Simplify: (i) (a + b + c)2 + (a – b + c)2 (ii) (a + b + c)2 – (a – b + c)2 (iii) (a + b + c)2 + (a – b + c)2 + (a + b – c)2 (iv) (2x + p – c)2 – (2x – p + c)2 (v) (x2 + y2 – z2)2 – (x2 – y2 + z2)2 -Maths 9th

Simplify: (i) (a + b + c)2 + (a – b + c)2 (ii) (a + b + c)2 – (a – b + c)2 (iii) (a + b + c)2 + (a – b + c)2 + (a + b – c)2 (iv) (2x + p – c)2 – (2x – p + c)2 (v) (x2 + y2 – z2)2 – (x2 – y2 + z2)2 -Maths 9th
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Description : Find the value of x3 + y3 + z3 – 3xyz if x2 + y2 + z2 = 83 and x + y + z = 15 -Maths 9th

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Last Answer : this is the correct answer!

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Last Answer : NEED ANSWER

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Description : What is the point of contact when the line y equals 2x meets the circle x2 plus y2 -8x -y plus 5 equals 0?

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Description : What are the solutions to the simultaneous equations of y equals -2x and x2 plus y2 equals 80?

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Description : Find (2x – y + 3z) (4x2 + y2 + 9z2 + 2xy + 3yz – 6xz). -Maths 9th

Last Answer : Solution of this question

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Description : Find (2x – y + 3z) (4x2 + y2 + 9z2 + 2xy + 3yz – 6xz). -Maths 9th

Last Answer : Solution of this question

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Description : If both (x+1) and (x -1) are factors of ax3 + x2 - 2x + b , find a and b. -Maths 9th

Last Answer : Let p(x) = ax3 + x2 - 2x + b Since (x+1) and (x-1) are the factors of p(x), ∴ p(-1) = 0 and p(1) = 0 ∴ p(-1) = a(-1)3 + (-1)2 - 2 (-1) + b = 0 ⇒ - a + 1 + 2 + b = 0 ⇒ a - b = 3 ---- (i) ... 0 ⇒ a + 1 - 2 + b = 0 ⇒ a + b = 1 ----- (ii) solving equations (i) and (ii) we get a = 2 and b = -1

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Description : If x +1 is a factor of ax3 +x2 -2x + 4a - 9, then find the value of a. -Maths 9th

Last Answer : The value of a

If both (x+1) and (x -1) are factors of ax3 + x2 - 2x + b , find a and b. -Maths 9th

Description : If both (x+1) and (x -1) are factors of ax3 + x2 - 2x + b , find a and b. -Maths 9th

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If x +1 is a factor of ax3 +x2 -2x + 4a - 9, then find the value of a. -Maths 9th

Description : If x +1 is a factor of ax3 +x2 -2x + 4a - 9, then find the value of a. -Maths 9th

Last Answer : The value of a

Simplify (2x- 5y)3 – (2x+ 5y)3. -Maths 9th

Description : Simplify (2x- 5y)3 – (2x+ 5y)3. -Maths 9th

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Simplify (2x- 5y)3 – (2x+ 5y)3. -Maths 9th

Description : Simplify (2x- 5y)3 – (2x+ 5y)3. -Maths 9th

Last Answer : (2x -5y)3 - (2x + 5y)3 = [(2x)3 - (5y)3 - 3(2x)(5y)(2x - 5y)] -[(2x)3 + (5y)3 + 3(2x)(5y)(2x+5y)] [using identity, (a - b)3 = a3 -b3 - 3ab and (a + b)3 =a3 +b3 + 3ab] = (2x)3 - (5y)3 - ... 2x)3- (5y)3 - 30xy (2x + 5y) = -2 (5y)3 - 30xy(2x - 5y + 2x + 5y) = -2 x 125y3 - 30xy(4x) = -250y3 -120x2y

