The positive solutions of the equation ax + by + c = 0 always lie in the -Maths 9th

The positive solutions of the equation ax + by + c = 0 always lie in the -Maths 9th
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(a) We know that, if a line passes through the Ist quadrant, then all solution lying on the line in first quadrant must be positive because the coordinate of all points in the Ist quadrant are positive.
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The positive solutions of the equation ax + by + c = 0 always lie in the -Maths 9th

Description : The positive solutions of the equation ax + by + c = 0 always lie in the -Maths 9th

Last Answer : (a) We know that, if a line passes through the Ist quadrant, then all solution lying on the line in first quadrant must be positive because the coordinate of all points in the Ist quadrant are positive.

Is ax + by + c = 0, where a, b and c are real numbers, a linear equation in two variables? Give reason. -Maths 9th

Description : Is ax + by + c = 0, where a, b and c are real numbers, a linear equation in two variables? Give reason. -Maths 9th

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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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In quadratic equation ax^2 + bx + c = 0 , Identities the sum and product of Roots? -Maths 9th

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Last Answer : If the two roots of the quadratic equation ax2 + bx + c = 0 obtained by the quadratic formula be denoted by a and b, then we have α = \(\frac{-b+\sqrt{b^2-4ac}}{2a},\) β = \(\frac{-b-\sqrt{b^2-4ac}}{2a}\) ∴ Sum of roots ... then α + β = - (- \(\frac{5}{6}\)) = \(\frac{5}{6}\), ab = \(\frac{7}{6}\).

If one of the roots of the equation x^2 + ax + 3 = 0 is 3 and one of the roots of the equation x2 + ax + b = 0 is three -Maths 9th

Description : If one of the roots of the equation x^2 + ax + 3 = 0 is 3 and one of the roots of the equation x2 + ax + b = 0 is three -Maths 9th

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If the roots of the equation ax^2 + bx + c = 0 are equal in magnitude but opposite in sign, then which one of the following is correct ? -Maths 9th

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Last Answer : Let the roots of the equation x3 – ax2 + bx – c = 0 be (α – 1), α, (α + 1) ∴ S2 = (α – 1)α + α(α + 1) + (α + 1) ( ... ; 1 = b ⇒ 3α2 – 1 = b ∴ Minimum value of b = – 1, when α = 0.

If one root of the equation ax^2 + x – 3 = 0 is –1, then what is the other root ? -Maths 9th

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If the sum of the roots of the equation ax^2 + bx + c = 0 is equal to the sum of their squares, then which one of the following is correct ? -Maths 9th

Description : If the sum of the roots of the equation ax^2 + bx + c = 0 is equal to the sum of their squares, then which one of the following is correct ? -Maths 9th

Last Answer : Given equation: ax2+bx+c=0 Let α and β be the roots of given quadratic equation Sum of the roots i.e. α+β=a−b Product of roots i.e. αβ=ac It is given that, Sum of the roots = Sum of squares of the roots i ... )2−2αβ i.e. a−b =(a−b )2−a2c i.e. −ab=b2−2ac i.e. ab+b2=2ac Hence, C is the correct option.

The solutions of a linear equation in two variables always take integral values .True / false -Maths 9th

Description : The solutions of a linear equation in two variables always take integral values .True / false -Maths 9th

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If a^2 + b^2 + c^2 = 1, x^2 + y^2 + z^2 = 1, where a, b, c, x, y, z are positive reals then ax + by + cz is -Maths 9th

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If a linear equation has solutions (-2, 2), (0, 0) and (2, – 2), then it is of the form. -Maths 9th

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Last Answer : (b) Let us consider a linear equation ax + by + c = 0 (i) Since, (-2,2), (0, 0) and (2, -2) are the solutions of linear equation therefore it satisfies the Eq. (i), we get At point(-2,2), -2a + 2b + c ... b(x + y)= 0 ⇒ x + y = 0, b ≠ 0 Hence, x + y= 0 is the required form of the linear equation.

If a linear equation has solutions (-2, 2), (0, 0) and (2, – 2), then it is of the form. -Maths 9th

Description : If a linear equation has solutions (-2, 2), (0, 0) and (2, – 2), then it is of the form. -Maths 9th

Last Answer : (b) Let us consider a linear equation ax + by + c = 0 (i) Since, (-2,2), (0, 0) and (2, -2) are the solutions of linear equation therefore it satisfies the Eq. (i), we get At point(-2,2), -2a + 2b + c ... b(x + y)= 0 ⇒ x + y = 0, b ≠ 0 Hence, x + y= 0 is the required form of the linear equation.

The number of real solutions of the equation 2|x|^2 – 5|x| + 2 = 0 is : -Maths 9th

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In which quadrant does a point both of whose coordinates are positive lie? -Maths 9th

Description : In which quadrant does a point both of whose coordinates are positive lie? -Maths 9th

Last Answer : Solution :- I quadrant.

