Example Question - coordinate plane
Here are examples of questions we've helped users solve.
Evaluating a Given Function Value from a Graph
<p>To find \( f(4) \), we locate the point where \( x = 4 \) on the graph and find the corresponding \( y \) value.</p> <p>At \( x = 4 \), the graph passes through the point \((4, 3)\).</p> <p>Therefore, \( f(4) = 3 \).</p>
Graphing a Linear Equation on a Coordinate Plane
<p>Para resolver esta pregunta, necesitamos graficar la ecuación lineal \(y = 3x - 2\) en el plano de coordenadas. Primero, utilizando la tabla de valores, podemos ver algunos puntos que ya se han calculado y que se pueden trazar en la gráfica.</p> <p>Paso 1: Trazar los puntos (3, 7), (2, 4) y (-2, -8) en el plano de coordenadas. Cada punto corresponde a un valor de 'x' y el valor de 'y' resultante tras aplicar la ecuación \(y = 3x - 2\).</p> <p>Paso 2: Dibujar una línea recta que pase por estos puntos, ya que representan la solución a la ecuación lineal y cualquier punto en esta línea satisfará la ecuación \(y = 3x - 2\).</p> <p>El punto donde la línea cruza el eje 'y' es el intercepto en y, que para esta ecuación es -2, y esto indica que cuando \(x=0\), \(y=-2\).</p> <p>La pendiente de la línea es 3, indicando que por cada aumento en 1 en la dirección de 'x', 'y' aumentará en 3 unidades.</p>
Graphing Linear Equations
<p>Para graficar la ecuación lineal \( y = 3x - 2 \), primero evaluamos \( y \) para cada valor de \( x \) en la tabla proporcionada.</p> <p>Si \( x = 3 \):</p> <p>\( y = 3(3) - 2 = 9 - 2 = 7 \)</p> <p>Si \( x = 2 \):</p> <p>\( y = 3(2) - 2 = 6 - 2 = 4 \)</p> <p>Si \( x = -2 \):</p> <p>\( y = 3(-2) - 2 = -6 - 2 = -8 \)</p> <p>Y cuando \( x = 0 \), que ya está dado en la tabla:</p> <p>\( y = 3(0) - 2 = 0 - 2 = -2 \)</p> <p>Ahora podemos graficar los puntos (3,7), (2,4), (-2,-8), y (0,-2) en el plano coordenado y dibujar la línea que los une, que será la gráfica de la ecuación lineal \( y = 3x - 2 \).</p>
Graphing a Linear Equation
\[ \begin{array}{c} \text{Para el valor de} \ x = 3: \\ y = 3(3) - 2 = 9 - 2 = 7 \\ \text{Por lo tanto, el par ordenado es} \ (3,7). \\ \text{Para el valor de} \ x = 2: \\ y = 3(2) - 2 = 6 - 2 = 4 \\ \text{Por lo tanto, el par ordenado es} \ (2,4). \\ \text{Para el valor de} \ x = -2: \\ y = 3(-2) - 2 = -6 - 2 = -8 \\ \text{Por lo tanto, el par ordenado es} \ (-2,-8). \\ \text{Para el valor de} \ x = 0: \\ y = 3(0) - 2 = 0 - 2 = -2 \\ \text{Por lo tanto, el par ordenado es} \ (0,-2). \\ \end{array} \] Con estos pares ordenados, se pueden marcar los puntos correspondientes en el plano coordenado y trazar la recta que los une para representar la ecuación lineal \( y = 3x - 2 \).
