If a, b, c are positive real numbers such that a + b + c = p, then 1/a+1/b+1/c is greater than -Maths 9th

If a, b, c are positive real numbers such that a + b + c = p, then 1/a+1/b+1/c is greater than -Maths 9th
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Description : If a1, a2, ....., an are distinct positive real numbers such that a1 + a2 + ..... + an = 1, then -Maths 9th

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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​.

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Description : If a, b, c, d are positive reals such that a + b + c + d = 2, then M = (a + b) (c + d) satisfies the relation -Maths 9th

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On the set R of all real numbers, a relation R is defined by R = {(a, b) : 1 + ab > 0}. Then R is -Maths 9th

Description : On the set R of all real numbers, a relation R is defined by R = {(a, b) : 1 + ab > 0}. Then R is -Maths 9th

Last Answer : (a) Reflexive and symmetric only(a, a) ∈ R ⇒ 1 + a . a = 1 + a2 > 0 V real numbers a ⇒ R is reflexive (a, b) ∈ R ⇒ 1 + ab > 0 ⇒ 1 + ba > 0 ⇒ (b, a) ∈ R ⇒ R is symmetricWe observe that \(\big(1,rac{1}{2}\big) ... }{2},-1\big)\) ∈ Rbut (1, - 1) ∉ R as 1 + 1 (-1) = 0 \( ot>\) 0 ⇒ R is not transitive.

If a, b, c, x, y, z are all positve real numbers, then -Maths 9th

Description : If a, b, c, x, y, z are all positve real numbers, then -Maths 9th

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If R is a relation defined on the set of natural numbers N such that (a, b) R (c, d) if and only if a + d = b + c, then R is -Maths 9th

Description : If R is a relation defined on the set of natural numbers N such that (a, b) R (c, d) if and only if a + d = b + c, then R is -Maths 9th

Last Answer : (d) An equivalence relationWe can check the given properties as follows: Reflexive: Let (a, b) ∈ N x N. Then (a, b) ∈ N ⇒ a + b = b + a (Communtative law of Addition) ⇒ (a, b) R (b, a) ⇒ (a, b) R (a, ... , f) ⇒ (a, b) R (e, f) on N x N so R is transitive.Hence R is an equivalence relation on N N.

If a is a positive rational number and n is a positive integer greater than 1, prove that an is a rational number . -Maths 9th

Description : If a is a positive rational number and n is a positive integer greater than 1, prove that an is a rational number . -Maths 9th

Last Answer : We know that product of two rational number is always a rational number. Hence if a is a rational number then a2 = a x a is a rational number, a3 = 4:2 x a is a rational number. ∴ an = an-1 x a is a rational number.

If a is a positive rational number and n is a positive integer greater than 1, prove that an is a rational number . -Maths 9th

Description : If a is a positive rational number and n is a positive integer greater than 1, prove that an is a rational number . -Maths 9th

Last Answer : We know that product of two rational number is always a rational number. Hence if a is a rational number then a2 = a x a is a rational number, a3 = 4:2 x a is a rational number. ∴ an = an-1 x a is a rational number.

How many different numbers greater than 60000 can be formed with the digits 0, 2, 2, 6, 8? -Maths 9th

Description : How many different numbers greater than 60000 can be formed with the digits 0, 2, 2, 6, 8? -Maths 9th

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If a point O lies between two points P and R such that PO=OR then prove that PO= 1/2PR. -Maths 9th

Description : If a point O lies between two points P and R such that PO=OR then prove that PO= 1/2PR. -Maths 9th

Last Answer : THINGS WHICH ARE COINCIDE WITH EACH OTHER ARE EQUAL TO ONE ANOTHER PO+OR=PR 2PO=PR PO=OR PO=1/2PR HENCE PROVED

If r is real number such that |r| < 1 and if a = 5 (1 – r), then -Maths 9th

Description : If r is real number such that |r| < 1 and if a = 5 (1 – r), then -Maths 9th

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Let ABCD be a parallellogram. Let m and n be positive integers such that n < m < 2n. Let AC = 2 mn -Maths 9th

Description : Let ABCD be a parallellogram. Let m and n be positive integers such that n < m < 2n. Let AC = 2 mn -Maths 9th

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Express the following numbers in the form of p/q.(1)0.15 (2)0.00026 (3)23.434343434343...... (4)0.6666666.... -Maths 9th

Description : Express the following numbers in the form of p/q.(1)0.15 (2)0.00026 (3)23.434343434343...... (4)0.6666666.... -Maths 9th

Last Answer : (i) 0.15 = 15/100 = 3/20 (ii) 0.00026 = 26/100000 = 13/50000 (iii) 23.43434343.... Let p/q = 23.434343...... - (i) Multiply both side by 100:- 100 * p/q = 100 * 23.434343...... 100p/q = 2343. ... .434343...... 99p/q = 2320.0 p/q = 2320/99 (iv) 0.6666.... = 6/9 Proceed same as question no. (iii)

Express the following numbers in the form of p/q.(1)0.15 (2)0.00026 (3)23.434343434343...... (4)0.6666666.... -Maths 9th

Description : Express the following numbers in the form of p/q.(1)0.15 (2)0.00026 (3)23.434343434343...... (4)0.6666666.... -Maths 9th