Find any four solutions of the equation 4x+3y=12. -Maths 9th

Description : Find any four solutions of the equation 4x+3y=12. -Maths 9th

Last Answer : Given equation is 4x + 3y =12 On putting x = 0 in Eq. (i), we get 4(0) +3y =12 ⇒ 3y =12 ⇒ y = 12 / 3 = 4 So, (0, 4) is a solution of given equation. On putting y = 0 in Eq. (i), we ... given equation. Hence the four solutions of given equation are (0, 4), (3, 0), (1, 8 / 3) and (2,4 / 3).

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

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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.

How many solutions of the equation 2x + 1 = x – 3 are there on the Cartesian plane? -Maths 9th

Description : How many solutions of the equation 2x + 1 = x – 3 are there on the Cartesian plane? -Maths 9th

Last Answer : 2x + 1 = x - 3 2x-x = -3-1 ∴ x = - 4 ..(i) and it can be written as 1.x + 0. y = - 4 ..(ii) (i) Number line represent the all real values of x on the X ... the equation x + 4 = 0 represent a straight line parallel to Y-axis and infinitely many points lie on a line in the cartesian plane.

Find any four solutions of the equation 4x+3y=12. -Maths 9th

Description : Find any four solutions of the equation 4x+3y=12. -Maths 9th

Last Answer : Given equation is 4x + 3y =12 On putting x = 0 in Eq. (i), we get 4(0) +3y =12 ⇒ 3y =12 ⇒ y = 12 / 3 = 4 So, (0, 4) is a solution of given equation. On putting y = 0 in Eq. (i), we ... given equation. Hence the four solutions of given equation are (0, 4), (3, 0), (1, 8 / 3) and (2,4 / 3).

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.

How many solutions of the equation 2x + 1 = x – 3 are there on the Cartesian plane? -Maths 9th

Description : How many solutions of the equation 2x + 1 = x – 3 are there on the Cartesian plane? -Maths 9th

Last Answer : 2x + 1 = x - 3 2x-x = -3-1 ∴ x = - 4 ..(i) and it can be written as 1.x + 0. y = - 4 ..(ii) (i) Number line represent the all real values of x on the X ... the equation x + 4 = 0 represent a straight line parallel to Y-axis and infinitely many points lie on a line in the cartesian plane.

How many solutions does the equation 2x +5y=8 has? -Maths 9th

Description : How many solutions does the equation 2x +5y=8 has? -Maths 9th

Last Answer : Solution :- Infinitely many solutions.

Write any two solutions of the linear equation 3x + 2y =9. -Maths 9th

Description : Write any two solutions of the linear equation 3x + 2y =9. -Maths 9th

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Write two solutions of the equation 4x -5 y = 15. -Maths 9th

Description : Write two solutions of the equation 4x -5 y = 15. -Maths 9th

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NCERT Solutions for class 9 Maths Chapter 4 Linear Equation in Two Variables Exercise 4.1 -Maths 9th

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Last Answer : (i) (x + 4)(x + 10): Using the identity (x + a)(x + b) = x2 + (a + b)x + ab, we have: (x + 4)(x + 10) = x2 + (4 + 10)x + (4 x 10) = x2 + 14x + 40 (ii) (x + 8)(x - 10): Here, a = 8 and b = ( ... have 64m3 - 343n3 = (4m)3 - (7n)3 = (4m - 7n)[(4m)2 + (4m)(7n) + (7n)2] = (4m - 7n)(16m2 + 28mn + 49n2)

NCERT Solutions for class 9 Maths Chapter 4 Linear Equation in Two Variables Exercise 4.2 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 4 Linear Equation in Two Variables Exercise 4.2 -Maths 9th

Last Answer : 1. How will you describe the position of a table lamp on your study table to another person ? To describe the position of a table lamp placed on the table, let us consider the table lamp as P and the table as a plane ... the point A(4, 3). (ii) A unique cross street as shown by the point B(3, 4).

NCERT Solutions for class 9 Maths Chapter 4 Linear Equation in Two Variables Exercise 4.3 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 4 Linear Equation in Two Variables Exercise 4.3 -Maths 9th

Last Answer : 1: Draw the graph of each of the following linear equations in two variables: (i) x + y = 4 (ii) x - y = 2 (iii) y = 3x (iv) 3 = 2x + y Solution: (i)x + y= 4 ⇒ y = 4 - x If ... graph paper and joining them, we get a straight line PQ. Thus, PQ is the required graph of the linear equation y = 5x + 3.

