If 2x^2 – 7xy + 3y^2 = 0, then the value of x : y is -Maths 9th

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Description : If (2, 0) is a solution of the linear equation 2x + 3y = k, then the value of k is -Maths 9th

Last Answer : (a) Since, (2, 0) is a solution of the given linear equation 2x + 3y = k, then put x =2 and y= 0 in the equation. ⇒ 2 (2) + 3 (0) = k ⇒ k = 4 Hence, the value of k is 4.

Description : If (2, 0) is a solution of the linear equation 2x + 3y = k, then the value of k is -Maths 9th

Last Answer : (a) Since, (2, 0) is a solution of the given linear equation 2x + 3y = k, then put x =2 and y= 0 in the equation. ⇒ 2 (2) + 3 (0) = k ⇒ k = 4 Hence, the value of k is 4.

Description : If the point (2k – 3, k + 2) lies on the graph of the equation 2x + 3y +15 = 0, find the value of k. -Maths 9th

Last Answer : Solution :-

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.

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.

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.

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.

Description : At what point does the graph of the linear equation 2x + 3y = 9 meet a line which is parallel to the y-axis, at a distance of 4 units from the origin and on the right of the y-axis? -Maths 9th

Last Answer : hope its clear

Description : If 2x + 3y = 13 and xy = 6, find the value of 8x3 + 21y3. -Maths 9th

Last Answer : 2x +3y = 13-----(1) xy =6-----(2) 8x³ +27y³ = (2x)³ +(3y)³ = (2x+3y)³ - 3*2x*3y(2x+3y) [using (a+b)³ = a³+b³+3ab(a+b)] = (13)³- 18 * 6 *13 [ using (1) and (2)] = 2197 - 1404 =793

Description : if 2x + 3y = 8 and xy = 2, find the value of 4X2 + 9y2. -Maths 9th

Last Answer : Given 2x+3y=8 and xy=2, formula, (a+b)2=a2+b2+2ab ∴(2x+3y)2=4x2+9y2+2(2x)(3y) (2x+3y)2=4x2+9y2+12xy 82=4x2+9y2+12(2) ∴4x2+9y2=64−24=40

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).

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).

Description : Find the co-ordinate where the equation 2x + 3y = 6 intersects x-axis. -Maths 9th

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Description : Determine the ratio in which 2x +3y – 30 = 0 divides the join of A(3, 4) and B(7, 8) and at what point? -Maths 9th

Last Answer : Let A(1, 2) and B(11, 9) be the given points. Let the points of trisection be P and Q. Then,AP = PQ = QB = k (say)⇒ AQ = AP + PQ = 2k and PB = PQ + QB = 2k ∴ AP : PB = k : 2k = 1 : 2 and AQ ... two points of trisection are \(\big(rac{13}{3},rac{13}{3}\big)\) and \(\big(rac{23}{3},rac{20}{3}\big)\).

Description : Express the given equation as linear equation in two variables in standard form: 3y = 2x. -Maths 9th

Last Answer : Solution :- 2x - √3y + 0 = 0

Description : (2x – 3y)^3 + (3y – 4z)^3 + (4z – 2x)^3 can be factorised into which one of the following ? -Maths 9th

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Description : If the points (2, 1) and (1, – 2) are equidistant from the point (x, y), show that x + 3y = 0. -Maths 9th

Last Answer : (a) The distance d between any two points say P(x1, y1) and Q(x2, y2) is given by:d = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)⇒ d2 = (x2 - x1)2 + (y2 - y1)2 ⇒ d = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)( ... distance of a point P(x1, y1) form the origin= \(\sqrt{(x_2-0)^2+(y_2-0)^2}\) = \(\sqrt{x^2_1+y^2_1}\)

Description : Find the value of k,if y+3 is a factor of 3y(to the power square) + ky + 6. -Maths 9th

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Description : If (x^4 – 2x^2y^2 + y^2)^(a –1) = (x – y)^2a (x + y) ^–2, then the value of a is -Maths 9th

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Description : x^4 + xy^3 + x^3y + xz^3 + y^4 + yz^3 is divisible by : -Maths 9th

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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.

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.

Description : If x + y + z = 0, then x^2/(2x^2+yz)+y^2/(2y^2+zx)+z^2/(2z^2+xy) = -Maths 9th

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Description : if (1.-2) is a solution of the equation 2x-y=p,then find the value of p. -Maths 9th

Last Answer : x = 1 y = -2 2x-y = p Therefore, p = 2(1)-(-2) = 2 + 2 = 4

Description : if (1.-2) is a solution of the equation 2x-y=p,then find the value of p. -Maths 9th

Last Answer : 2x-y=p put x-1,y=-2 =2(1)-(-2)=p p=4

Description : A pair of linear equations which has a unique solution x = 2, y = -3 is (a) x + y = -1 ; 2x – 3y = -5 (b) 2x + 5y = -11 ; 4x + 10y = -22(c) 2x – y = 1 ; 3x + 2y = 0 (d) x – 4y – 14 = 0 ;5x – y – 13 = 0

Last Answer : (d) x – 4y – 14 = 0 ;5x – y – 13 = 0

Description : For what value of c, the linear equation 2x + cy = 8 has equal values of x and y for its solution? -Maths 9th

Last Answer : The given linear equation is 2x + cy= 8. …(i) Now, by condition, x and y-coordinate of given linear equation are same, i.e., x = y. Put y = x in Eq. (i), we get

Description : For what value of c, the linear equation 2x + cy = 8 has equal values of x and y for its solution? -Maths 9th

Last Answer : The given linear equation is 2x + cy= 8. …(i) Now, by condition, x and y-coordinate of given linear equation are same, i.e., x = y. Put y = x in Eq. (i), we get

Description : For what value of c, the linear equation 2x + cy = 8 has equal values of x and y as its solution? -Maths 9th

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Description : Express the equation –x + 3y = -2/3 in the form of ax + by + c =0 and identify the values of a,b and c. -Maths 9th

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Description : Let R be a relation on the set N, defined by {(x, y) : 2x – y = 10} then R is -Maths 9th

Last Answer : (a) ReflexiveGiven, {(\(x\), y) : 2\(x\) – y = 10} Reflexive, \(x\) R \(x\) = 2\(x\) – \(x\) = 10 ⇒ \(x\) = 10 ⇒ y = 10 ∴ Point (10, 10) ∈ N ⇒ R is reflexive.

