What is the greatest number of points of intersection that can occur when 2 different circles and 2 different straight lines are drawn on the same piece of paper?

What is the greatest number of points of intersection that can occur when 2 different circles and 2 different straight lines are drawn on the same piece of paper?
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If circles are drawn taking two sides of a triangle as diameter, prove that the point of intersection of these circles lie on the third side. -Maths 9th

Description : If circles are drawn taking two sides of a triangle as diameter, prove that the point of intersection of these circles lie on the third side. -Maths 9th

Last Answer : Solution :- Given: Two circles are drawn on sides AB and AC of a △ABC as diameters. The circles intersects at D. To prove: D lies on BC Construction: Join A and D Proof: ∠ADB = 90° (Angle in the semi-circle ... + 90° => ∠ADB + ∠ADC = 180° => BDC is a straight line. Hence, D lies On third side BC.

India's latitudinal and longitudinal extent measured in degrees are almost the same, but its north-south extent measured in km is greater than its east-west extent. This is due to the fact that (a) longitudes are not parallel lines (b) the distance between latitudes remains the same but the distance between longitudes is greatest at the Equator and nil at the poles where all longitudes join (c) all longitudes with their opposites form great circles (d) the earth is not a perfect sphere

Description : India's latitudinal and longitudinal extent measured in degrees are almost the same, but its north-south extent measured in km is greater than its east-west extent. This is due to the fact that (a) ... (c) all longitudes with their opposites form great circles (d) the earth is not a perfect sphere

Last Answer : Ans: (b)

30. Radical centre of circles drawn on the sides as a diameter of triangle formed by the lines the `3x-4y+6=0`, `x-y+2=0` and `4x+3y-17=0` is

Description : 30. Radical centre of circles drawn on the sides as a diameter of triangle formed by the lines the `3x-4y+6=0`, `x-y+2=0` and `4x+3y-17=0` is

Last Answer : 30. Radical centre of circles drawn on the sides as a diameter of triangle formed by the lines the `3x-4y+6=0`, `x-y+ ... . (-2,3) C. (-2,-3) D. (2,3)

Two circles with centre O and O' intersect at two points A and B. A line PQ is drawn parallel to OO' through B intersecting the circles at P and Q. Prove that PQ = 2 OO'. -Maths 9th

Description : Two circles with centre O and O' intersect at two points A and B. A line PQ is drawn parallel to OO' through B intersecting the circles at P and Q. Prove that PQ = 2 OO'. -Maths 9th

Last Answer : Solution :- Construction: Draw two circles having centres O and O' intersecting at points A and B. Draw a parallel line PQ to OO' ... iii) Again, OO' = MN [As OO' NM is a rectangle] ...(iv) ⇒ 2OO' = PQ Hence proved.

Prove that the tangent lines of any three circles a,b,and c will converge along a straight line.

Description : Prove that the tangent lines of any three circles a,b,and c will converge along a straight line.

Last Answer : The vanishing points of any three spheres will converge at the horizon.

The magnetic field produced by a long straight wire carrying a current points: w) in the direction of the current x) toward the wire y) away from the wire z) circles around the wire

Description : The magnetic field produced by a long straight wire carrying a current points: w) in the direction of the current x) toward the wire y) away from the wire z) circles around the wire

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While working on a plane table, the correct rule is: (A) Draw continuous lines from all instrument stations (B) Draw short rays sufficient to contain the points sought (C) Intersection should be obtained by actually drawing second rays (D) Take maximum number of sights as possible from each station to distant objects

Description : While working on a plane table, the correct rule is: (A) Draw continuous lines from all instrument stations (B) Draw short rays sufficient to contain the points sought (C) Intersection ... drawing second rays (D) Take maximum number of sights as possible from each station to distant objects

Last Answer : (B) Draw short rays sufficient to contain the points sought

what- A triangle is formed by the intersection of the lines y = 0, y = -3x + 3, and y = 3x + 3.Is the triangle equilateral, isosceles, or scalene Graph the lines on grid paper to find the vertices of the triangle?

