Example Question - slope
Here are examples of questions we've helped users solve.
Determining the Slope of a Tangent to a Curve Involving Radicals
<p>Given \( f(x) = x^{\frac{7}{5}} + \frac{4}{\sqrt[5]{x^3}} \), to find the slope of the tangent line at any point \( x \), we need to find the derivative of \( f(x) \) with respect to \( x \).</p> <p>First, we rewrite the function in a more derivative-friendly form:</p> <p>\( f(x) = x^{\frac{7}{5}} + 4x^{-\frac{3}{5}} \)</p> <p>Now we differentiate \( f(x) \) with respect to \( x \) using the power rule \( \frac{d}{dx} x^n = nx^{n-1} \):</p> <p>\( f'(x) = \frac{7}{5}x^{\frac{7}{5}-1} - \frac{12}{5}x^{-\frac{3}{5}-1} \)</p> <p>Simplifying, we get:</p> <p>\( f'(x) = \frac{7}{5}x^{\frac{2}{5}} - \frac{12}{5}x^{-\frac{8}{5}} \)</p> <p>So, the slope of the tangent line to the curve at any point \( x \) is given by \( f'(x) \).</p>
Motion Graph Analysis
<p>The solution involves analyzing the given velocity vs. time graph to determine the acceleration. The slope of the velocity vs. time graph, which is a straight line, represents the acceleration.</p> <p>To find the slope (\(a\)) of the velocity vs. time graph, which is a straight line, we use the formula:</p> <p>\[ a = \frac{\Delta v}{\Delta t} \]</p> <p>Inspecting the graph, we can estimate the change in velocity (\(\Delta v\)) from 0 to approximately 3000 m/s and the change in time (\(\Delta t\)) from 0 to 5 s.</p> <p>\[ a \approx \frac{3000 \text{ m/s}}{5 \text{ s}} \]</p> <p>\[ a \approx 600 \text{ m/s}^2 \]</p> <p>The slope of the velocity vs. time graph is approximately 600 m/s², and this value represents the acceleration.</p>
Analysis of Coordinates and Linear Equation of Points in Cartesian Plane
<p>Given point Q(8, k) and point S(-6, 0), it is known that line PQ is parallel to the x-axis.</p> <p>(a) Since PQ is parallel to the x-axis, the y-coordinates of points P and Q are equal, which means the y-coordinate of point Q is k = 0 (the same as point S).</p> <p>\(k = 0\)</p> <p>(b) To find the equation of line PS, we can calculate the slope (m) using the coordinates of points P and S:</p> <p>\(m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 0}{-6 - 8}\)</p> <p>Since the change in y is 0 (the line is horizontal), the slope is:</p> <p>\(m = 0\)</p> <p>The general equation of a line is \(y = mx + b\). Since the slope m = 0, the equation simplifies to</p> <p>\(y = b\)</p> <p>To find b, we use the fact that the line passes through point S(-6, 0):</p> <p>\(0 = 0 \cdot (-6) + b\)</p> <p>\(b = 0\)</p> <p>The equation of line PS is therefore \(y = 0\).</p>
Equation of a Parallel Line Passing Through a Given Point
Como la pregunta es sobre matemáticas, procedo como Profesor de Matemáticas para abordar el problema. La pregunta requiere que se encuentre la ecuación de una recta que pase por un punto dado y que sea paralela a una recta dada. La recta dada es \( \frac{2y-3}{x-1} = 3 \). Para encontrar la pendiente, primero reescribimos la ecuación en la forma \( y = mx + b \), donde \( m \) es la pendiente. \[ \frac{2y-3}{x-1} = 3 \rightarrow 2y - 3 = 3(x - 1) \rightarrow 2y = 3x - 3 + 3 \] \[ 2y = 3x \rightarrow y = \frac{3}{2}x \] Por lo tanto, la pendiente de la recta dada es \( \frac{3}{2} \). Como la recta que buscamos debe ser paralela, tendrá la misma pendiente \( \frac{3}{2} \). La ecuación de la recta con pendiente \( \frac{3}{2} \) que pasa por el punto \( (-2, -1) \) se encontrará utilizando la forma punto-pendiente de la ecuación de una recta: \( y - y_1 = m(x - x_1) \), donde \( (x_1, y_1) \) es un punto en la recta y \( m \) es la pendiente. \[ y - (-1) = \frac{3}{2}(x - (-2)) \rightarrow y + 1 = \frac{3}{2}(x + 2) \] \[ \Rightarrow y + 1 = \frac{3}{2}x + 3 \] Restando 1 a ambos lados nos da la ecuación final de la recta: \[ y = \frac{3}{2}x + 2 \] Esta es la ecuación de la recta que pasa por el punto \( (-2, -1) \) y es paralela a la recta dada \( \frac{2y-3}{x-1} = 3 \).
