plot the graph -Maths 9th

plot the graph -Maths 9th
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F = ma This equation can be written in the form of y = mx, where m is the slope In this case m = m = 3 kg The graph will be a straight line passing through the origin. When a = 2, F = 6 N When a = 3, F = 9 N
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Related questions

Taking 0.5 cm as 1 unit, plot the following points on the graph paper. -Maths 9th

Description : Taking 0.5 cm as 1 unit, plot the following points on the graph paper. -Maths 9th

Last Answer : Here, in point 4(1, 3) both x and y-coordinates are positive, so it lies in I quadrant. In point 8(-3, -1),both x and y-coordinates are negative, so it lies in III quadrant. In point C(1, -4), x- ... is zero, so it lies on Y-axis and in point F(1,0) y-coordinate is zero, so it lies on X-axis.

plot the graph -Maths 9th

Description : plot the graph -Maths 9th

Last Answer : F = ma This equation can be written in the form of y = mx, where m is the slope In this case m = m = 3 kg The graph will be a straight line passing through the origin. When a = 2, F = 6 N When a = 3, F = 9 N

Taking 0.5 cm as 1 unit, plot the following points on the graph paper. -Maths 9th

Description : Taking 0.5 cm as 1 unit, plot the following points on the graph paper. -Maths 9th

Last Answer : Here, in point 4(1, 3) both x and y-coordinates are positive, so it lies in I quadrant. In point 8(-3, -1),both x and y-coordinates are negative, so it lies in III quadrant. In point C(1, -4), x- ... is zero, so it lies on Y-axis and in point F(1,0) y-coordinate is zero, so it lies on X-axis.

Plot the following points and write the name of the figure obtained by joining, them in order -Maths 9th

Description : Plot the following points and write the name of the figure obtained by joining, them in order -Maths 9th

Last Answer : Let X’ OX and Y’ OY be the coordinate axes and mark point on it. Here, point P(-3,2) lies in II quadrant, Q(-7,-3) lies in III quadrant, R(6, -3) lies in IV quadrant and S(2,2) lies in I quadrant. Plotting the points on the graph paper, the figure obtained is trapezium PQRS.

Plot the points (x, y) given by the following table. -Maths 9th

Description : Plot the points (x, y) given by the following table. -Maths 9th

Last Answer : On plotting the given points on the graph, we get the points P(2,4), Q(4,2) R (-3, 0), S (-2, 5), T (3, – 3)and O (0 0).

Plot the following points and check whether they are collinear or not -Maths 9th

Description : Plot the following points and check whether they are collinear or not -Maths 9th

Last Answer : (i) Plotting the points P (1, 3), Q (-1, -1) and R (-2, - 3) on the graph paper and join these points, we get a straight line. Hence, these points are collinear. (ii) Plotting the points ... 6 (5, 5)on the graph paper and join these points, we get a straight line. Hence, given points are collinear.

Plot the points (x, y) given by the following table. Use scale 1 cm= 0.25 unit. -Maths 9th

Description : Plot the points (x, y) given by the following table. Use scale 1 cm= 0.25 unit. -Maths 9th

Last Answer : Let X’OX and X’ OX be the coordinate axes. Plot the given points (1.25, -0.5), (0.25, 1), (1.5,1.5) and (-1.75, – 0.25) on the graph paper.

Plot the points P(1, 0), Q(4, 0) and 5(1, 3). Find the coordinates of the point R such that PQRS is a square. -Maths 9th

Description : Plot the points P(1, 0), Q(4, 0) and 5(1, 3). Find the coordinates of the point R such that PQRS is a square. -Maths 9th

Last Answer : see the below answer

Plot the points A (1, – 1) and B (4, 5). -Maths 9th

Description : Plot the points A (1, – 1) and B (4, 5). -Maths 9th

Last Answer : In point A(1, -1), x-coordinate is positive and y-coordinate is negative, so it lies in IV quadrant. In point B(4, 5), both coordinates are positive, so it lies in I quadrant. On plotting these point, we ... Y-axis at y = 7. Thus, we get the point Q(5,7) which lies outside the line segment AB.