Description : what- A triangle is formed by the intersection of the lines y = 0, y = -3x + 3, and y = 3x + 3.Is the triangle equilateral, isosceles, or scalene Graph the lines on grid paper to find the vertices of the triangle?

Last Answer : isosceles

what- A triangle is formed by the intersection of the lines y = 2x + 4, y = -x – 2, and x = 1.Is the triangle equilateral, isosceles, or scalene Graph the lines on grid paper to find the vertices of the triangle?

Description : what- A triangle is formed by the intersection of the lines y = 2x + 4, y = -x – 2, and x = 1.Is the triangle equilateral, isosceles, or scalene Graph the lines on grid paper to find the vertices of the triangle?

Last Answer : scalene

What is antenna bandwidth? A. Antenna length divided by the number of elements B. The angle between the half-power radiation points C. The angle formed between two imaginary lines drawn through D. The frequency range over which an antenna can be expected to operate satisfactorily

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Last Answer : D. The frequency range over which an antenna can be expected to operate satisfactorily

In a purely cohesive soil, the critical centre lies at the intersection of (A) Perpendicular bisector of slope and the locus of the centre (B) Perpendicular drawn at 1/3rd slope from toe and the locus of the centre (C) Perpendicular drawn at 2/3rd slope from toe and the locus of the centre (D) Directional angles

Description : In a purely cohesive soil, the critical centre lies at the intersection of (A) Perpendicular bisector of slope and the locus of the centre (B) Perpendicular drawn at 1/3rd slope from toe and the ... C) Perpendicular drawn at 2/3rd slope from toe and the locus of the centre (D) Directional angles

Last Answer : Answer: Option D

The greatest induced EMF will occur in a straight wire moving at constant speed through a uniform magnetic field when the angle between the direction of the wire's motion and the direction of the magnetic field is w) 0 degrees x) 30 degrees y) 60 degrees z) 90 degrees

Description : The greatest induced EMF will occur in a straight wire moving at constant speed through a uniform magnetic field when the angle between the direction of the wire's motion and the direction of the magnetic field is w) 0 degrees x) 30 degrees y) 60 degrees z) 90 degrees

Last Answer : ANSWER: Z -- 90 DEGREES

Two congruent circles intersect each other at point A and B.Through A any line segment PAQ is drawn so that P,Q lie on the two circles.Prove that BP = BQ. -Maths 9th

Description : Two congruent circles intersect each other at point A and B.Through A any line segment PAQ is drawn so that P,Q lie on the two circles.Prove that BP = BQ. -Maths 9th

Last Answer : Solution :- Let, O and O' be the centres of two congruent circles. As, AB is the common chord of these circles. ∴ ACB = ADB As congruent arcs subtent equal angles at the centre. ∠AOB = ∠AO'B ⇒ 1/2∠AOB = 1/2∠AO'B ⇒ ∠BPA = ∠BQA ⇒ BP = BQ (Sides opposite to equal angles)

How many convex circles can be drawn in a triangle ?

Description : How many convex circles can be drawn in a triangle ?

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1. Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points? -Maths 9th

Description : 1. Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points? -Maths 9th

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If all the lines in the image below are of the same length but of different thickness and are drawn in perspective with respect to their distance from the viewer, which of the following statements will be TRUE?

Description : If all the lines in the image below are of the same length but of different thickness and are drawn in perspective with respect to their distance from the viewer, which of the following statements will be TRUE ... and line 3 have same thickness D. Line 3 is the closest line of all to the viewer 

Last Answer : A D

In figure X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. -Maths 9th

Description : In figure X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. -Maths 9th

Last Answer : Given X and Y are the mid-points of AC and AB respectively. Also, QP|| BC and CYQ, BXP are straight lines. To prove ar (ΔABP) = ar (ΔACQ) Proof Since, X and Y are the mid-points of AC and AB respectively. So, ... ar (ΔBYX) + ar (XYAP) = ar (ΔCXY) + ar (YXAQ) ⇒ ar (ΔABP) = ar (ΔACQ) Hence proved.

