is -49 for the given equation \( y = 7x + 49 \), you will need to substitute -49 for \( y \) in the equation and solve for \( x \). Starting with the equation: \[ y = 7x + 49 \] Now, substitute \( y \) with -49: \[ -49 = 7x + 49 \] To solve for \( x \), subtract 49 from both sides of the equation to isolate the term containing \( x \): \[ -49 - 49 = 7x \] \[ -98 = 7x \] Now, divide both sides by 7 to solve for \( x \): \[ \frac{-98}{7} = x \] \[ x = -14 \] Thus, the value of \( x \) when \( y \) is -49 is \( x = -14 \).
Finding the Value of x When y is Given
The equation given in the image is \( y = \frac{1}{5}x - 5 \). We are asked to find the value of \(x\) when \(y\) is \(-14\). To solve for \(x\), substitute \(-14\) for \(y\) in the equation: \( -14 = \frac{1}{5}x - 5 \) Now, solve for \(x\) by isolating it on one side of the equation. First, add 5 to both sides: \( -14 + 5 = \frac{1}{5}x - 5 + 5 \) \( -9 = \frac{1}{5}x \) Now, to get \(x\) by itself, multiply both sides of the equation by 5: \( 5(-9) = \left(\frac{1}{5}x\right) \times 5 \) \( -45 = x \) Therefore, when \(y\) is \(-14\), the value of \(x\) is \(-45\).
Determinant Calculation for 3x3 Matrix
The image contains a determinant of a 3x3 matrix equal to -30. We need to calculate the value(s) of x for which this determinant holds true. The determinant is: | 3 1 x | | -1 3 4 | | x 1 0 | To find the determinant, we will expand along the third column (as it seems to be the easiest due to the zero present): det = x * | -1 3 | | x 1 | - x * | 3 1 | | x 0 | + 4 * | 3 1 | | -1 3 | Solving each of these determinants: det(x) = x((-1)(1) - (3)(x)) - x((3)(0) - (1)(x)) + 4((3)(3) - (1)(-1)) det(x) = x(-1 - 3x) - x(0 - x) + 4(9 + 1) det(x) = x(-1 - 3x) - x(-x) + 4(10) det(x) = -x - 3x^2 + x^2 + 40 det(x) = -3x^2 + x^2 - x + 40 Combine like terms: det(x) = -2x^2 - x + 40 Now we set this determinant equal to the given value (-30) and solve for x: -2x^2 - x + 40 = -30 Move -30 to the other side: -2x^2 - x + 40 + 30 = 0 -2x^2 - x + 70 = 0 This is a quadratic equation, which can often be solved by factoring, completing the square, or using the quadratic formula. To use the quadratic formula (\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)) we identify the coefficients a = -2, b = -1, c = 70. Before using the quadratic formula, it might be helpful to check if the quadratic can be factored easily, but in this case, it cannot be factored into integers, so let's apply the quadratic formula: \( x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(-2)(70)}}{2(-2)} \) \( x = \frac{1 \pm \sqrt{1 + 560}}{-4} \) \( x = \frac{1 \pm \sqrt{561}}{-4} \) The two solutions for x are: \( x = \frac{1 + \sqrt{561}}{-4} \) and \( x = \frac{1 - \sqrt{561}}{-4} \) These are the values of x for which the determinant of the matrix equals -30.
Determinant calculation for 2x2 matrix
The given matrix M is: \[ M = \begin{pmatrix} 2 & x \\ -1 & 3 \end{pmatrix} \] And we are told that the determinant of matrix M is 9. The determinant of a 2x2 matrix \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\) is calculated by \(ad - bc\). Therefore, for the given matrix M, the determinant is calculated by: \( (2)(3) - (-1)(x) = 6 + x \) Since we know that the determinant is 9, we can set up the following equation: \( 6 + x = 9 \) Now, let's solve for \( x \): \( x = 9 - 6 \) \( x = 3 \) Hence, the value of \( x \) that satisfies the given determinant value is 3.
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