For what value of k,(x+1) is a factor of p(x) - kx(square) - x - 4 ? -Maths 9th

For what value of k,(x+1) is a factor of p(x) - kx(square) - x - 4 ? -Maths 9th
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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

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Description : The value(s) of 'k' for which the following pair of linear equations will represent a pair of intersecting lines graphically is/are: 3x - 4y + 7 = 0 kx - 8y = 5 (a) k = 6 (b) k ≠ 6(c) any real number (d) any real number except 6

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Description : If K is a constant depending upon the ratio of the width of the slab to its effective span l, x is the distance of the concentrated load from the nearer support, bw is the width of the area of contact of the concentrated load measured ... ) Kx (1 - x/l) + bw (C) Kx (1 + x/l) + bw (D) All the above

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Last Answer : x3+px+q 3x2+p have common root Let common root =a a3+ap+q=0 a2=3−p​a=3−p​​4p3+27q2=? a3(a2+p)+q=0 ⇒3−p​​[(3−p​)+p]+q=0 3−p​​(3+2p​)=−q ⇒27−p(4p2)​=q2

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In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP. -Maths 9th

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Last Answer : In △PAD, ∠A = 90° and DA = PA = PB ⇒ ∠ADP = ∠APD = 90° / 2 = 45° Similarly, in △QBC, ∠B = 90° and BQ = BC = AB ⇒∠BCQ = ∠BQC = 90° / 2 = 45° In △PAD and △QBC , we have PA = QB [given] ∠A = ... [each = 90° + 45° = 135°] ⇒ △PDC = △QCD [by SAS congruence rule] ⇒ PC = QD or DQ = CP

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P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square. -Maths 9th

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Last Answer : Given In quadrilateral ABCD, P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively. Also, AC = BD and AC ⊥ BD. To prove PQRS is a square. Proof Now, in ΔADC, S and R are the mid-points of the sides AD and DC respectively, then by mid-point theorem,

In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP. -Maths 9th

Description : In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP. -Maths 9th

Last Answer : In △PAD, ∠A = 90° and DA = PA = PB ⇒ ∠ADP = ∠APD = 90° / 2 = 45° Similarly, in △QBC, ∠B = 90° and BQ = BC = AB ⇒∠BCQ = ∠BQC = 90° / 2 = 45° In △PAD and △QBC , we have PA = QB [given] ∠A = ... [each = 90° + 45° = 135°] ⇒ △PDC = △QCD [by SAS congruence rule] ⇒ PC = QD or DQ = CP

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

P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square. -Maths 9th

Description : P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square. -Maths 9th

Last Answer : Given In quadrilateral ABCD, P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively. Also, AC = BD and AC ⊥ BD. To prove PQRS is a square. Proof Now, in ΔADC, S and R are the mid-points of the sides AD and DC respectively, then by mid-point theorem,

WXYZ is a square of side length 30. V is a point on XY and P is a point inside the square with PV perpendicular to XY. PW = PZ = PV – 5. Find PV. -Maths 9th

Description : WXYZ is a square of side length 30. V is a point on XY and P is a point inside the square with PV perpendicular to XY. PW = PZ = PV – 5. Find PV. -Maths 9th

Last Answer : answer:

ABCD is a square. P, Q, R, S are the mid-points of AB, BC, CD and DA respectively. By joining AR, BS, CP, DQ, we get a quadrilateral which is a -Maths 9th

Description : ABCD is a square. P, Q, R, S are the mid-points of AB, BC, CD and DA respectively. By joining AR, BS, CP, DQ, we get a quadrilateral which is a -Maths 9th

Last Answer : According to the given statement, the figure will be a shown alongside; using mid-point theorem: In △ABC,PQ∥AC and PQ=21 AC .......(1) In △ADC,SR∥AC and SR=21 AC .... ... are perpendicular to each other) ∴PQ⊥QR(angle between two lines = angle between their parallels) Hence PQRS is a rectangle.

Let P(–3, 2), Q(–5, –5), R(2, –3) and S(4, 4) be four points in a plane. Then show that PQRS is a rhombus. Is it a square ? -Maths 9th

Description : Let P(–3, 2), Q(–5, –5), R(2, –3) and S(4, 4) be four points in a plane. Then show that PQRS is a rhombus. Is it a square ? -Maths 9th

Last Answer : Let P(1, -1), Q \(\big(rac{-1}{2},rac{1}{2}\big)\) and R(1,2) be the vertices of the ΔPQR.Then, PQ = \(\sqrt{\big(rac{-1}{2}-1\big)^2+\big(rac{1}{2}+1\big)^2}\) = \(\sqrt{rac{9}{4}+rac{9}{4}} ... {3\sqrt2}{2}\)PR = \(\sqrt{(1-1)^2+(2+1)^2}\) = \(\sqrt9\) = 3∵ PQ = QR, the triangle PQR is isosceles.

Find the value of k for 5x +2ky =3k, if x =1 and y =1 is its solution. -Maths 9th

Description : Find the value of k for 5x +2ky =3k, if x =1 and y =1 is its solution. -Maths 9th

Last Answer : Given, equation is 5x + 2ky = 3k. On putting x =1 and y =1 in this equation, we get 5(1) + 2k(1) =3k ⇒ 5 + 2k =3k ⇒ 5 = 3k - 2k ⇒ k = 5 Hence, required value of k is 5.

Find the value of k for 5x +2ky =3k, if x =1 and y =1 is its solution. -Maths 9th

Description : Find the value of k for 5x +2ky =3k, if x =1 and y =1 is its solution. -Maths 9th

Last Answer : Given, equation is 5x + 2ky = 3k. On putting x =1 and y =1 in this equation, we get 5(1) + 2k(1) =3k ⇒ 5 + 2k =3k ⇒ 5 = 3k - 2k ⇒ k = 5 Hence, required value of k is 5.

FOR WHAT VALUE OF K, THE POLYNOMIAL X2+(4-K)X+2 IS DIVISIBLE BY X-2 -Maths 9th

Description : FOR WHAT VALUE OF K, THE POLYNOMIAL X2+(4-K)X+2 IS DIVISIBLE BY X-2 -Maths 9th

Last Answer : As the given polynomial divisible by x-2 means the polynomial satisfies for the value x=2 So putting x=2 in x²+(4-k)x+2 yields 0 ⇒2²+(4-k)2+2=0 ⇒4+8-2k+2=0 ⇒ 2k=14 ⇒ k= ... ;-3x+2 if factorized yields (x-1)(x-2). Thus is divisible by x-2 as well as divisible by x-1.

FOR WHAT VALUE OF K, THE POLYNOMIAL X2+(4-K)X+2 IS DIVISIBLE BY X-2 -Maths 9th

Description : FOR WHAT VALUE OF K, THE POLYNOMIAL X2+(4-K)X+2 IS DIVISIBLE BY X-2 -Maths 9th

Last Answer : The value of 'k' is 4

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Description : If x = 0 and y = k is a solution of the equation 5x - 3 y = 0, find the value of k. -Maths 9th

Last Answer : Solution :-

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

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.

If (x + k) is the HCF of ax^2 + ax + b and x^2 + cx + d, then what is the value of k ? -Maths 9th

Description : If (x + k) is the HCF of ax^2 + ax + b and x^2 + cx + d, then what is the value of k ? -Maths 9th

Last Answer : answer:

Definition: In a direct proportion y = kx, the number k is called the constant of ______.?

Description : Definition: In a direct proportion y = kx, the number k is called the constant of ______.?

Last Answer : proportion

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

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.