In figure X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. -Maths 9th

Description : In figure X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. -Maths 9th

Last Answer : Given X and Y are the mid-points of AC and AB respectively. Also, QP|| BC and CYQ, BXP are straight lines. To prove ar (ΔABP) = ar (ΔACQ) Proof Since, X and Y are the mid-points of AC and AB respectively. So, ... ar (ΔBYX) + ar (XYAP) = ar (ΔCXY) + ar (YXAQ) ⇒ ar (ΔABP) = ar (ΔACQ) Hence proved.

Straight lines shown in a linear programming graph indicates ........................... a. Objective function b. Constraints c. Points d. All of the above

Description : Straight lines shown in a linear programming graph indicates ........................... a. Objective function b. Constraints c. Points d. All of the above

Last Answer : b. Constraints

53. While plotting constraints on a graph paper, terminal points on both the axes are connected by straight line because a. The resources are limited in supply b. The objective function as a linear function c. The constraints are linear equations or inequalities d. All of the above

Description : 53. While plotting constraints on a graph paper, terminal points on both the axes are connected by straight line because a. The resources are limited in supply b. The objective function as a linear function c. The constraints are linear equations or inequalities d. All of the above

Last Answer : c. The constraints are linear equations or inequalities

I go around in circles But always straight ahead, Never complain No matter where I am led. -Riddles

Description : I go around in circles But always straight ahead, Never complain No matter where I am led. -Riddles

Last Answer : wagon wheel

I go around in circles, But always straight ahead Never complain, No matter where I am led. What am I? -Riddles

Description : I go around in circles, But always straight ahead Never complain, No matter where I am led. What am I? -Riddles

Last Answer : A wheel.

In the diagram three circles in a straight line must add to 100 write in the missing numbers?

Description : In the diagram three circles in a straight line must add to 100 write in the missing numbers?

Last Answer : does anyone know this mathswatch answer please help me

Let S be the set of circles `x^(2)+y^(2)=r^(2)" for "r=1,` 2 and 3 and T be the set of lines `y= x+ksqrt(2)" for "k=0, ne 1 and pm 2`. Find the number

Description : Let S be the set of circles `x^(2)+y^(2)=r^(2)" for "r=1,` 2 and 3 and T be the set of lines `y= x+ksqrt(2)" for "k=0, ne 1 and pm 2`. Find the number

Last Answer : Let S be the set of circles `x^(2)+y^(2)=r^(2)" for "r=1,` 2 and 3 and T be ... of distinct points of intersection of the graphs of set S and T.

AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (in the given figure). Show that AP || BQ. -Maths 9th

Description : AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (in the given figure). Show that AP || BQ. -Maths 9th

Last Answer : Given In the figure l || m, AP and BQ are the bisectors of ∠EAB and ∠ABH, respectively. To prove AP|| BQ Proof Since, l || m and t is transversal. Therefore, ∠EAB = ∠ABH [alternate interior ... ∠PAB and ∠ABQ are alternate interior angles with two lines AP and BQ and transversal AB. Hence, AP || BQ.

AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (in the given figure). Show that AP || BQ. -Maths 9th

Description : AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (in the given figure). Show that AP || BQ. -Maths 9th

Last Answer : Given In the figure l || m, AP and BQ are the bisectors of ∠EAB and ∠ABH, respectively. To prove AP|| BQ Proof Since, l || m and t is transversal. Therefore, ∠EAB = ∠ABH [alternate interior ... ∠PAB and ∠ABQ are alternate interior angles with two lines AP and BQ and transversal AB. Hence, AP || BQ.

What is the equation of the line joining the origin with the point of intersection of the lines 4x + 3y = 12 and 3x + 4y = 12 ? -Maths 9th

Description : What is the equation of the line joining the origin with the point of intersection of the lines 4x + 3y = 12 and 3x + 4y = 12 ? -Maths 9th

Last Answer : (b) (5, 6)Let the foot of the perpendicular be M(x1, y1) Slope of line AB, i.e., y = -x + 11 = -1 Slope of line PM = \(rac{y_1-3}{x_1-2}\)Now, PM ⊥ AB⇒ \(\bigg(rac{y_1-3}{x_1-2}\bigg)\) x - ... get 2x1 = 10 ⇒ x1 = 5 Putting x1 in (ii), we get y1 = 6. ∴ Required foot of the perpendicular M is (5, 6).