NCERT Solutions for class 9 Maths Chapter 4 Linear Equation in Two Variables Exercise 4.4 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 4 Linear Equation in Two Variables Exercise 4.4 -Maths 9th

Last Answer : 1. Give the geometric representations of y = 3 as an equation (i) in one variable (ii) in two variables (i) y = 3 [An equation in one variable] ∵ y = 3 is an equation in one variable, i.e. y only. ∴ It has ... equation in two variables] We can write 2x + 9 = 0 as 2x + 0y + 9 = 0 or 2x = -9 + 0y

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Description : A linear equation in two variables has infinite solutions. True/false. -Maths 9th

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Write any four solutions of the linear equation y = 4 x – 11. -Maths 9th

Description : Write any four solutions of the linear equation y = 4 x – 11. -Maths 9th

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The number of solutions satisfying the given equation -Maths 9th

Description : The number of solutions satisfying the given equation -Maths 9th

Last Answer : (d) 3Taking log of both the sides to base 3, we have,\(\big[(log_3\,x)^2-rac{9}{2}log_3\,x+5\big]\) log3x = log333/2 = \(rac{3}{2}\) (∵ log33 = 1)⇒ 2(log3x)3 - 9(log3x)2 + 10 log3x - 3 = 0 ⇒ ... log3x = 3, 2 log3x = 1 ⇒ x = 31, x = 33, x2 = 31⇒ \(x\) = (3, 27, √3)∴ There are three solutions.

What are the number of solutions for real x, which satisfy the equation -Maths 9th

Description : What are the number of solutions for real x, which satisfy the equation -Maths 9th

Last Answer : log2 x>0 2log2 log2 x+log21 log2 (22 x)=1 ⇒2log2 log2 x−log2 log2 (22 x)=1[∵loga1 b=−loga b] $$\Rightarrow \log _{ 2 }{ \left[ \dfrac { { \left( \log _{ 2 }{ x } \right) }^{ 2 } }{ \log_{ 2 } ... log2 (22 )2=log2 8=log2 23=3] ⇒t=3,−1=log2 x ⇒x=2−1or 23 That is x=21 or 8 Hence, the answer is 8.

Express X+2=0 in the form of ax+by+C=0 -Maths 9th

Description : Express X+2=0 in the form of ax+by+C=0 -Maths 9th

Last Answer : 1[x] + 0[y] + 2=0

Express x+2=0 in ax+by+c=0 -Maths 9th

Description : Express x+2=0 in ax+by+c=0 -Maths 9th

Last Answer : x+2y-c=0 is the answer of x+2

Express X+2=0 in the form of ax+by+C=0 -Maths 9th

Description : Express X+2=0 in the form of ax+by+C=0 -Maths 9th

Last Answer : 1[x] + 0[y] + 2=0

Express x+2=0 in ax+by+c=0 -Maths 9th

Description : Express x+2=0 in ax+by+c=0 -Maths 9th

Last Answer : x+2y-c=0 is the answer of x+2

express in the form of ax+by+c=0 and identify a,b,c if 2x=5 -Maths 9th

Description : express in the form of ax+by+c=0 and identify a,b,c if 2x=5 -Maths 9th

Last Answer : NEED ANSWER

express in the form of ax+by+c=0 and identify a,b,c if 2x=5 -Maths 9th

Description : express in the form of ax+by+c=0 and identify a,b,c if 2x=5 -Maths 9th

Last Answer : Let y=0. So, the equation will be 2x+0y-5=0. Then from the equation, a=2; b=0; c=-5

If x(square) - 1 is a factor of ax(cube) + bx(square) + cx + d,show that a+c=0. -Maths 9th

Description : If x(square) - 1 is a factor of ax(cube) + bx(square) + cx + d,show that a+c=0. -Maths 9th

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Find the condition that one root of ax^2 + bx + c = 0 may be four times the other. -Maths 9th

Description : Find the condition that one root of ax^2 + bx + c = 0 may be four times the other. -Maths 9th

Last Answer : answer:

If the expression ax^2 + bx + c is equal to 4, when x = 0, leaves a remainder 4 when divided by x + 1 and leaves a -Maths 9th

Description : If the expression ax^2 + bx + c is equal to 4, when x = 0, leaves a remainder 4 when divided by x + 1 and leaves a -Maths 9th

Last Answer : Given exp. f(x) = ax2 + bx + c ∴ When x = 0, a.0 + b.0 + c = 4 ⇒ c = 4. The remainders when f(x) is divided by (x + 1) and (x + 2) respectively are f(–1) and f(–2). ∴ f( ... 2b = 2 ...(ii) Solving (i) and (ii) simultaneously we get, a = 1, b = 1.

The values of a, b and c respectively for the expression f(x) = x^3 + ax^2 + bx + c, if f(1) = f(2) = 0 and f(4) = f(0) are : -Maths 9th

Description : The values of a, b and c respectively for the expression f(x) = x^3 + ax^2 + bx + c, if f(1) = f(2) = 0 and f(4) = f(0) are : -Maths 9th

Last Answer : f′(x)=3x2+2ax+6 f(x)⇒f′(x)≥0 3x2+2ax+6≥0 ⇒D≤0 4[a2−3b]≤0 a2≤3b ∴P=6×6×6(16)×6​=3616​=94​ Answer.

The straight line ax + by + c = 0 and the co-ordinate axes form an isosceles triangle under which of the following conditions ? -Maths 9th

Description : The straight line ax + by + c = 0 and the co-ordinate axes form an isosceles triangle under which of the following conditions ? -Maths 9th

Last Answer : (a) | a | = | b | The equation of line AB, i.e., ax + by + c = 0 in intercept form is ax + by = - c⇒ \(rac{x}{\big(-rac{c}{a}\big)}\) + \(rac{x}{\big(-rac{c}{b}\big)}\) = 1Δ AOB is isosceles Δ if OA = OB, i.e., ... \(rac{-c}{a}\) = \(rac{-c}{a}\) ⇒ \(rac{1}{a}\) = \(rac{1}{a}\) ⇒ | a | = | b |.

If the equation x2 minus ax plus 2b equals 0 has prime roots where a and b are positive integers then a minus b is equal to?

Description : If the equation x2 minus ax plus 2b equals 0 has prime roots where a and b are positive integers then a minus b is equal to?

Last Answer : I think you mean the roots are prime numbers.Let the two roots be primes p and qThen the equation factorises to (x - p)(x - q) = 0 which can beexpanded to give:x² - (p + q)x + pq = 0Which comparing ... = 2(It doesn't matter if the other prime is even (2) or not as itcancels out from a - b.)

If p, q, r are positive and are in A.P., the roots of quadratic equation px^2 + qx + r = 0 are real for : -Maths 9th

Description : If p, q, r are positive and are in A.P., the roots of quadratic equation px^2 + qx + r = 0 are real for : -Maths 9th

Last Answer : Given p,q,r are in A.P. then q=2p+r​.....(1). Now px2+qx+r=0 will have real root then q2−4pr≥0. or, 4(p+r)2​−4pr≥0 or, p2+r2−14pr≥0 or, r2−14rp+49p2≥48p2 or, (r−7p)2≥(43​p)2 or, (pr​−7)2≥(43​)2 [ Since p=0 for the given equation to be quadratic] or, ∣∣∣∣∣​pr​−7∣∣∣∣∣​≥43​.

Find the equation of the straight line with a positive gradient which passes through the point (–5, 0) -Maths 9th

Description : Find the equation of the straight line with a positive gradient which passes through the point (–5, 0) -Maths 9th

Last Answer : (d) Both (a) and (c)Since the line passes through A(a, 0) and B(0, b), it makes intercepts a and b on x-axis and y-axis respectively. Let the equation of this line in the intercept from be \(rac{x}{a}\) + \(rac{y}{a}\) ... \(rac{x}{-12}\) + \(rac{y}{-5}\) = 1⇒ 5x + 12y = 60 and 5x + 12y + 60 = 0.

When f(x) = x4 - 2x3 + 3x2 - ax is divided by x + 1 and x - 1 , we get remainders as 19 and 5 respectively . -Maths 9th

Description : When f(x) = x4 - 2x3 + 3x2 - ax is divided by x + 1 and x - 1 , we get remainders as 19 and 5 respectively . -Maths 9th

Last Answer : When f(x) is divided by (x+1) and (x-1) , the remainders are 19 and 5 respectively . ∴ f(-1) = 19 and f(1) = 5 ⇒ (-1)4 - 2 (-1)3 + 3(-1)2 - a (-1) + b = 19 ⇒ 1 +2 + 3 + a + b = 19 ∴ a + b = 13 ------- ... + 3x2 - 5x + 8 ⇒ f(3) = 34 - 2 33 + 3 32 - 5 3 + 8 = 81 - 54 + 27 - 15 + 8 = 47

ABCD is a parallelogram and line segments AX, CY bisect the angles A and C, respectively. -Maths 9th

Description : ABCD is a parallelogram and line segments AX, CY bisect the angles A and C, respectively. -Maths 9th

Last Answer : Since opposite angles are equal in a parallelogram . Therefore , in parallelogram ABCD , we have ∠A = ∠C ⇒ 1 / 2 ∠A = 1 / 2 ∠C ⇒ ∠1 = ∠2 ---- i) [∵ AX and CY are bisectors of ∠A and ∠C ... intersects AX and YC at A and Y such that ∠1 = ∠3 i.e. corresponding angles are equal . ∴ AX | | CY .