Description : If 2x-y=4 then calculate 6x-3y=? (a)15 (b)12 (c)18 (d)10

Last Answer : (b)12

Description : If x + 2a is a factor of a5 -4a2x3 +2x + 2a +3, then find the value of a. -Maths 9th

Last Answer : Let p(x) = a5 -4a2x3 +2x + 2a +3 Since, x + 2a is a factor of p(x), then put p(-2a) = 0 (-2a)5 – 4a2 (-2a)3 + 2(-2a) + 2a + 3 = 0 ⇒ -32a5 + 32a5 -4a + 2a+ 3 = 0 ⇒ -2a + 3 = 0 2a =3 a = 3/2. Hence, the value of a is 3/2.

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

Description : If x + 2a is a factor of a5 -4a2x3 +2x + 2a +3, then find the value of a. -Maths 9th

Last Answer : Let p(x) = a5 -4a2x3 +2x + 2a +3 Since, x + 2a is a factor of p(x), then put p(-2a) = 0 (-2a)5 – 4a2 (-2a)3 + 2(-2a) + 2a + 3 = 0 ⇒ -32a5 + 32a5 -4a + 2a+ 3 = 0 ⇒ -2a + 3 = 0 2a =3 a = 3/2. Hence, the value of a is 3/2.

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

Description : If tan x = b/a , then what is the value of a cos 2x + b sin 2x? -Maths 9th

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Description : If sin^4x + sin^2x = 1 then what is 1 are the value of cot^4 x + cot^2 x? -Maths 9th

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Description : If the expressions (px^3 + 3x^2 – 3) and (2x^3 – 5x + p) when divided by (x – 4) leave the same remainder, then what is the value of p ? -Maths 9th

Last Answer : Given that the following polynomials leave the same remainder when divided by (x - 4) : We are to find the value of a. Remainder theorem: When (x - b) divides a polynomial p(x), then the remainder is p(b). So, from (i) and (ii), we get Thus, the required value of a is 1.

Description : If the HCF of (x^2 + x – 12) and (2x^2 – kx – 9) is (x – k), then what is the value of k ? -Maths 9th

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Description : If x^3 + 5x^2 + 10k leaves remainder – 2x when divided by x^2 + 2, then what is the value of k ? -Maths 9th

Last Answer : x3+5x2+10k =(x2+2)(x+5)+10k−2x−10 ⇒10k−2x−10=−2x ⇒10k−10=0 or k=1.

Description : If the sum of the zeroes of the polynomial p(x) = (k2 – 14) x2 – 2x – 12 is 1, then find the value of k. -Maths 9th

Last Answer : p(x) = (k2 – 14) x2 – 2x – 12 Here a = k2 – 14, b = -2, c = -12 Sum of the zeroes, (α + β) = 1 …[Given] ⇒ − = 1 ⇒ −(−2)2−14 = 1 ⇒ k2 – 14 = 2 ⇒ k2 = 16 ⇒ k = ±4

Description : The acute angle which the perpendicular from the origin on the line 7x –3y = 4 makes with the x-axis is: -Maths 9th

Last Answer : (c) negativeAs the line from the origin is perpendicular to the line 7x - 3y = 4, so its slope = \(rac{-1}{ ext{slope of }\,7x-3y=4}\)Slope of 7x - 3y - 4 = \(rac{7}{3}\)∴ Slope of line from origin = \(rac{-1} ... of x-axis⇒ θ = tan-1 \(\big(rac{-3}{7}\big)\) = - tan-1 \(\big(rac{3}{7}\big)\)

Description : Find the value of k if the line on 2x + y = k passes through the point (3,5). -Maths 9th

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Description : For what value of p is the coefficient of x^2 in the product (2x – 1) (x – k) (px + 1) equal to 0 and the constant term equal to 2 ? -Maths 9th

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Description : The image of the origin with reference to the line 4x + 3y – 25 = 0 is -Maths 9th

Last Answer : (c) 25x + 25y - 4 = 0 The point of intersection of the given lines can be obtained by solving the equations of the two lines simultaneously. 100x + 50y = 1 ...(i) 75x + 25y = -3 ... = \(rac{4}{25}\)∴ Eqn of required line: x + y = \(rac{4}{25}\) ⇒ 25 + 25y - 4 = 0.

Description : The equation 2x+ 5y = 7 has a unique solution, if x and y are -Maths 9th

Last Answer : (a) In natural numbers, there is only one pair i.e., (1, 1) which satisfy the given equation but in positive real numbers, real numbers and rational numbers there are many pairs to satisfy the given linear equation.

Description : The equation 2x+ 5y = 7 has a unique solution, if x and y are -Maths 9th

Last Answer : (a) In natural numbers, there is only one pair i.e., (1, 1) which satisfy the given equation but in positive real numbers, real numbers and rational numbers there are many pairs to satisfy the given linear equation.

Description : Draw a graph of the equation x - Y = 4 & 2x+ 2y =4 on the same graph paper find the coordinates of the point whose two lines intersect. -Maths 9th

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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°.