Parallel Line Equations
La primera pregunta es encontrar la ecuación de una recta que pasa por el punto \( (\frac{1}{2}, \frac{1}{3}) \) y tiene pendiente \( m = -2 \). <p>\( y - y_1 = m(x - x_1) \)</p> <p>\( y - \frac{1}{3} = -2(x - \frac{1}{2}) \)</p> <p>\( y - \frac{1}{3} = -2x + 1 \)</p> <p>\( y = -2x + 1 + \frac{1}{3} \)</p> <p>\( y = -2x + \frac{4}{3} \)</p> La segunda pregunta es encontrar la ecuación de una recta que pasa por el punto \( (1, 0) \) y es paralela a la recta \( y = 2x - 3 \). <p>Dado que la recta es paralela, tiene la misma pendiente, \( m = 2 \).</p> <p>\( y - y_1 = m(x - x_1) \)</p> <p>\( y - 0 = 2(x - 1) \)</p> <p>\( y = 2x - 2 \)</p> La tercera pregunta es encontrar la ecuación de una recta que pasa por el punto \( (-2, -1) \) y es paralela a la recta \( \frac{y-1}{2} = \frac{2x-3}{3} \). <p>Primero, encontramos la pendiente de la recta dada resolviendo para \( y \).</p> <p>\( \frac{y-1}{2} = \frac{2x-3}{3} \)</p> <p>\( 3(y - 1) = 2(2x - 3) \)</p> <p>\( 3y - 3 = 4x - 6 \)</p> <p>\( 3y = 4x - 3 \)</p> <p>\( y = \frac{4}{3}x - 1 \)</p> <p>Entonces, la pendiente de la recta a encontrar es \( m = \frac{4}{3} \).</p> <p>Usamos la forma punto-pendiente para encontrar la ecuación de la nueva recta.</p> <p>\( y - y_1 = m(x - x_1) \)</p> <p>\( y + 1 = \frac{4}{3}(x + 2) \)</p> <p>\( y + 1 = \frac{4}{3}x + \frac{8}{3} \)</p> <p>\( y = \frac{4}{3}x + \frac{8}{3} - 1 \)</p> <p>\( y = \frac{4}{3}x + \frac{5}{3} \)</p>
Explanation of Slope-Intercept Form
The given equation is the slope-intercept form of a line, which is represented as \( y = mx + b \), where: - \( m \) is the slope of the line - \( b \) is the y-intercept, i.e., the value of \( y \) when \( x = 0 \) There is no specific question to solve, as the image simply provides the general form of a linear equation in slope-intercept form.