A line passes through the point of intersection of the lines 100x + 50y – 1 = 0 and 75x + 25y + 3 = 0 and makes equal intercepts on the axes. -Maths 9th

Description : A line passes through the point of intersection of the lines 100x + 50y – 1 = 0 and 75x + 25y + 3 = 0 and makes equal intercepts on the axes. -Maths 9th

Last Answer : (d) x + 2y = 2Let the required equation make intercept on x-axis = 2a ⇒ intercept made on y-axis = a ∴ Eqn of the given line in the intercept from:\(rac{x}{2a}+rac{y}{a}=1\) ...(i)Since the line ... 1 ⇒ a = 1.∴ Required equation of line : \(rac{x}{2 imes1}+rac{y}{1}=1\) ⇒ x + 2y = 2.

Find the equation of the line which passes through the point of intersection of the lines 2x – y + 5 = 0 -Maths 9th

Description : Find the equation of the line which passes through the point of intersection of the lines 2x – y + 5 = 0 -Maths 9th

Last Answer : (a) 45º 3x + y - 7 = 0 ⇒ y = -3x + 7 ⇒ Slope (m1) = -3 x + 2y + 9 = 0 ⇒ y = \(rac{-x}{2}\) - \(rac{9}{2}\) ⇒ Slope (m2) = \(-rac{1}{2}\)If θ is the angle between the given lines, then tan θ = \(\ ... \bigg|rac{-rac{5}{2}}{1+rac{3}{2}}\bigg|\)= \(\bigg|rac{-rac{5}{2}}{rac{5}{2}}\bigg|\) = 1∴ θ = 45°.

Center of gravity of an irregular body lies on the A. edgeB. center of body C. point of intersection of lines D. along the axis of rotation

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An intersection is a point where two or more lines/curves meet or cross. How many intersections are there in the figure given below?

Description : An intersection is a point where two or more lines/curves meet or cross. How many intersections are there in the figure given below? 

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STL file format is represented by interaction of ______ a.lines and hexagons b.lines and rectangles c.lines and triangles d.lines and circles

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P and O are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. -Maths 9th

Description : P and O are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. -Maths 9th

Last Answer : According to question PQ passes through the point of intersection O of its diagonals AC and BD.

P and O are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. -Maths 9th

Description : P and O are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. -Maths 9th

Last Answer : According to question PQ passes through the point of intersection O of its diagonals AC and BD.

Draw the graph of the equation 3x + 4y = 12 and find the co-ordinates of the points of intersection of the equation with the co-ordinate axes. -Maths 9th

Description : Draw the graph of the equation 3x + 4y = 12 and find the co-ordinates of the points of intersection of the equation with the co-ordinate axes. -Maths 9th

Last Answer : Solution :-

The locus of points of intersection of the tangents to `x^(2)+y^(2)=a^(2)` at the extremeties of a chord of circle `x^(2)+y^(2)=a^(2)` which touches t

Description : The locus of points of intersection of the tangents to `x^(2)+y^(2)=a^(2)` at the extremeties of a chord of circle `x^(2)+y^(2)=a^(2)` which touches t

Last Answer : The locus of points of intersection of the tangents to `x^(2)+y^(2)=a^(2)` at the extremeties of a chord of circle ` ... ` C. `(a,0)` D. `((a)/(2),0)`

What are the points of intersection of the line 2x plus 5y equals 4 with the curve y squared equals x plus 4?

Description : What are the points of intersection of the line 2x plus 5y equals 4 with the curve y squared equals x plus 4?

Last Answer : If: 2x +5y = 4 then 25y^2 = 4x^2 -16x +16If: y^2 = x +4 then 25y^2 = 25x +100So: 4x^2 -16x +16 = 25x +100Transposing terms: 4x^2 -41x -84 = 0Factorizing the above: (4x+7)(x-12) = 0 meaning x = -7/4 or x =12By substitution into original equation points of intersection:(-7/4, 3/2) and (12, -4)

What are the points of intersection between the line 3x -y equals 5 and the curve 2x squared plus y squared equals 129?

Description : What are the points of intersection between the line 3x -y equals 5 and the curve 2x squared plus y squared equals 129?

