Example Question - coefficients
Here are examples of questions we've helped users solve.
Multiplying Powers with Coefficients
<p>First, multiply the coefficients: \( 12 \times 8 \times \frac{3}{4} = 24 \times 3 = 72 \)</p> <p>Next, add the exponents of \( p \): \( 6 + 3 + 7 = 16 \)</p> <p>The final answer is \( 72p^{16} \)</p>
Determining Coefficients in Quadratic Expressions
<p>For each quadratic expression, identify the coefficients a, b, and c using the standard form of a quadratic equation, which is \( ax^2 + bx + c \).</p> <p>(a) \( 2x^2 - 5x + 1 \): \( a = 2, b = -5, c = 1 \)</p> <p>(b) \( x^2 - 2x \): \( a = 1, b = -2, c = 0 \)</p> <p>(c) \( 2y^2 + 1 \): \( a = 2, b = 0, c = 1 \)</p> <p>(d) \( -\frac{1}{2}p^2 + 4p \): \( a = -\frac{1}{2}, b = 4, c = 0 \)</p> <p>(e) \( -x - 2x^2 \): \( a = -2, b = -1, c = 0 \)</p> <p>(f) \( 4x^2 \): \( a = 4, b = 0, c = 0 \)</p> <p>(g) \( h^2 + \frac{3}{2}h - 4 \): \( a = 1, b = \frac{3}{2}, c = -4 \)</p> <p>(h) \( \frac{1}{3}k^2 - 2 \): \( a = \frac{1}{3}, b = 0, c = -2 \)</p> <p>(i) \( 2(r - 3) \): expand to get \( 2r - 6 \): \( a = 2, b = 0, c = -6 \)</p>
Determine Values for Quadratic Expressions
<p>For each quadratic expression, we identify coefficients a, b, and c in the standard form \( ax^2 + bx + c \):</p> <p>(a) \( 2x^2 - 5x + 1 \): a = 2, b = -5, c = 1</p> <p>(b) \( x^2 - 2x \): a = 1, b = -2, c = 0</p> <p>(c) \( 2x^2 + 1 \): a = 2, b = 0, c = 1</p> <p>(d) \( -\frac{1}{2}p^2 + 4p \): a = -\frac{1}{2}, b = 4, c = 0</p> <p>(e) \( -1 - x - 2x^2 \): a = -2, b = -1, c = -1</p> <p>(f) \( 4x^2 \): a = 4, b = 0, c = 0</p> <p>(g) \( h^2 + \frac{3}{2}h - 4 \): a = 1, b = \frac{3}{2}, c = -4</p> <p>(h) \( \frac{1}{3}k^2 - 2 \): a = \frac{1}{3}, b = 0, c = -2</p> <p>(i) \( 2r(r - 3) \): a = 2, b = -6, c = 0</p>
Determine Coefficients of Quadratic Expressions
<p>a = 2, b = -5, c = 1</p> <p>a = 1, b = -2, c = 0</p> <p>a = -\frac{1}{2}, b = 4, c = 0</p> <p>a = -2, b = 0, c = 0</p> <p>a = 1, b = \frac{3}{2}, c = -4</p>
Determine the Values of Coefficients
<p>Given the quadratic expression \(2x^2 - 5x + 1\), we can identify the coefficients as follows:</p> <p>Let \(a = 2\), \(b = -5\), and \(c = 1\).</p>
Finding Coefficients of Quadratic Expressions
<p>For the expression \(2x^2 - 5x + 1\), we identify the coefficients as follows:</p> <p><b>a = 2</b></p> <p><b>b = -5</b></p> <p><b>c = 1</b></p> <p>For the expression \(x^2 - 2x\), we identify the coefficients as:</p> <p><b>a = 1</b></p> <p><b>b = -2</b></p> <p><b>c = 0</b></p>
Completing Mathematical Expressions
Dado que la imagen muestra una tarea en la que se debe completar expresiones matemáticas, pero falta el contexto de los números específicos o las expresiones a utilizar, proporcionaré una solución genérica para la segunda expresión proporcionada en español, ya que la primera frase está incompleta y no se puede resolver tal como se pide. La segunda afirmación es: "La suma del producto de dos números y el triple del sustrato de los números." Para la segunda expresión, consideremos dos números arbitrarios \( x \) e \( y \). La expresión en términos de \( x \) e \( y \) sería: <p>\( x \cdot y + 3(x - y) \)</p> Esta es la expresión matemática que representa la afirmación dada, sin conocer los números específicos involucrados.