Plot the following points and write the name of the figure obtained by joining, them in order -Maths 9th

Description : Plot the following points and write the name of the figure obtained by joining, them in order -Maths 9th

Last Answer : Let X’ OX and Y’ OY be the coordinate axes and mark point on it. Here, point P(-3,2) lies in II quadrant, Q(-7,-3) lies in III quadrant, R(6, -3) lies in IV quadrant and S(2,2) lies in I quadrant. Plotting the points on the graph paper, the figure obtained is trapezium PQRS.

Plot the points (x, y) given by the following table. -Maths 9th

Description : Plot the points (x, y) given by the following table. -Maths 9th

Last Answer : On plotting the given points on the graph, we get the points P(2,4), Q(4,2) R (-3, 0), S (-2, 5), T (3, – 3)and O (0 0).

Plot the following points and check whether they are collinear or not -Maths 9th

Description : Plot the following points and check whether they are collinear or not -Maths 9th

Last Answer : (i) Plotting the points P (1, 3), Q (-1, -1) and R (-2, - 3) on the graph paper and join these points, we get a straight line. Hence, these points are collinear. (ii) Plotting the points ... 6 (5, 5)on the graph paper and join these points, we get a straight line. Hence, given points are collinear.

Plot the points (x, y) given by the following table. Use scale 1 cm= 0.25 unit. -Maths 9th

Description : Plot the points (x, y) given by the following table. Use scale 1 cm= 0.25 unit. -Maths 9th

Last Answer : Let X’OX and X’ OX be the coordinate axes. Plot the given points (1.25, -0.5), (0.25, 1), (1.5,1.5) and (-1.75, – 0.25) on the graph paper.

Plot the points A (1, – 1) and B (4, 5). -Maths 9th

Description : Plot the points A (1, – 1) and B (4, 5). -Maths 9th

Last Answer : In point A(1, -1), x-coordinate is positive and y-coordinate is negative, so it lies in IV quadrant. In point B(4, 5), both coordinates are positive, so it lies in I quadrant. On plotting these point, we ... Y-axis at y = 7. Thus, we get the point Q(5,7) which lies outside the line segment AB.

Plot the points P(1, 0), Q(4, 0) and 5(1, 3). Find the coordinates of the point R such that PQRS is a square. -Maths 9th

Description : Plot the points P(1, 0), Q(4, 0) and 5(1, 3). Find the coordinates of the point R such that PQRS is a square. -Maths 9th

Last Answer : see the below answer

A rectangular plot is given for constructing a house having a measurement of 40 m long and 15 m in the front. -Maths 9th

Description : A rectangular plot is given for constructing a house having a measurement of 40 m long and 15 m in the front. -Maths 9th

Last Answer : Let ABCD is a rectangular plot having a measurement of 40 m long and 15 m front. ∴ Length of inner-rectangle, EF = 40 - 3 - 3 = 34 m and breadth of inner-rectangle, FG =15 - 2 - 2 = ... [∴ area of a rectangle = length x breadth] Hence, the largest area where the house can be constructed in 374 m2

For drawing a frequency polygon of a continuous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and abscissae are, respectively -Maths 9th

Description : For drawing a frequency polygon of a continuous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and abscissae are, respectively -Maths 9th

Last Answer : NEED ANSWER

A rectangular plot is given for constructing a house having a measurement of 40 m long and 15 m in the front. -Maths 9th

Description : A rectangular plot is given for constructing a house having a measurement of 40 m long and 15 m in the front. -Maths 9th

Last Answer : Let ABCD is a rectangular plot having a measurement of 40 m long and 15 m front. ∴ Length of inner-rectangle, EF = 40 - 3 - 3 = 34 m and breadth of inner-rectangle, FG =15 - 2 - 2 = ... [∴ area of a rectangle = length x breadth] Hence, the largest area where the house can be constructed in 374 m2.