Converting Function to Slope-Intercept Form
The function given in the image is: \[ f(x) = -2 \left( x + \frac{1}{4} \right) - \frac{49}{8} \] To express the function in the form of y-intercept, we need to simplify it while keeping it in the form of \( y = mx + b \), where \( m \) stands for the slope and \( b \) is the y-intercept. Let's simplify the function: \[ f(x) = -2x - \frac{1}{2} - \frac{49}{8} \] To combine the constant terms, we need a common denominator. The common denominator for \( \frac{1}{2} \) and \( \frac{49}{8} \) is 8. So let's convert \( \frac{1}{2} \) to have the denominator of 8: \[ - \frac{1}{2} = - \frac{4}{8} \] Now rewritting the equation we have: \[ f(x) = -2x - \frac{4}{8} - \frac{49}{8} = -2x - \frac{53}{8} \] So the function in the form of intercept is: \[ f(x) = -2x - \frac{53}{8} \] This function is now in slope-intercept form with \( m = -2 \) (the slope) and \( b = -\frac{53}{8} \) (the y-intercept).
Finding Tangent Line Equation of a Circle
Câu hỏi trong hình là một câu hỏi về việc tìm phương trình tiếp tuyến của đường tròn tại một điểm. Đường tròn có phương trình là \((x - 2)^2 + (y + 1)^2 = 25\) và tiếp điểm là \(M(5;3)\). Đầu tiên, chúng ta cần xác định tâm và bán kính của đường tròn. Tâm của đường tròn là \(O(2;-1)\) và bán kính \(R\) là căn bậc hai của 25, \(R = 5\). Tiếp theo, chúng ta sẽ tìm hệ số góc của tiếp tuyến tại \(M(5;3)\). Đường kính qua tiếp điểm M và tâm O là pháp tuyến của tiếp tuyến tại M. Hệ số góc của đường kính này sẽ là đối số của hệ số góc của tiếp tuyến. Đường kính có hướng từ M đến O có hệ số góc: \[ k_{MO} = \frac{y_O - y_M}{x_O - x_M} = \frac{-1 - 3}{2 - 5} = \frac{-4}{-3} = \frac{4}{3} \] Hệ số góc của tiếp tuyến tại M, ký hiệu là \(k_{TM}\), là đối của hệ số góc đường kính, do đó: \[ k_{TM} = -\frac{1}{k_{MO}} = -\frac{1}{\frac{4}{3}} = -\frac{3}{4} \] Bây giờ, sử dụng hệ số góc \(k_{TM}\) và tiếp điểm M(5;3), chúng ta có thể viết phương trình của tiếp tuyến như sau: \[ y - y_M = k_{TM}(x - x_M) \] \[ y - 3 = -\frac{3}{4}(x - 5) \] Ta có thể biến đổi phương trình này để có được một phương trình theo định dạng chuẩn: \[ y + \frac{3}{4}x = 3 + \frac{15}{4} \] \[ y + \frac{3}{4}x = 3 + \frac{15}{4} = \frac{12}{4} + \frac{15}{4} = \frac{27}{4} \] \[ y = - \frac{3}{4}x + \frac{27}{4} \] \[ 4y = -3x + 27 \] \[ 3x + 4y - 27 = 0 \] Như vậy, phương trình tiếp tuyến cần tìm là \(3x + 4y - 27 = 0\), do đó đáp án đúng là D.
Parallel Lines and Inconsistent Systems of Equations
It looks like the image shows a system of linear equations, and you might be asked to find the solution for this system. The equations in the image are: 1) \( y = -\frac{5}{4}x + 8 \) 2) \( y = -\frac{5}{4}x - 9 \) To solve these equations, we usually look for a point (x, y) that satisfies both equations. However, if you look closely at both equations, you'll notice that they both have the same slope, -5/4, which means they are parallel lines. Since parallel lines never intersect, they don't have a point in common. Therefore, this system of equations has no solution. In mathematical terms, this is known as an inconsistent system.