Plotting Linear Equations on a Graph
<p>Para resolver la ecuación lineal \( y = 3x - 2 \) y trazarla en un plano coordenado, primero necesitamos calcular los valores de \( y \) para los valores dados de \( x \):</p> <p>\( x = 3 \) \\ \( y = 3(3) - 2 \) \\ \( y = 9 - 2 \) \\ \( y = 7 \)</p> <p>\( x = 2 \) \\ \( y = 3(2) - 2 \) \\ \( y = 6 - 2 \) \\ \( y = 4 \)</p> <p>\( x = 1 \) \\ \( y = 3(1) - 2 \) \\ \( y = 3 - 2 \) \\ \( y = 1 \)</p> <p>\( x = 0 \) \\ \( y = 3(0) - 2 \) \\ \( y = 0 - 2 \) \\ \( y = -2 \)</p> <p>Con estos puntos, podemos trazar la línea en el gráfico.</p>
Graphical Analysis of Inequalities and Regions
<p>Langkah pertama adalah mengenal pasti persamaan garis dan ketaksamaan yang ditunjukkan dalam graf.</p> <p>Garis yang terdapat dalam graf termasuk:</p> <p>1. \( y = 2 \) : Garis mendatar yang melalui \( y = 2 \).</p> <p>2. \( y = -x + 1 \) : Garis miring dengan gradien -1 yang melalui titik potong y pada \( (0, 1) \).</p> <p>3. \( y = x + 1 \) : Garis miring dengan gradien 1 yang melalui titik potong y pada \( (0, 1) \).</p> <p>Ketaksamaan yang terdapat pada graf:</p> <p>1. \( y \leq 2 \) : Ini berarti bahwa daerah yang diarsir berada pada atau di bawah garis \( y = 2 \).</p> <p>2. \( y \geq -x + 1 \) : Ini berarti bahwa daerah yang diarsir berada pada atau di atas garis \( y = -x + 1 \).</p> <p>3. \( y \leq x + 1 \) : Ini berarti bahwa daerah yang diarsir berada pada atau di bawah garis \( y = x + 1 \).</p> <p>Daerah yang diarsir menunjukkan daerah penyelesaian untuk sistem ketaksamaan yang disajikan oleh garis-garis ini. Daerah ini adalah sempadan yang dibentuk oleh garis-garis ini dan termasuk titik di mana ketiga-tiga garis bertemu, iaitu titik (0.5, 1.5).</p>
Deciphering a Blurry Graph in Mathematics
The image you've provided is quite blurry, and it's challenging to discern the details of the graph accurately. However, I can see a coordinate plane with some points plotted on it and what appears to be a line passing through them. To solve a question related to such an image, you would typically need to: 1. Determine the coordinates of the plotted points if the question requires it. 2. Understand the slope of the line if it's a question about the equation of the line. 3. Possibly write an equation if the line represents a linear relationship. Without a specific question or clearer image, I can't provide an exact answer, but I'm ready to help with any mathematical concepts or procedures you might need assistance with once the details are clearer.
Interpreting a Graph in a Coordinate Plane
It seems that you would like to solve a problem related to the graph in the provided image. However, the image alone doesn't give me a specific question to solve. The graph shows a coordinate plane with the x-axis and y-axis ranging from -1,000 to 1,000. There is a line that seems to pass through the origin (0,0) and extends into the first quadrant, suggesting a positive linear relationship between x and y. There is a red dot on the line, but without more context, it is impossible to determine the exact nature of the question. If your question is to find the coordinates of that red dot, we can estimate its position based on the axes. Assuming that the scale is uniform and each grid division represents the same increment, we can try to approximate the coordinates of the red dot by counting the grid lines: By eyeballing the grid, the red dot seems to be around the point (200, 300), although this is an approximation. If you have a specific question regarding this graph that you need help with, such as finding the equation of the line, determining the coordinates of a specific point, or calculating the slope, please provide the question, and I would be glad to help you solve it.
Graphing a Linear Equation in Point-Slope Form
The equation provided in the image is a linear equation in point-slope form, which is \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is a point on the line and \( m \) is the slope of the line. The given equation is: \[ y - 5 = -\frac{8}{3}(x - 6) \] From this equation, we can identify the slope \( m = -\frac{8}{3} \) and a point on the line \( (x_1, y_1) = (6, 5) \). To graph this line, follow these steps: 1. Plot the point \( (6, 5) \) on a coordinate plane. 2. Use the slope to find additional points. The slope \( -\frac{8}{3} \) means that for every 3 units you move to the right (positive direction along the x-axis), you move 8 units down (negative direction along the y-axis). 3. From \( (6, 5) \), move 3 units to the right to \( (9, 5) \) and then 8 units down to \( (9, -3) \). Plot this second point. 4. Draw a straight line through the two points to extend it across the graph. Now you have graphed the line based on the provided linear equation.
Solving for x in Intersecting Lines Adjacent Angles Problem
The image you provided shows a pair of intersecting straight lines on a coordinate plane, creating adjacent angles marked as "3x - 14" and "56°." To solve for "x," you need to use the fact that the sum of these adjacent angles forms a straight line, which is 180 degrees. Let's set up the equation: \( 3x - 14 + 56 = 180 \) Now, we'll combine like terms: \( 3x + 42 = 180 \) Next, subtract 42 from both sides of the equation: \( 3x = 138 \) Finally, divide both sides by 3 to solve for x: \( x = \frac{138}{3} \) \( x = 46 \) Therefore, x equals 46 degrees.