For drawing a frequency polygon of a continuous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and abscissae are, respectively -Maths 9th

Description : For drawing a frequency polygon of a continuous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and abscissae are, respectively -Maths 9th

Last Answer : (c) Class marks i.e., the mid-point of the classes are abscissa of the points, which we plot for frequency polygon.

Plot the points A (5, 5) and B (–5, 5) in cartesian plane. Join AB, OA and OB. Name the type of triangle so obtained. -Maths 9th

Description : Plot the points A (5, 5) and B (–5, 5) in cartesian plane. Join AB, OA and OB. Name the type of triangle so obtained. -Maths 9th

Last Answer : Solution :- The obtained triangle is an isosceles triangle.

Plot the following points and check whether they are collinear or not: -Maths 9th

Description : Plot the following points and check whether they are collinear or not: -Maths 9th

Last Answer : Solution :-

Plot the point A(2,0), B(5,0) and C(5,3). Find the coordinates of the point D such that ABCD is a square. -Maths 9th

Description : Plot the point A(2,0), B(5,0) and C(5,3). Find the coordinates of the point D such that ABCD is a square. -Maths 9th

Last Answer : Solution :-

Plot the points A(1,-3) and B(5,4). -Maths 9th

Description : Plot the points A(1,-3) and B(5,4). -Maths 9th

Last Answer : Solution :-

Plot the points a(5,5) and b(-5,5) in the cartesian plane .join OA AB and OB name the figure obtained and find its area -Maths 9th

Description : Plot the points a(5,5) and b(-5,5) in the cartesian plane .join OA AB and OB name the figure obtained and find its area -Maths 9th

Last Answer : This answer was deleted by our moderators...

Plot the points A(3, 2), B(-2, 2), C(-2, -2) and D(3, -2) in the cartesian plane. Join these points and name the figure so formed. -Maths 9th

Description : Plot the points A(3, 2), B(-2, 2), C(-2, -2) and D(3, -2) in the cartesian plane. Join these points and name the figure so formed. -Maths 9th

Last Answer : answer:

graph of 2x+9=0 in one variable -Maths 9th

Description : graph of 2x+9=0 in one variable -Maths 9th

Last Answer : 2x+9=0 2x=-9 x=-9/2 x=-4.5

The point (2,3) lies on the graph of the linear equation 3x - (a -1)y =2a -1. If the same point also lies on the graph of the linear equation 5x + (1-2a)y = 3b, then find the value of b. -Maths 9th

Description : The point (2,3) lies on the graph of the linear equation 3x - (a -1)y =2a -1. If the same point also lies on the graph of the linear equation 5x + (1-2a)y = 3b, then find the value of b. -Maths 9th

Last Answer : Given, point (2,3) lies on the line. So, the point (2, 3) is the solution of 3x - (a -1) y = 2a - 1 On putting x = 2 and y = 3 in given solution. ∴ 3 2 - (a-1) 3 = 2a - 1 ⇒ 6 - 3a + 3 = 2a - 1 ⇒ - 3a ... 2 2) 3 = 3b ⇒ 10 - 9 = 3b ⇒ 1 = 3b ⇒ 1 / 3 = b Hence, the value of b is 1 / 3.

A student wrote the equations of the lines a and b drawn in the following graph as y =1 and 2x + 3y =6. Is he right? -Maths 9th

Description : A student wrote the equations of the lines a and b drawn in the following graph as y =1 and 2x + 3y =6. Is he right? -Maths 9th

Last Answer : Clearly, line a is parallel to X-axis at a distance of 1 unit in positive direction of Y-axis, therefore its equation is y = 1. Also, if we draw the graph of line 2x + 3y = 6, then its graph should intersect X - axis at (3,0 ... Base Height = 1/2 BC AC = 1/2 1 3 / 2 = 3 / 4 sq unit.