Parallel Lines and Inconsistent Equations
The equations provided are: 1) \( y = -\frac{5}{4}x + 8 \) 2) \( y = -\frac{5}{4}x - 9 \) To solve these equations, we need to find the values of \(x\) and \(y\) where both equations are satisfied, meaning where the lines intersect if these were graphed. However, when we observe the equations, we notice that the coefficients of \(x\) in both equations are identical, and the constants are different. This means that the lines are parallel and never intersect. These equations represent parallel lines because they have the same slope, which is \(-\frac{5}{4}\), but different y-intercepts. The y-intercept of the first equation is 8, and the y-intercept of the second equation is -9. Since the lines never meet, there is no solution to this system of equations—they are inconsistent. Therefore, there are no specific values of \(x\) and \(y\) that would solve both equations simultaneously.
Parallel Lines in Geometry
Dựa vào hình ảnh bạn cung cấp, tôi sẽ giải câu hỏi số 2, vì câu hỏi số 3 không liên quan đến việc giải toán. Câu 2. Cho đường thẳng (d): y = 3x + 4 và (d'): y = ax - 19 Để (d) song song với (d') thì a = ... Đường thẳng (d) có dạng y = mx + b, trong đó m là hệ số góc và b là hệ số tự do. Hai đường thẳng song song với nhau khi và chỉ khi chúng có cùng hệ số góc. Trong trường hợp của (d) và (d'), hệ số góc của (d) là 3 ( với m = 3 từ phương trình y = 3x + 4). Để (d) song song với (d') thì hệ số góc của (d') cũng phải bằng 3 (hệ số a phải bằng m). Vậy a = 3. Chọn đáp án C: a = 3.
Writing Linear Equations in Slope-Intercept Form
Para escribir la ecuación de la línea en la forma pendiente-intersección, que es \( y = mx + b \), necesitamos encontrar dos piezas de información: la pendiente (m) y la intersección con el eje y (b). Observando la imagen, podemos identificar dos puntos por los cuales pasa la línea. En este caso, parece que podemos usar los puntos (0, 4) y (-2, 0). Si no fueran claros en la imagen, deberíamos buscar dos puntos donde la línea cruza exactamente a través de intersecciones de la cuadrícula para ser eso precisos. La pendiente (m) se calcula con la fórmula \( m = \frac{y_2 - y_1}{x_2 - x_1} \). Utilizando los puntos (0, 4) y (-2, 0): \[ m = \frac{0 - 4}{-2 - 0} = \frac{-4}{-2} = 2 \] Ahora que tenemos la pendiente, necesitamos la intersección con el eje y (b), que es simplemente el valor de y donde la línea cruza el eje y. Observando el punto que ya hemos identificado como (0, 4), podemos ver que b = 4. Por lo tanto, la ecuación de la línea en la forma pendiente-intersección es: \[ y = 2x + 4 \]
Understanding a Line Equation in Slope-Intercept Form
It seems you've provided an image with an equation of a line: \( y = \frac{1}{2}x - 4 \). This equation is already in slope-intercept form, which is given by \( y = mx + b \), where \( m \) is the slope of the line and \( b \) is the y-intercept. From the given equation: - The slope (\( m \)) of the line is \( \frac{1}{2} \). - The y-intercept (\( b \)) of the line is \( -4 \). If you need to solve for a specific value, you would need to substitute that value for \( x \) and solve for \( y \), or vice versa. However, since there is no specific request on what to solve for, this is as much as can be said about the given equation. If you have a specific question about this equation, such as finding a point on the line or graphing the line, please provide further information.
Understanding Linear Equations: Slope-Intercept Form
The image you provided shows the equation y = 1/3x - 4. This is a linear equation in the slope-intercept form y = mx + b, where: - m stands for the slope of the line, which in this case is 1/3, indicating that for every increase of x by 3 units, y increases by 1 unit. - b stands for the y-intercept, which is the value of y when x is 0. In this case, b is -4, meaning the line crosses the y-axis at (0, -4). This equation does not have a specific solution since it represents a line with infinitely many points that satisfy the equation. To graph this line, you can plot the y-intercept at (0, -4) on a coordinate plane and use the slope to determine another point. Starting at (0, -4), move up 1 unit and to the right 3 units to find the next point at (3, -3). Connect these points with a straight line to represent the equation.
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