Last Answer : If: 3x-y = 5 then y^2 = (3x_5)^2 => 9x^2 -30x+25If: 2x^2 + y^2 = 129 then y^2 = 129-2x^2So: 9x^2 -30x+25 = 129-2x^2Transposing terms: 11x^2 -30x -104 = 0Factorizing the above: (11x- ... x = 52/11 or x= -2By substituting x into the original equation intersections areat: (52/11, 101/11) and (-2, -11)

What are the points of intersection between the equations of 3x -5y equals 16 and xy equals 7?

Description : What are the points of intersection between the equations of 3x -5y equals 16 and xy equals 7?

Last Answer : If 3x -5y = 16 and xy = 7 then by combining both equations into a single quadratic equation and solving it then the points of intersection are at (-5/3, -21/5) and (7, 1)

The distances AC and BC are measured from two fixed points A and B whose distance AB is known. The point C is plotted by intersection. This method is generally adopted in (A) Chain surveying (B) Traverse method of surveys (C) Triangulation (D) None of these

Description : The distances AC and BC are measured from two fixed points A and B whose distance AB is known. The point C is plotted by intersection. This method is generally adopted in (A) Chain surveying (B) Traverse method of surveys (C) Triangulation (D) None of these

Last Answer : (A) Chain surveying

Locating the position of a plane table station with reference to three known points, is known as (A) Intersection method (B) Radiation method (C) Resection method (D) Three point problem

Description : Locating the position of a plane table station with reference to three known points, is known as (A) Intersection method (B) Radiation method (C) Resection method (D) Three point problem

Last Answer : (D) Three point problem

To orient a plane table at a point with two inaccessible points, the method generally adopted, is (A) Intersection (B) Resection (C) Radiation (D) Two point problem

Description : To orient a plane table at a point with two inaccessible points, the method generally adopted, is (A) Intersection (B) Resection (C) Radiation (D) Two point problem

Last Answer : (D) Two point problem

In a plane-table survey, the process of determining the plotted position of a station occupied by the plane-table by means of sights taken towards known points, the locations of which have already been plotted, is known as (a) Radiation (b) Resection (c) Intersection (d) Traversing

Description : In a plane-table survey, the process of determining the plotted position of a station occupied by the plane-table by means of sights taken towards known points, the locations of which have already been plotted, is known as (a) Radiation (b) Resection (c) Intersection (d) Traversing

Last Answer : (b) Resection

The plotting of inaccessible points in a plane-table survey can be done by the method of (a) Interpolation (b) Radiation (c) Intersection (d) Traversing

Description : The plotting of inaccessible points in a plane-table survey can be done by the method of (a) Interpolation (b) Radiation (c) Intersection (d) Traversing

Last Answer : (c) Intersection

I messed up both my hands at a young age from repetitive injuries can I fix my hands over time from drawing circles on paper?

Description : I messed up both my hands at a young age from repetitive injuries can I fix my hands over time from drawing circles on paper?

Last Answer : RedDeerGuy1 Edit I will practice signing my name over and over in pencil and pen.

Two circles with centres O and O’ intersect at two points A and B. -Maths 9th

Description : Two circles with centres O and O’ intersect at two points A and B. -Maths 9th

Last Answer : Given, draw two circles having centres O and O’ intersect at points A and 8.

Two circles with centres O and O’ intersect at two points A and B. -Maths 9th

Description : Two circles with centres O and O’ intersect at two points A and B. -Maths 9th

Last Answer : Given, draw two circles having centres O and O’ intersect at points A and 8.

In Fig. 10.20, two circles intersects at two points A and B.AD and AC are diameters to the circles. Prove that B lies on the line A segment DC. -Maths 9th

Description : In Fig. 10.20, two circles intersects at two points A and B.AD and AC are diameters to the circles. Prove that B lies on the line A segment DC. -Maths 9th

Last Answer : Solution :- Jion AB ∠ABD = 90° (Angle in a semicircle) Similarly, ∠ABC = 90° So, ∠ABD + ∠ABC = 90° + 90° = 180° Therefore,DBC is a line i.e.,B lies on the segment DC.