If P(-l, 1), Q(3, -4), R(1, -1), S(-2, -3) and T(-4, 4) are plotted on the graph paper, then the point(s) in the fourth quadrant is/are -Maths 9th

Description : If P(-l, 1), Q(3, -4), R(1, -1), S(-2, -3) and T(-4, 4) are plotted on the graph paper, then the point(s) in the fourth quadrant is/are -Maths 9th

Last Answer : (b) In point P (-1, 1), x-coordinate is -1 unit and y-coordinate is 1 unit, so it lies in llnd quadrant. Similarly, we can plot all the points Q (3, -4), R (1, -1), S (-2, -3) and T (-4, 4), It is clear from the graph that points R and Q lie in fourth quadrant.

If P (5,1), Q (8, 0), R(0, 4), S(0, 5) and O(0, 0) are plotted on the graph paper, then the points on the X-axis is/are -Maths 9th

Description : If P (5,1), Q (8, 0), R(0, 4), S(0, 5) and O(0, 0) are plotted on the graph paper, then the points on the X-axis is/are -Maths 9th

Last Answer : (d) We know that, a point lies on X-axis, if its y-coordinate is zero. So, on plotting the given points on graph paper, we get Q and O lie on the X-axis.

The graph of the linear equation 2x + 3y = 6 cuts the Y-axis at the point. -Maths 9th

Description : The graph of the linear equation 2x + 3y = 6 cuts the Y-axis at the point. -Maths 9th

Last Answer : (d) Since, the graph of linear equation 2x + 3y = 6 cuts the Y-axis. So, we put x = 0 in the given equation 2x+ 3y = 6, we get 2 x 0+ 3y = 6 ⇒ 3y = 6 y = 2. Hence, at the point (0, 2), the given linear equation cuts the Y-axis.

The graph of y = 6 is a Line -Maths 9th

Description : The graph of y = 6 is a Line -Maths 9th

Last Answer : (a) Given equation of line can be written as, a . x + 1 . y = 6 To draw the graph of above equation, we need at least two solutions. When x = 0, then y = 6 When x =2, then y = 6. Here, we find two points A (0, 6) and B (2, 6). So, draw the graph, by plotting the points and joining the line AB.

The graph of the linear equation 2x+ 3y = 6 is a line which meets the X-axis at the point. -Maths 9th

Description : The graph of the linear equation 2x+ 3y = 6 is a line which meets the X-axis at the point. -Maths 9th

Last Answer : (c) Since, the graph of linear equation 2x + 3y = 6 meets the X-axis. So, we put y = 0 in 2x + 3y = 6 ⇒ 2x + 3(0) = 6 = 2x + 0 = 6 ⇒ x = 6/2 ⇒ x = 3 Hence, the coordinate on X-axis is (3, 0).

The graph of the linear equation y = x passes through the point. -Maths 9th

Description : The graph of the linear equation y = x passes through the point. -Maths 9th

Last Answer : (c) The linear equation y = x has same value of x and y-coordinates are same. Therefore, the point (1,1) must lie on the line y = x.

Draw the graph of the equation represented by a straight Line which is parallel to the X-axis and at a distance 3 units below it. -Maths 9th

Description : Draw the graph of the equation represented by a straight Line which is parallel to the X-axis and at a distance 3 units below it. -Maths 9th

Last Answer : Any straight line parallel to X-axis in negative direction of Y-axis is given by y = - k, where k is the distance of the line from the X-axis. Here, k = 3. Therefore, the equation of the line is y = -3. To ... , plot the points (1,-3), (2, -3) and (3, -3) and join them. This is the required graph.

Draw the graph of the linear equation whose solutions are represented by the points having the sum of the coordinates as 10 units. -Maths 9th

Description : Draw the graph of the linear equation whose solutions are represented by the points having the sum of the coordinates as 10 units. -Maths 9th

Last Answer : As per question, the sum of the coordinates is 10 units. Let x and y be two coordinates, then we get x + y = 10. For x = 5, y = 5, therefore, (5, 5) lies on the graph of x + y = 10. For x = ... and (3, 7) on the graph paper and joining them by a line, we get graph of the linear equation x + y = 10.