Simplify: (2x–5y)3 – (2x + 5y)3 -Maths 9th

Description : Simplify: (2x–5y)3 – (2x + 5y)3 -Maths 9th

Last Answer : Solution :-

In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22). Show that (i) quadrilateral ABED is a parallelogram (ii) quadrilateral BEFC is a parallelogram (iii) AD || CF and AD = CF (iv) quadrilateral ACFD is a parallelogram (v) AC = DF (vi) ΔABC ≅ ΔDEF. -Maths 9th

Description : In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22). Show that (i) quadrilateral ABED is a parallelogram ( ... CF and AD = CF (iv) quadrilateral ACFD is a parallelogram (v) AC = DF (vi) ΔABC ≅ ΔDEF. -Maths 9th

Last Answer : . Solution: (i) AB = DE and AB || DE (Given) Two opposite sides of a quadrilateral are equal and parallel to each other. Thus, quadrilateral ABED is a parallelogram (ii) Again BC = EF and BC || EF ... (Given) BC = EF (Given) AC = DF (Opposite sides of a parallelogram) , ΔABC ≅ ΔDEF [SSS congruency]

Evaluate each of the following: (i) (103)3 (ii) (98)3 (iii) (9.9)3 (iv) (10.4)3 (v) (598)3 (vi) (99)3 -Maths 9th

Description : Evaluate each of the following: (i) (103)3 (ii) (98)3 (iii) (9.9)3 (iv) (10.4)3 (v) (598)3 (vi) (99)3 -Maths 9th

Last Answer : answer:

If z2+ 1/z2 = 14, find the value of z3+ 1/z3. -Maths 9th

Description : If z2+ 1/z2 = 14, find the value of z3+ 1/z3. -Maths 9th

Last Answer : Solution :-

Solve the following in equalities `(i) |x+7| gt 5` `(ii) |x+3| lt 10` `(iii) (x+2) lt |x^(2)+3x+5|``(iv) |(2x-1)/(x-1)| gt 2` `(v) |x-6| le x^(2)-5x+9

Description : Solve the following in equalities `(i) |x+7| gt 5` `(ii) |x+3| lt 10` `(iii) (x+2) lt |x^(2)+3x+5|``(iv) |(2x-1)/(x-1)| gt 2` `(v) |x-6| le x^(2)-5x+9

Last Answer : Solve the following in equalities `(i) |x+7| gt 5` `(ii) |x+3| lt 10` `(iii) (x+2) lt |x^(2)+3x+5|``(iv) ... -5) lt 0` `(viii) |x-1|+|x-2|+|x-3| le 6`

Solve for `x` `(i) |x+1|=4x+3` `(ii) |x+1|=|x+3|` `(iii) 7|x-2|-|x-7|=5` `(iv) ||x-1|-2|=6x+8` `(v) |2x^(2)-3x+1|=|x^(2)+x-3|`

Description : Solve for `x` `(i) |x+1|=4x+3` `(ii) |x+1|=|x+3|` `(iii) 7|x-2|-|x-7|=5` `(iv) ||x-1|-2|=6x+8` `(v) |2x^(2)-3x+1|=|x^(2)+x-3|`

Last Answer : Solve for `x` `(i) |x+1|=4x+3` `(ii) |x+1|=|x+3|` `(iii) 7|x-2|-|x-7|=5` `(iv) ||x-1|-2|=6x+8` `(v) |2x^(2)-3x+1|=|x^(2)+x-3|`

If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p (1/2). -Maths 9th

Description : If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p (1/2). -Maths 9th

Last Answer : Solution of this question

If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p (1/2). -Maths 9th

Description : If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p (1/2). -Maths 9th

Last Answer : Solution of this question

The centroid of the area between the circle x2 + y2 = 16 and the line x + y = 4 will have the coordinates a.(4, 2) b.(2, 4) c.(2.34, 2.34) d.(1.16, 1.16) e.None of the above

Description : The centroid of the area between the circle x2 + y2 = 16 and the line x + y = 4 will have the coordinates a.(4, 2) b.(2, 4) c.(2.34, 2.34) d.(1.16, 1.16) e.None of the above

Last Answer : c. (2.34, 2.34)

The x coordinate of the centroid of the area enclosed by the parabolas y = x2 and x = y2 will be a.107 dynes b.0.44 c.0.43 d.0.45 e.0.46

Description : The x coordinate of the centroid of the area enclosed by the parabolas y = x2 and x = y2 will be a.107 dynes b.0.44 c.0.43 d.0.45 e.0.46

Last Answer : d. 0.45

What is the length of a tangent line from the point 7 -2 to a point when it touches the circle x2 plus y2 -10 equals 49?

