Example Question - slope calculation
Here are examples of questions we've helped users solve.
Equation of Perpendicular Line Passing Through a Point
To find the equation of the line h, which is perpendicular to line g, and passes through the point (3, −9), we first need to determine the slope of line g. The equation for line g is given by: \[ y = \frac{3}{10}x - \frac{8}{10} \] The slope of line g is the coefficient of x, which is \(\frac{3}{10}\). Two lines that are perpendicular to each other have slopes that are negative reciprocals of one another. So, the slope (m) of line h will be: \[ m = -\frac{1}{\text{slope of g}} = -\frac{1}{\frac{3}{10}} = -\frac{10}{3} \] Now we know the slope of line h, we can use the point-slope form of the equation of a line to find its equation. The point-slope form is: \[ y - y_1 = m(x - x_1) \] Plugging in our values for the slope and the coordinates of the given point (3, −9), we get: \[ y - (-9) = -\frac{10}{3}(x - 3) \] \[ y + 9 = -\frac{10}{3}x + 10 \] To write this in slope-intercept form, which is \( y = mx + b \), we solve for y: \[ y = -\frac{10}{3}x + 10 - 9 \] \[ y = -\frac{10}{3}x + 1 \] Here \( b \), the y-intercept, is 1. The numbers in the equation are already simplified proper fractions, improper fractions, or integers. So, the equation for line h in slope-intercept form is: \[ y = -\frac{10}{3}x + 1 \]
Determining Equation of Trend Line in Scatter Plot with Two Points
To determine the equation of the trend line in a scatter plot, you usually need two points on the line to find the slope and then use one of these points to solve for the y-intercept. From the image provided, we can identify two points on the trend line. Let's select two points that clearly lie on the line and are easy to read: Point 1 (on the left end of the line): It looks like it crosses the y-axis at y = 1. This would make the point (0,1). Point 2 (somewhere on the line where it crosses a grid intersection): It's hard to tell exactly from this image, but let's assume it's (5,5). Next, we calculate the slope (m) using the formula: \( m = \frac{(y_2 - y_1)}{(x_2 - x_1)} \) Plugging in the point values: \( m = \frac{(5 - 1)}{(5 - 0)} \) \( m = \frac{4}{5} = 0.8 \) With the slope of 0.8 and knowing the line crosses the y-axis at (0,1), we can write the equation of the line in slope-intercept form, which is y = mx + b, where m is the slope and b is the y-intercept. Thus, the equation of the trend line is: y = 0.8x + 1 Please note that the accuracy of this equation depends on how precisely we can read the points from the image. For a more accurate equation, a clear reading of exact points on the line or the actual data used to generate the scatter plot would be necessary.
Determining Equation of Trend Line on Scatter Plot
To determine the equation of the trend line on a scatter plot, we typically look for two points through which the line passes. Once we have those points, we can use them to calculate the slope of the line and then use one of the points to find the equation in the slope-intercept form, which is y = mx + b, where m is the slope and b is the y-intercept. From the image, it's not possible to determine the exact coordinates of the points on the trend line with absolute precision, but we can estimate based on the grid. Let's identify two points that the trend line appears to pass through. For example, one point on the trend line seems to be approximately at (1, 1). The other point that we can infer intersects the trend line and grid line might be at (10, 9). Let's use these two estimated points to calculate the slope (m). The slope is calculated by taking the difference in the y-values divided by the difference in the x-values: \( m = (y_2 - y_1) / (x_2 - x_1) \) Using our estimated points (1, 1) and (10, 9): \( m = (9 - 1) / (10 - 1) \) \( m = 8 / 9 \) Now that we know the slope, we can determine the y-intercept (b) of the trend line using the y = mx + b format and one of the points. Let's use point (1, 1): \( y = mx + b \) \( 1 = (8/9)(1) + b \) \( b = 1 - 8/9 \) \( b = 1/9 \) So, the equation of the trend line, based on our estimations, would be approximately: \( y = (8/9)x + 1/9 \) Keep in mind that since the exact coordinates of the points the trend line passes through are not clearly provided, the equation obtained is based on visual estimation and might not be the precise equation used to generate the trend line in the scatter plot. However, without more accurate data, this is the best estimation we can make based on the provided image.