Write the linear equation such that each point on its graph has an ordinate 3 times its abscissa. -Maths 9th

Description : Write the linear equation such that each point on its graph has an ordinate 3 times its abscissa. -Maths 9th

Last Answer : Let the abscissa of the point be x, According to the question, Ordinate (y) = 3 x Abscissa ⇒ y=3x When x = 1, then y = 3 x 1 = 3 and when x = 2, then y = 3 x 2 = 6. Here, ... the line AB. Hence, y = 3x is the required equation such that each point on its graph has an ordinate 3 times its abscissa.

If the point (3, 4) lies on the graph of 3y = ax + 7, then find the value of a. -Maths 9th

Description : If the point (3, 4) lies on the graph of 3y = ax + 7, then find the value of a. -Maths 9th

Last Answer : Since, the point (x = 3, y = 4) lies on the equation 3y = ax + 7, then the equation will be , satisfied by the point. Now, put x = 3 and y = 4 in given equation, we get 3(4) = a (3)+7 ⇒ 12 = 3a+7 ⇒ 3a = 12 – 7 ⇒ 3a = 5 Hence, the value of a is 5/3.

graph of 2x+9=0 in one variable -Maths 9th

Description : graph of 2x+9=0 in one variable -Maths 9th

Last Answer : 2x+9=0 2x=-9 x=-9/2 x=-4.5

The point (2,3) lies on the graph of the linear equation 3x - (a -1)y =2a -1. If the same point also lies on the graph of the linear equation 5x + (1-2a)y = 3b, then find the value of b. -Maths 9th

Description : The point (2,3) lies on the graph of the linear equation 3x - (a -1)y =2a -1. If the same point also lies on the graph of the linear equation 5x + (1-2a)y = 3b, then find the value of b. -Maths 9th

Last Answer : Given, point (2,3) lies on the line. So, the point (2, 3) is the solution of 3x - (a -1) y = 2a - 1 On putting x = 2 and y = 3 in given solution. ∴ 3 2 - (a-1) 3 = 2a - 1 ⇒ 6 - 3a + 3 = 2a - 1 ⇒ - 3a ... 2 2) 3 = 3b ⇒ 10 - 9 = 3b ⇒ 1 = 3b ⇒ 1 / 3 = b Hence, the value of b is 1 / 3.

A student wrote the equations of the lines a and b drawn in the following graph as y =1 and 2x + 3y =6. Is he right? -Maths 9th

Description : A student wrote the equations of the lines a and b drawn in the following graph as y =1 and 2x + 3y =6. Is he right? -Maths 9th

Last Answer : Clearly, line a is parallel to X-axis at a distance of 1 unit in positive direction of Y-axis, therefore its equation is y = 1. Also, if we draw the graph of line 2x + 3y = 6, then its graph should intersect X - axis at (3,0 ... Base Height = 1/2 BC AC = 1/2 1 3 / 2 = 3 / 4 sq unit.

If P(-l, 1), Q(3, -4), R(1, -1), S(-2, -3) and T(-4, 4) are plotted on the graph paper, then the point(s) in the fourth quadrant is/are -Maths 9th

Description : If P(-l, 1), Q(3, -4), R(1, -1), S(-2, -3) and T(-4, 4) are plotted on the graph paper, then the point(s) in the fourth quadrant is/are -Maths 9th

Last Answer : (b) In point P (-1, 1), x-coordinate is -1 unit and y-coordinate is 1 unit, so it lies in llnd quadrant. Similarly, we can plot all the points Q (3, -4), R (1, -1), S (-2, -3) and T (-4, 4), It is clear from the graph that points R and Q lie in fourth quadrant.