Description : What is the length of a tangent line from the point 7 -2 to a point when it touches the circle x2 plus y2 -10 equals 49?

Last Answer : x² + y² - 10 = 49→ x² + y² = 59= (x - 0)² + (y - 0)² = (√59)²→ circle has centre (0, 0) - the origin - and radius √59The point (7, -2) has a distance from the centre of the circleof:√((7 - 0)² + (-2 - 0)²) = √(7² + (-2)²) = √(49 + 4) = √53

What is the length of the circle radius if the circle is x2 plus y2 equals 1?

Description : What is the length of the circle radius if the circle is x2 plus y2 equals 1?

Last Answer : The center of the circle is at (0, 0) and its radius is thesquare root of 1 which is 1

What is the length of the circles radius for the circle x2 plus y2 169.?

Description : What is the length of the circles radius for the circle x2 plus y2 169.?

Last Answer : If x^2 plus y^2 = 169 then the center of the circle is at (0, 0)and its radius is the square root of 169 which is 13

What is the point of contact between the line y equals x plus 4 and the circle x2 plus y2 -8x plus 4y equals 30?

Description : What is the point of contact between the line y equals x plus 4 and the circle x2 plus y2 -8x plus 4y equals 30?

Last Answer : Equations: y = x+4 and x^2 +y^2 -8x +4y = 30It appears that the given line is a tangent line to the givencircle and the point of contact works out as (-1, 3)

What is the length of the tangent line from the point 9 0 to the circle x2 plus 8x plus y2 -9 equals 0?

Description : What is the length of the tangent line from the point 9 0 to the circle x2 plus 8x plus y2 -9 equals 0?

Last Answer : Equation of circle: x^2 +8x +y^2 -9 = 0Completing the square: (x+4)^2 +y^2 = 25Radius of circle: 5Center of circle: (-4, 0)Distance from (9, 0) to (-4, 0) = 13 which is ... the length of the tangent line is 12 unitsNote that the tangent line of a circle meets the radius of thecircle at right angles

What is the equation of the tangent line that meets the circle x2 -4x plus y2 -6y equals 4 at the point 6 4?

Description : What is the equation of the tangent line that meets the circle x2 -4x plus y2 -6y equals 4 at the point 6 4?

Last Answer : Circle equation: x^2 -4x +y^2 -6y = 4Completing the squares: (x-2)^2 +(y-3)^2 = 17Point of contact: (6, 4)Center of circle: (2, 3)Slope of radius: 1/4Slope of tangent line: -4Tangent equation: y-4 = -4(x-6) => y = -4x+28Tangent line equation in its general form: 4x+y-28 = 0

What is the length of the tangent line from the point 8 2 to a point where it meets the circle x2 plus y2 -4x -8y -5 equals 0?

Description : What is the length of the tangent line from the point 8 2 to a point where it meets the circle x2 plus y2 -4x -8y -5 equals 0?

Last Answer : Equation of circle: x^2 +y^2 -4x -8y -5 = 0Completing the squares: (x-2)^2 +(y-4)^2 = 25 which is radiussquaredCenter of circle: (2, 4)Tangent line originates from: (8, ... angle triangleUsing Pythagoras theorem: distance^2 minus radius^2 = 15Therefore length of tangent line is the square root of 15

What is the center and the radius of the circle x2 plus y2 -4x -6y -3 equals 0?

Description : What is the center and the radius of the circle x2 plus y2 -4x -6y -3 equals 0?