Analyzing a Linear Relationship on a Cartesian Coordinate Plane
The image depicts a Cartesian coordinate plane with an x-axis and a y-axis. There is a straight line graphed that passes through the origin (0,0) and extends into the first quadrant, indicating a positive linear relationship between x and y. Additionally, there is a red dot marked on the line. To solve a question related to this graph, I would need to know the specific question you have in mind, as there could be several questions related to finding the equation of the line, calculating the slope, determining the y-intercept, or interpreting the significance of the point marked with the red dot. However, based on the common tasks associated with such graphs, I can provide a general overview: 1. To find the slope of the line, you could pick two points on the line, find the change in y (rise) over the change in x (run), and use the slope formula: slope (m) = rise / run. 2. Since the line passes through the origin, it's clear that the y-intercept is 0, so the equation of the line would be in the form of y = mx. 3. If the question is about the red dot, we would need to determine its coordinates. Unfortunately, I can't measure the precise location on this image, so you would have to provide the coordinates or the context around what is needed regarding that point. Please provide the specific question or details regarding what you need to solve using this graph, and I'll be glad to assist further.
Finding Values for Parallel and Intersecting Lines
Bài toán này cho chúng ta hai phương trình đường thẳng d1 là y = mx - 5 và d2 là y = (2m + 1)x + 3. Chúng ta cần tìm giá trị của m để: a) Hai đường thẳng song song với nhau Hai đường thẳng sẽ song song với nhau khi và chỉ khi chúng có cùng hệ số góc nhưng khác phần tử cắt trên trục y. Hệ số góc của d1 là m, và hệ số góc của d2 là 2m + 1. Để chúng song song với nhau, m = 2m + 1. Giải phương trình này, ta có: m = 2m + 1 ⇒ 2m - m = -1 ⇒ m = -1 Vậy m = -1 là giá trị cần tìm để hai đường thẳng song song với nhau. b) Hai đường thẳng cắt nhau Hai đường thẳng sẽ cắt nhau khi và chỉ khi chúng có các hệ số góc khác nhau. Tức là m ≠ 2m + 1. Giả sử m = 2m + 1 để tìm điều kiện mâu thuẫn và sau đó xác định điều kiện thực sự cho m: m = 2m + 1 ⇒ m - 2m = 1 ⇒ -m = 1 ⇒ m = -1 Điều này cho thấy khi m = -1 thì hai đường thẳng song song, nghĩa là đối với tất cả các giá trị của m khác -1, hai đường thẳng sẽ cắt nhau. Vậy tất cả giá trị của m khác -1 đều thoả mãn điều kiện để hai đường thẳng cắt nhau.
Finding Perpendicular and Parallel Line Slopes
Para resolver esta pregunta, primero necesitamos encontrar la pendiente de la recta dada. La ecuación está en la forma general de una ecuación lineal Ax + By = C. Para hallar la pendiente, primero queremos despejar y y poner la ecuación en la forma pendiente-intercepto, que es y = mx + b, donde m es la pendiente. La ecuación dada es: \-8x - 3y = -7 Primero sumamos 8x a ambos lados para despejar los términos de y: \-3y = 8x - 7 Ahora dividimos cada término por -3 para obtener y por sí solo: \(y = \dfrac{8x}{-3} - \dfrac{-7}{-3}\) Simplificando, tenemos: \(y = -\dfrac{8}{3}x + \dfrac{7}{3}\) La pendiente (m) de la recta dada es \(-\dfrac{8}{3}\). La pendiente de una recta que es perpendicular a otra es el negativo recíproco de la pendiente de la recta original. Si m es la pendiente de la recta original, entonces la pendiente de la recta perpendicular será \( -\dfrac{1}{m}\). Por lo tanto, la pendiente de la recta perpendicular a la recta dada será: \(m_{perpendicular} = -\dfrac{1}{-\dfrac{8}{3}} = \dfrac{3}{8}\) La pendiente de una recta paralela a la recta dada, por otro lado, sería la misma que la de la recta original, ya que las rectas paralelas tienen pendientes iguales. Por lo tanto, la pendiente de la recta paralela a la recta dada es: \(m_{paralela} = -\dfrac{8}{3}\) En resumen, la pendiente de una recta perpendicular es \(\dfrac{3}{8}\) y la pendiente de una recta paralela es \(-\dfrac{8}{3}\).
Calculating Slope and Y-Intercept from a Line Graph
To find the slope of the line represented in the graph, you need to pick two points on the line that you can clearly identify the coordinates for. Then, use the formula for slope, which is: \[ slope (m) = \frac{change \, in \, y}{change \, in \, x} = \frac{y_2 - y_1}{x_2 - x_1} \] Looking at the graph, let's pick the two points where the line intersects the grid lines. One point could be at \( (1, 1) \) and another point could be at \( (3, 7) \). Now using these points: \[ x_1 = 1, y_1 = 1 \] \[ x_2 = 3, y_2 = 7 \] Let's calculate the slope: \[ slope (m) = \frac{7 - 1}{3 - 1} = \frac{6}{2} = 3 \] Next, to find the y-intercept (the point where the line crosses the y-axis), look at where the line intersects the y-axis. From the graph, it appears that the line crosses the y-axis at \( (0, -1) \). Therefore, the y-intercept (b) is -1. So the slope is 3, and the y-intercept is -1.
Finding the Derivative of a Function Expression
The image shows two graphs, y = f(x) and y = g(x), intersecting on a Cartesian coordinate plane, and there are lines \( d_1 \) and \( d_2 \) which appear to be parallel to the x-axis at different y-values. The function \( h(x) \) is defined such that \( h(x + 1) = f(x) - g(x + 2) \). We are tasked with finding the value of \( h'(2) \). The first step to find \( h'(2) \) is to differentiate the equation for \( h(x + 1) \) with respect to \( x \). So, let's differentiate \( h(x + 1) \) using the chain rule: If \( u = x + 1 \), then \( du/dx = 1 \) and so, \[ h'(u) = \frac{dh}{du} \cdot \frac{du}{dx} = \frac{dh}{du} \] Now differentiate \( f(x) \) and \( g(x + 2) \) separately with respect to \( x \). Using the property \( \frac{d}{dx}f(x) = f'(x) \) and \( \frac{d}{dx}g(x+2) \) with the chain rule (\( v = x + 2 \), \( dv/dx = 1 \)), we get: \[ h'(u) = f'(x) - g'(v) \] Knowing that \( h'(u) = h'(x + 1) \), we can convert the differentiation point from \( x \) to \( u \) to find \( h'(2) \): \[ h'(2) = f'(1) - g'(4) \] To get the values of \( f'(1) \) and \( g'(4) \), we need to find the slopes of \( f(x) \) and \( g(x) \) at \( x = 1 \) and \( x = 4 \), respectively. Looking at the graph, \( f(x) \) passes through point A(1,0), and since \( d_1 \) and \( d_2 \) are parallel to the x-axis, it means the slope is 0 at \( x = 1 \) for \( f(x) \). Therefore, the slope \( f'(1) = 0 \). The graph doesn't give us \( g(4) \) directly, but we can use the fact that the line through the point B(-1, 2) is at a 45-degree angle to infer that this line, representing \( g(x) \), has a slope of 1, as a 45-degree line in the first quadrant will have a slope of 1. Hence, \( g'(x) = 1 \) for all \( x \) in the domain of \( g \), and specifically \( g'(4) = 1 \). Now, substitute the slope values into the equation for \( h'(2) \): \[ h'(2) = 0 - 1 = -1 \] Thus, the value of \( h'(2) \) is \(-1\).
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