If P (5,1), Q (8, 0), R(0, 4), S(0, 5) and O(0, 0) are plotted on the graph paper, then the points on the X-axis is/are -Maths 9th

Description : If P (5,1), Q (8, 0), R(0, 4), S(0, 5) and O(0, 0) are plotted on the graph paper, then the points on the X-axis is/are -Maths 9th

Last Answer : (d) We know that, a point lies on X-axis, if its y-coordinate is zero. So, on plotting the given points on graph paper, we get Q and O lie on the X-axis.

The graph of the linear equation 2x + 3y = 6 cuts the Y-axis at the point. -Maths 9th

Description : The graph of the linear equation 2x + 3y = 6 cuts the Y-axis at the point. -Maths 9th

Last Answer : (d) Since, the graph of linear equation 2x + 3y = 6 cuts the Y-axis. So, we put x = 0 in the given equation 2x+ 3y = 6, we get 2 x 0+ 3y = 6 ⇒ 3y = 6 y = 2. Hence, at the point (0, 2), the given linear equation cuts the Y-axis.

The graph of y = 6 is a Line -Maths 9th

Description : The graph of y = 6 is a Line -Maths 9th

Last Answer : (a) Given equation of line can be written as, a . x + 1 . y = 6 To draw the graph of above equation, we need at least two solutions. When x = 0, then y = 6 When x =2, then y = 6. Here, we find two points A (0, 6) and B (2, 6). So, draw the graph, by plotting the points and joining the line AB.

The graph of the linear equation 2x+ 3y = 6 is a line which meets the X-axis at the point. -Maths 9th

Description : The graph of the linear equation 2x+ 3y = 6 is a line which meets the X-axis at the point. -Maths 9th

Last Answer : (c) Since, the graph of linear equation 2x + 3y = 6 meets the X-axis. So, we put y = 0 in 2x + 3y = 6 ⇒ 2x + 3(0) = 6 = 2x + 0 = 6 ⇒ x = 6/2 ⇒ x = 3 Hence, the coordinate on X-axis is (3, 0).

The graph of the linear equation y = x passes through the point. -Maths 9th

Description : The graph of the linear equation y = x passes through the point. -Maths 9th

Last Answer : (c) The linear equation y = x has same value of x and y-coordinates are same. Therefore, the point (1,1) must lie on the line y = x.

Draw the graph of the equation represented by a straight Line which is parallel to the X-axis and at a distance 3 units below it. -Maths 9th

Description : Draw the graph of the equation represented by a straight Line which is parallel to the X-axis and at a distance 3 units below it. -Maths 9th

Last Answer : Any straight line parallel to X-axis in negative direction of Y-axis is given by y = - k, where k is the distance of the line from the X-axis. Here, k = 3. Therefore, the equation of the line is y = -3. To ... , plot the points (1,-3), (2, -3) and (3, -3) and join them. This is the required graph.

Draw the graph of the linear equation whose solutions are represented by the points having the sum of the coordinates as 10 units. -Maths 9th

Description : Draw the graph of the linear equation whose solutions are represented by the points having the sum of the coordinates as 10 units. -Maths 9th

Last Answer : As per question, the sum of the coordinates is 10 units. Let x and y be two coordinates, then we get x + y = 10. For x = 5, y = 5, therefore, (5, 5) lies on the graph of x + y = 10. For x = ... and (3, 7) on the graph paper and joining them by a line, we get graph of the linear equation x + y = 10.

Write the linear equation such that each point on its graph has an ordinate 3 times its abscissa. -Maths 9th

Description : Write the linear equation such that each point on its graph has an ordinate 3 times its abscissa. -Maths 9th

Last Answer : Let the abscissa of the point be x, According to the question, Ordinate (y) = 3 x Abscissa ⇒ y=3x When x = 1, then y = 3 x 1 = 3 and when x = 2, then y = 3 x 2 = 6. Here, ... the line AB. Hence, y = 3x is the required equation such that each point on its graph has an ordinate 3 times its abscissa.