Calculating Distance Between Two Points on a Coordinate Plane
The question is asking you to calculate the distance between two points on a coordinate plane, which essentially is asking you to find the length of the hypotenuse of a right triangle formed with these two points as its endpoints. The points provided are (3, 7) and (8, 5). To determine the distance between these two points, you can use the distance formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] Where (x1, y1) and (x2, y2) are the coordinates of the two points. Substitute the coordinates into the formula: \[ d = \sqrt{(8 - 3)^2 + (5 - 7)^2} \] \[ d = \sqrt{(5)^2 + (-2)^2} \] \[ d = \sqrt{25 + 4} \] \[ d = \sqrt{29} \] So, the distance between the two points to the nearest tenth is approximately \(\sqrt{29}\), which is about 5.4 when rounded to the nearest tenth.
Calculating Lengths in a Right Triangle
The image shows a coordinate plane with two points plotted on it: (3, 7) and (8, 5). The question asks to graph a right triangle with those two points forming the hypotenuse and calculate the distance between the two points (the length of the hypotenuse), as well as the lengths of the other sides. To form a right triangle, we need to find a point that is perpendicular to the hypotenuse and intersects at one of the axes. Given the coordinates, we can easily see that the right angle will be at the point where x or y coordinate is the same as one of the given points. Let's create the right triangle by plotting a third point that has the same x-coordinate as one point and the same y-coordinate as the other. We can take the x-coordinate from (8, 5) and the y-coordinate from (3, 7) to get the new point, which will be (8, 7). Now, to calculate the distances: 1. The horizontal side (base) of the triangle is the difference between the x-coordinates of (3, 7) and (8, 7), which is 8 - 3 = 5 units. 2. The vertical side (height) of the triangle is the difference between the y-coordinates of (3, 7) and (8, 5), which is 7 - 5 = 2 units. To find the distance between the two original points (the length of the hypotenuse), we use the Pythagorean theorem: \( c = \sqrt{a^2 + b^2} \) where c is the hypotenuse, a is one leg, and b is the other leg of the triangle. Plugging in our values: \( c = \sqrt{5^2 + 2^2} = \sqrt{25 + 4} = \sqrt{29} \) \( \sqrt{29} \) is already in its simplest radical form, but if we want to find the decimal value, it is approximately 5.4. So the length of the hypotenuse is approximately 5.4 units (to the nearest tenth). The lengths of the other two sides are 5 units and 2 units, respectively.
Analyzing a Line on a Coordinate Plane
La imagen muestra un plano de coordenadas con una línea recta dibujada a través de él. Para resolver la pregunta que se haga referencia a esta línea recta, necesito que me digas específicamente qué necesitas resolver. ¿Estás buscando encontrar la ecuación de la línea recta, identificar puntos particulares a lo largo de la línea, o algún otro tipo de análisis? Por favor proporciona más detalles.
Completing Ordered Pairs for a Linear Equation
The given equation is: y = (4/5)x + 1 We need to complete the given ordered pairs for x values -5, 0, and 5. For x = -5: y = (4/5)(-5) + 1 y = -4 + 1 y = -3 So the ordered pair is (-5, -3). For x = 0: y = (4/5)(0) + 1 y = 0 + 1 y = 1 So the ordered pair is (0, 1). For x = 5: y = (4/5)(5) + 1 y = 4 + 1 y = 5 So the ordered pair is (5, 5). To graph the solution set, you can take any two of the completed ordered pairs and plot them on the coordinate plane. Then draw a straight line through those points, which represents the equation y = (4/5)x + 1.
Calculating Distance Between Two Points in a Coordinate Plane
To find the distance between two points, we can use the formula for the distance \( d \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) in the coordinate plane: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] The points given are \( (-3, -4) \) and \( (-10, -9) \). Let's plug these coordinates into the formula: \[ d = \sqrt{(-10 - (-3))^2 + (-9 - (-4))^2} \] \[ d = \sqrt{(-10 + 3)^2 + (-9 + 4)^2} \] \[ d = \sqrt{(-7)^2 + (-5)^2} \] \[ d = \sqrt{49 + 25} \] \[ d = \sqrt{74} \] Now, let's find the square root of 74. \[ d \approx 8.602 \] Rounded to the nearest tenth, the distance is approximately 8.6 units.
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