Polynomial Equation Analysis
<p>Bạn chưa cung cấp đầy đủ thông tin hoặc câu hỏi cụ thể cần giải quyết liên quan đến các phương trình đa thức B và D trong hình ảnh vì vậy tôi không thể đưa ra lời giải cụ thể. Nếu bạn cần tìm đạo hàm, tích phân, hoặc giải bất kỳ bài toán cụ thể nào liên quan đến B hoặc D, vui lòng cung cấp thông tin chi tiết hơn.</p>
Algebraic Expressions
1. 8p 2. ab 3. b^2 4. 6x + 6
Solving Quadratic Equations with Coefficients
La ecuación que se muestra en la imagen es una ecuación cuadrática y se ve así: \[ 2x^2 - 4x + 1 = 0 \] Para resolver esta ecuación cuadrática, se puede utilizar la fórmula general para las raíces de una ecuación cuadrática, la cual es: \[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} \] En donde \( a \), \( b \) y \( c \) son los coeficientes de la ecuación cuadrática. En este caso, tenemos que \( a = 2 \), \( b = -4 \), y \( c = 1 \). Primero calculamos el discriminante (\( \Delta \)), que es \( b^2 - 4ac \): \[ \Delta = (-4)^2 - 4(2)(1) = 16 - 8 = 8 \] Ahora usamos la fórmula general para encontrar los valores de \( x \): \[ x = \frac{{-(-4) \pm \sqrt{8}}}{{2(2)}} \] \[ x = \frac{{4 \pm \sqrt{8}}}{{4}} \] \[ x = \frac{{4 \pm 2\sqrt{2}}}{{4}} \] Podemos simplificar dividiendo numerador y denominador entre 2: \[ x = \frac{{2 \pm \sqrt{2}}}{{2}} \] De aquí obtenemos dos soluciones para la ecuación: \[ x_1 = \frac{{2 + \sqrt{2}}}{2} \] \[ x_2 = \frac{{2 - \sqrt{2}}}{2} \] Estas son las dos soluciones de la ecuación cuadrática dada.
Solving a Simplification Expression with Variables and Coefficients
To solve the expression \( \frac{10x^8}{5x^4} \), you can simplify it by dividing both the coefficients (the numerical parts) and the variables separately. First, divide the coefficients: \( \frac{10}{5} = 2 \) Next, apply the laws of exponents for dividing like bases: \( x^8 / x^4 = x^{8-4} = x^4 \) Combine these two results to get the final answer: \( 2x^4 \)
Solving a Quadratic Equation Using the Quadratic Formula
To solve the quadratic equation \( a^2 + 4a + 4 = 0 \) using the quadratic formula, we need to identify the coefficients of the equation, which are in the standard form \( ax^2 + bx + c = 0 \). For this equation: \( a = 1 \) \( b = 4 \) \( c = 4 \) The quadratic formula is given by: \[ x = \frac{{-b \pm \sqrt{b^2 - 4ac}}}{{2a}} \] Plugging in the coefficients: \[ x = \frac{{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot 4}}}{{2 \cdot 1}} \] \[ x = \frac{{-4 \pm \sqrt{16 - 16}}}{{2}} \] \[ x = \frac{{-4 \pm \sqrt{0}}}{{2}} \] \[ x = \frac{{-4 \pm 0}}{{2}} \] \[ x = \frac{{-4}}{{2}} \] \[ x = -2 \] Since the discriminant (\( b^2 - 4ac \)) is zero, there is one real, repeated solution to the equation. Therefore, the solution to the equation \( a^2 + 4a + 4 = 0 \) is \( x = -2 \). This is the simplest form.
Solving Simultaneous Equations by Elimination Method
To solve the system of simultaneous equations: 2x - 5y = 4 ...(1) 3x + 2y = -13 ...(2) We can use either the substitution method or the elimination method. Here, I will use the elimination method. First, let's try to eliminate one of the variables by making the coefficients of either x or y the same in both equations. If we multiply equation (1) by 3 and equation (2) by 2, we will get the coefficients of x to be the same: 3*(2x - 5y) = 3*4 2*(3x + 2y) = 2*(-13) This results in: 6x - 15y = 12 ...(3) 6x + 4y = -26 ...(4) Now, we will subtract equation (4) from equation (3) to eliminate x and find y: (6x - 15y) - (6x + 4y) = 12 - (-26) 6x - 6x - 15y - 4y = 12 + 26 -19y = 38 Divide both sides by -19 to find y: y = 38 / -19 y = -2 Now we know the value of y; we can substitute it into either original equation (1) or (2) to find x. We'll use equation (1): 2x - 5*(-2) = 4 2x + 10 = 4 Subtract 10 from both sides: 2x = 4 - 10 2x = -6 Divide by 2 to solve for x: x = -6 / 2 x = -3 So the solution to the system of equations is: x = -3 y = -2
Quadratic Formula and Coefficients
The equation in the image appears to be the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] For a quadratic equation in the form \(ax^2 + bx + c = 0\). However, it appears that there is a missing term in the denominator (it should be \(2a\), but only \(2(1)\) is written), and the coefficients 'a', 'b', and 'c' are not specified in the image. The formula shown is: \[x = \frac{-1 \pm \sqrt{1^2 - 4(8)(20)}}{2(1)}\] This suggests 'a' = 8, 'b' = 1, and 'c' = 20 for the quadratic equation. However, given the typical structure of a quadratic equation \(ax^2 + bx + c = 0\), it's unusual to have an 'a' term larger than 'b' or 'c'. It's possible there's a mistake in the handwriting. Assuming the 'a', 'b', and 'c' values as 1, 1, and -160 respectively (if we take the coefficient of 'a' to be 1 instead of 8), the equation would be: \[x = \frac{-1 \pm \sqrt{(1)^2 - 4(1)(-160)}}{2(1)}\] Now let's solve it with these coefficients: \[x = \frac{-1 \pm \sqrt{1 + 640}}{2}\] \[x = \frac{-1 \pm \sqrt{641}}{2}\] We now have two possible solutions for x, which are: \[x_1 = \frac{-1 + \sqrt{641}}{2}\] \[x_2 = \frac{-1 - \sqrt{641}}{2}\] These are the two solutions to the quadratic equation based on the given formula and assuming 'a' = 1. If 'a' is actually 8 as shown, then the quadratic equation would be incorrect, and we would need the correct coefficients to proceed.
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.
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