Last Answer : Equation of circle: x^2 +y^2 -4x -6y -3 = 0Completing the squares: (x-2)^2 +(y-3)^2 = 16 square unitsTherefore center of circle is at (2, 3) and its radius is 4units

What is the center and the radius of the circle x2 plus y2 equals 12x -10y -12?

Description : What is the center and the radius of the circle x2 plus y2 equals 12x -10y -12?

Last Answer : Equation of circle: x^2 +y^2 = 12x-10y-12Completing the squares: (x-6)^2 +(y+5)^2 = 49So center of circle is at (6, -5) and its radius is 7 units

If z equals yf x2 - y2 show that ydz divided by dx plus xdz divided by dy equals xz divided by y?

Description : If z equals yf x2 - y2 show that ydz divided by dx plus xdz divided by dy equals xz divided by y?

Last Answer : 4

The point (-1,2) divides the line segment joining the points A(2,5) and B(x,y) in the ratio 3:4, then the value of x2 + y2 is : (a) 27 (b) 28 (c) 29 (d) 30

Description : The point (-1,2) divides the line segment joining the points A(2,5) and B(x,y) in the ratio 3:4, then the value of x2 + y2 is : (a) 27 (b) 28 (c) 29 (d) 30

Last Answer : (c) 29

Solve for x and y: 1/x - 1/y = 1/3, 1/x2 + 1/y2 = 5/9.

Description : Solve for x and y: 1/x - 1/y = 1/3, 1/x2 + 1/y2 = 5/9.

Last Answer : Answer: x = 3/2 or -3 and y = 3 or -3/2.

Find the remainder when y3 + y2 - 2y + 5 is divided by y - 5. -Maths 9th

Description : Find the remainder when y3 + y2 - 2y + 5 is divided by y - 5. -Maths 9th

Last Answer : Remainder = 145 Again, we should evaluate p(5) Let p(y) = y3 + y2 - 2y + 5 ∴ p(5) = 53 + 52 - 2 x 5 + 5 = 125 + 25 - 10 + 5 = 145 Thus , we find that p(5) is the remainder when p(y) is divided by y - 5 .

Find the remainder when y3 + y2 - 2y + 5 is divided by y - 5. -Maths 9th

Description : Find the remainder when y3 + y2 - 2y + 5 is divided by y - 5. -Maths 9th

Last Answer : Remainder = 145 Again, we should evaluate p(5) Let p(y) = y3 + y2 - 2y + 5 ∴ p(5) = 53 + 52 - 2 x 5 + 5 = 125 + 25 - 10 + 5 = 145 Thus , we find that p(5) is the remainder when p(y) is divided by y - 5 .

Find the value of 4x2 + y2 + 25z2 + 4xy – 10yz – 20zx when x = 4, y = 3 and z = 2. -Maths 9th

Description : Find the value of 4x2 + y2 + 25z2 + 4xy – 10yz – 20zx when x = 4, y = 3 and z = 2. -Maths 9th

Last Answer : 4x²+y²+25z²+4xy-10yz-20zx when x=4, y=3 &z=2 so =>4(4)²+(3)²+ 25(2)²+4(4)(3)-10(3)(2)-20(2)(4) =>64+9+100+48-60-160 =>221-220 =>1

ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that (i) ∠A = ∠B (ii) ∠C = ∠D (iii) ΔABC ≅ ΔBAD (iv) diagonal AC = diagonal BD [Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.] -Maths 9th

Description : ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that (i) ∠A = ∠B (ii) ∠C = ∠D (iii) ΔABC ≅ ΔBAD (iv) diagonal AC = diagonal BD [Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.] -Maths 9th

Last Answer : ] Solution: To Construct: Draw a line through C parallel to DA intersecting AB produced at E. (i) CE = AD (Opposite sides of a parallelogram) AD = BC (Given) , BC = CE ⇒∠CBE = ∠CEB also, ∠A+∠CBE = ... BC (Given) , ΔABC ≅ ΔBAD [SAS congruency] (iv) Diagonal AC = diagonal BD by CPCT as ΔABC ≅ ΔBA.