Last Answer : (i) 0.15 = 15/100 = 3/20 (ii) 0.00026 = 26/100000 = 13/50000 (iii) 23.43434343.... Let p/q = 23.434343...... - (i) Multiply both side by 100:- 100 * p/q = 100 * 23.434343...... 100p/q = 2343. ... .434343...... 99p/q = 2320.0 p/q = 2320/99 (iv) 0.6666.... = 6/9 Proceed same as question no. (iii)

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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Consider the following relations R = {(x, y) | x, y are real numbers and x = wy for some rational number w}; -Maths 9th

Description : Consider the following relations R = {(x, y) | x, y are real numbers and x = wy for some rational number w}; -Maths 9th

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The set of all real numbers x, for which x^2 – |x + 2| + x > 0, is -Maths 9th

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Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ . -Maths 9th

Description : Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ . -Maths 9th

Last Answer : Join AQ and PC . Since ABCD is a parallelogram . ⇒ AB | | DC ⇒ AP | | QC ∵ AP and QC are parts of AB and DC respectively] Also, AP = CQ [given] Thus, APCQ is a parallelogram . We know that diagonals of a parallelogram bisect each other . Hence AC and PQ bisect each other .

l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

Description : l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

Last Answer : Though E, draw a line parallel to p intersecting L at G and n at H respectively. Since l | | m ⇒ AG | | BE and AB | | GE [by construction] ∴ Opposite sides of quadrilateral AGEB are ... ∠DGE = ∠FHE [alternate interior angles] By ASA congruence axiom, we have △DEG ≅ △FEH Hence, DE = EF

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

Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other. -Maths 9th

Description : Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other. -Maths 9th

Last Answer : According to question parallelogram ABCD such that AP = CQ.

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.

Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ . -Maths 9th

Description : Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ . -Maths 9th

Last Answer : Join AQ and PC . Since ABCD is a parallelogram . ⇒ AB | | DC ⇒ AP | | QC ∵ AP and QC are parts of AB and DC respectively] Also, AP = CQ [given] Thus, APCQ is a parallelogram . We know that diagonals of a parallelogram bisect each other . Hence AC and PQ bisect each other .

l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

Description : l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

Last Answer : Though E, draw a line parallel to p intersecting L at G and n at H respectively. Since l | | m ⇒ AG | | BE and AB | | GE [by construction] ∴ Opposite sides of quadrilateral AGEB are ... ∠DGE = ∠FHE [alternate interior angles] By ASA congruence axiom, we have △DEG ≅ △FEH Hence, DE = EF

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

Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other. -Maths 9th

Description : Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other. -Maths 9th

Last Answer : According to question parallelogram ABCD such that AP = CQ.

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.

In Fig. 8.37, ABCD is a parallelogram and P, Q are the points on the diagonal BD such that BQ = DP. Show what APCQ is a parallelogram. -Maths 9th

Description : In Fig. 8.37, ABCD is a parallelogram and P, Q are the points on the diagonal BD such that BQ = DP. Show what APCQ is a parallelogram. -Maths 9th

Last Answer : Solution :-

l,m and n are three parallel lines intersected by transversal p and q such that l,m and n cut-off equal intersepts AB and BC on p (Fig.8.55). Show that l,m and n cut - off equal intercepts DE and EF on q also. -Maths 9th

Description : l,m and n are three parallel lines intersected by transversal p and q such that l,m and n cut-off equal intersepts AB and BC on p (Fig.8.55). Show that l,m and n cut - off equal intercepts DE and EF on q also. -Maths 9th

Last Answer : Given:l∥m∥n l,m and n cut off equal intercepts AB and BC on p So,AB=BC To prove:l,m and n cut off equal intercepts DE and EF on q i.e.,DE=EF Proof:In △ACF, B is the mid-point of ... a triangle, parallel to another side, bisects the third side. Since E is the mid-point of DF DE=EF Hence proved.

AB and AC are two chords of a circle of radius r such that AB = 2AC. If p and q are the distances of AB and AC from the centre. Prove that -Maths 9th

Description : AB and AC are two chords of a circle of radius r such that AB = 2AC. If p and q are the distances of AB and AC from the centre. Prove that -Maths 9th

Last Answer : Draw OM perpendicular AB and ON perpendicular AC Join OA. In right △OAM, OA2 = OM2 + AM2 ⇒ r2 = p2 + (1/2AB)2 (Since,OM perpendicular AB, ∴ OM bisects AB ) ⇒ 1/4AB2 = r2 - p2 or AB2 = 4r2 - 4p2 ... ) and (ii), we have 4r2 - 4p2 = 16r2 - 16q2 or r2 - p2 = 4r2 - 4q2 or 4q2 = 3r2 + p2

In the given figure, (not drawn to scale), P is a point on AB such that AP : PB = 4 : 3. PQ is parallel to AC and QD -Maths 9th

Description : In the given figure, (not drawn to scale), P is a point on AB such that AP : PB = 4 : 3. PQ is parallel to AC and QD -Maths 9th

Last Answer : answer:

ABCD is a parallelogram. P is a point on AD such that AP = 1/3 AD and Q is a point on BC such that CQ = 1/3 BC. Prove that AQCP is a parallelogram. -Maths 9th

Description : ABCD is a parallelogram. P is a point on AD such that AP = 1/3 AD and Q is a point on BC such that CQ = 1/3 BC. Prove that AQCP is a parallelogram. -Maths 9th

Last Answer : answer: