the value f (249)2-(248)2 is -Maths 9th

the value f (249)2-(248)2 is -Maths 9th
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the value f (249)2-(248)2 is -Maths 9th

Description : the value f (249)2-(248)2 is -Maths 9th

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1 ,2 ,5 , 12 , 27 , 58 , 121, ? (a) 246 (b) 247 (c) 248 (d) 249

Description : 1 ,2 ,5 , 12 , 27 , 58 , 121, ? (a) 246 (b) 247 (c) 248 (d) 249

Last Answer : The sequence in this series is number into 2 plus the numbers in natural order starting . from 0.

2 . If the mean of the following distribution is 6 . Find the value of p ? x 2 4 6 10 P + 5 f 3 2 3 1 2 -Maths 9th

Description : 2 . If the mean of the following distribution is 6 . Find the value of p ? x 2 4 6 10 P + 5 f 3 2 3 1 2 -Maths 9th

Last Answer : Here's ur answer...

2 . If the mean of the following distribution is 6 . Find the value of p ? x 2 4 6 10 P + 5 f 3 2 3 1 2 -Maths 9th

Description : 2 . If the mean of the following distribution is 6 . Find the value of p ? x 2 4 6 10 P + 5 f 3 2 3 1 2 -Maths 9th

Last Answer : Here's ur answer...

Find the value of f(x) = 2x(square) + 7x + 3 at x= -2. -Maths 9th

Description : Find the value of f(x) = 2x(square) + 7x + 3 at x= -2. -Maths 9th

Last Answer : Solution :-

The molar mass of `CuSO_(4).5H_(2)O` is 249. Its equivalent mass in the reaction (a) and (b) would be (a) Reaction `CuSO_(4)+Kirarr"Product"` (b) Elec

Description : The molar mass of `CuSO_(4).5H_(2)O` is 249. Its equivalent mass in the reaction (a) and (b) would be (a) Reaction `CuSO_(4)+Kirarr"Product"` (b) Elec

Last Answer : The molar mass of `CuSO_(4).5H_(2)O` is 249. Its equivalent mass in the reaction (a) and (b) would be (a) ... ) 249 (b) 124.5 D. (s) 124.5 (b) 249

How minutes are there in 249 days?

Description : How minutes are there in 249 days?

Last Answer : 358560 minutes

Article 249 of the Indian Constitution is associated with the functions of - (1) The concurrent list (2) The state list (3) The president (4) The union list

Description : Article 249 of the Indian Constitution is associated with the functions of - (1) The concurrent list (2) The state list (3) The president (4) The union list

Last Answer : (2) The state list Explanation: Article 249 of the Indian Constitution is associated with the functions of the state list. If the Council of States has declared by resolution supported by not less ... in the resolution. Concurrent list of 52 items is mentioned in 7th schedule in article 246.

Under which article the parliament of India can legislate on any subject in the state list in national interest? (1) Article 229 (2) Article 230 (3) Article 247 (4) Article 249

Description : Under which article the parliament of India can legislate on any subject in the state list in national interest? (1) Article 229 (2) Article 230 (3) Article 247 (4) Article 249

Last Answer : (4) Article 249 Explanation: Under the Article 249, the Parliament of India can legislate on any subject in the state list in national interest. The Council of States has declared ... union territories. Article 247 gives power to the Parliament to provide establishment of certain additional courts.

Which article of the Indian Constitution empowers parliament to legislate on a subject of the state list? (1) Article 115 (2) Article 116 (3) Article 226 (4) Article 249

Description : Which article of the Indian Constitution empowers parliament to legislate on a subject of the state list? (1) Article 115 (2) Article 116 (3) Article 226 (4) Article 249

Last Answer : (4) Article 249 Explanation: Article 249 empowers the power of Parliament to legislate with respect to a matter in the state list in the national interest. Article 226 gives power of high courts to issue certain writs. Article 116 deals with votes on account, votes of credit and exceptional grants.

249/15 × 299/19 ÷ 14/99 = ? a) 1850 b) 1700 c) 1750 d) 1900 e) 2000

Description : 249/15 × 299/19 ÷ 14/99 = ? a) 1850 b) 1700 c) 1750 d) 1900 e) 2000

Last Answer : Answer: c)

259 256 ? 244 235 224 a) 250 b) 251 c) 253 d) 249 e) None of these

Description : 259 256 ? 244 235 224 a) 250 b) 251 c) 253 d) 249 e) None of these

Last Answer : Difference of odd numbers 3, 5, 7, 9, 11. Answer: b)

5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.31). Show that the line segments AF and EC trisect the diagonal BD. -Maths 9th

Description : 5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.31). Show that the line segments AF and EC trisect the diagonal BD. -Maths 9th

Last Answer : . Solution: Given that, ABCD is a parallelogram. E and F are the mid-points of sides AB and CD respectively. To show, AF and EC trisect the diagonal BD. Proof, ABCD is a parallelogram , AB || CD also, ... (i), DP = PQ = BQ Hence, the line segments AF and EC trisect the diagonal BD. Hence Proved.

4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.30). Show that F is the mid-point of BC. -Maths 9th

Description : 4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.30). Show that F is the mid-point of BC. -Maths 9th

Last Answer : . Solution: Given that, ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. To prove, F is the mid-point of BC. Proof, BD intersected EF at G. In ΔBAD, E is the ... point of BD and also GF || AB || DC. Thus, F is the mid point of BC (Converse of mid point theorem)

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]

a squar ABCD in which AC =BE when BC produced .A is joined to E prove that FG=GE when AE intersect BD at F and CD at G -Maths 9th

Description : a squar ABCD in which AC =BE when BC produced .A is joined to E prove that FG=GE when AE intersect BD at F and CD at G -Maths 9th

Last Answer : Please give the figure to get your answer, as it is necessary to have figure to answer the question related to geometry.

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

D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. -Maths 9th

Description : D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. -Maths 9th

Last Answer : Since the segment joining the mid points of any two sides of a triangle is half the third side and parallel to it. DE = 1 / 2 AC ⇒ DE = AF = CF EF = 1 / 2 AB ⇒ EF = AD = BD DF = 1 ... △DEF ≅ △AFD Thus, △DEF ≅ △CFE ≅ △BDE ≅ △AFD Hence, △ABC is divided into four congruent triangles.

In the fig, D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. -Maths 9th

Description : In the fig, D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. -Maths 9th

Last Answer : Since line segment joining the mid-points of two sides of a triangle is half of the third side. Therefore, D and E are mid-points of BC and AC respectively. ⇒ DE = 1 / 2 AB --- (i) E and F are the mid - ... CA ⇒ DE = EF = FD [using (i) , (ii) , (iii) ] Hence, DEF is an equilateral triangle .

ABCD is a square. E and F are respectively the mid - points of BC and CD. If R is the mid point of EF. -Maths 9th

Description : ABCD is a square. E and F are respectively the mid - points of BC and CD. If R is the mid point of EF. -Maths 9th

Last Answer : Since R is the mid point of EF . ∴ AR is the median in △AEF. As, a median of a triangle divides it into two triangles of equal area . ∴ ar(△AER) = ar(△AFR)

ABCD is a trapezium with parallel sides AB = a cm and DC = b cm . E and F are the mid - points of the non - parallel sides . -Maths 9th

Description : ABCD is a trapezium with parallel sides AB = a cm and DC = b cm . E and F are the mid - points of the non - parallel sides . -Maths 9th

Last Answer : Clearly, EF = AB + DC / 2 = a + b / 2 Let h be the height , then ar(Trap. ABFE) : ar(Trap. EFCD) ⇒ 1/2 [a + (a+b / 2)] × h : 1/2 [b + (a+b / 2)] × h ⇒ 2a + a + b / 2 : 2b + a + b / 2 ⇒ 3a + b : 3b + a

E and F are points on diagonal AC of a parallelogram ABCD such that AE = CF. -Maths 9th

Description : E and F are points on diagonal AC of a parallelogram ABCD such that AE = CF. -Maths 9th

Last Answer : According to question diagonal AC of a parallelogram ABCD such that AE = CF.

E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. -Maths 9th

Description : E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. -Maths 9th

Last Answer : According to question the mid-points of the non-parallel sides AD and BC of a trapezium ABCD.

ABCD is a square. E and F are respectively the mid-points of BC and CD. -Maths 9th

Description : ABCD is a square. E and F are respectively the mid-points of BC and CD. -Maths 9th

Last Answer : According to question prove that ar (ΔAER) = ar (ΔAFR).

ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F. -Maths 9th

Description : ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F. -Maths 9th

Last Answer : According to question find the area of the parallelogram ABCD.

a squar ABCD in which AC =BE when BC produced .A is joined to E prove that FG=GE when AE intersect BD at F and CD at G -Maths 9th

Description : a squar ABCD in which AC =BE when BC produced .A is joined to E prove that FG=GE when AE intersect BD at F and CD at G -Maths 9th

Last Answer : Please give the figure to get your answer, as it is necessary to have figure to answer the question related to geometry.

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

D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. -Maths 9th

Description : D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. -Maths 9th

Last Answer : Since the segment joining the mid points of any two sides of a triangle is half the third side and parallel to it. DE = 1 / 2 AC ⇒ DE = AF = CF EF = 1 / 2 AB ⇒ EF = AD = BD DF = 1 ... △DEF ≅ △AFD Thus, △DEF ≅ △CFE ≅ △BDE ≅ △AFD Hence, △ABC is divided into four congruent triangles.

In the fig, D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. -Maths 9th

Description : In the fig, D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. -Maths 9th

Last Answer : Since line segment joining the mid-points of two sides of a triangle is half of the third side. Therefore, D and E are mid-points of BC and AC respectively. ⇒ DE = 1 / 2 AB --- (i) E and F are the mid - ... CA ⇒ DE = EF = FD [using (i) , (ii) , (iii) ] Hence, DEF is an equilateral triangle .

ABCD is a square. E and F are respectively the mid - points of BC and CD. If R is the mid point of EF. -Maths 9th

Description : ABCD is a square. E and F are respectively the mid - points of BC and CD. If R is the mid point of EF. -Maths 9th

Last Answer : Since R is the mid point of EF . ∴ AR is the median in △AEF. As, a median of a triangle divides it into two triangles of equal area . ∴ ar(△AER) = ar(△AFR)

ABCD is a trapezium with parallel sides AB = a cm and DC = b cm . E and F are the mid - points of the non - parallel sides . -Maths 9th

Description : ABCD is a trapezium with parallel sides AB = a cm and DC = b cm . E and F are the mid - points of the non - parallel sides . -Maths 9th

Last Answer : Clearly, EF = AB + DC / 2 = a + b / 2 Let h be the height , then ar(Trap. ABFE) : ar(Trap. EFCD) ⇒ 1/2 [a + (a+b / 2)] × h : 1/2 [b + (a+b / 2)] × h ⇒ 2a + a + b / 2 : 2b + a + b / 2 ⇒ 3a + b : 3b + a

E and F are points on diagonal AC of a parallelogram ABCD such that AE = CF. -Maths 9th

Description : E and F are points on diagonal AC of a parallelogram ABCD such that AE = CF. -Maths 9th

Last Answer : According to question diagonal AC of a parallelogram ABCD such that AE = CF.

E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. -Maths 9th

Description : E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. -Maths 9th

Last Answer : According to question the mid-points of the non-parallel sides AD and BC of a trapezium ABCD.

ABCD is a square. E and F are respectively the mid-points of BC and CD. -Maths 9th

Description : ABCD is a square. E and F are respectively the mid-points of BC and CD. -Maths 9th

Last Answer : According to question prove that ar (ΔAER) = ar (ΔAFR).

ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F. -Maths 9th

Description : ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F. -Maths 9th

Last Answer : According to question find the area of the parallelogram ABCD.

Find the remainder when f(x)=4x(cube) - 12x(square) +14x - 3 is divided by g(x) = (2x-1). -Maths 9th

Description : Find the remainder when f(x)=4x(cube) - 12x(square) +14x - 3 is divided by g(x) = (2x-1). -Maths 9th

Last Answer : ____2x2-5x+4________________ 2x-1 ) 4x3-12x2+14x-3( 4x3-2x2 - + ____________ 0 -10x2+14x-3 ... + ___________ X+1

Find the remainder when f(x)=9x(cube) -x 3x(square) + 14x - 3 is divided by g(x)=(3x-1). -Maths 9th

Description : Find the remainder when f(x)=9x(cube) -x 3x(square) + 14x - 3 is divided by g(x)=(3x-1). -Maths 9th

Last Answer : Solution :-

In some countries temperature is measured in Fahrenheit, whereas in countries like India it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius: F = 9/5 C+ 32. If the temperature is 30°C, then what is the temperature in Fahrenheit? -Maths 9th

Description : In some countries temperature is measured in Fahrenheit, whereas in countries like India it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius: F = 9/5 C+ 32. If the temperature is 30°C, then what is the temperature in Fahrenheit? -Maths 9th

Last Answer : Solution :-

D,E and F are the mid-points of the sides BC,CA and AB,respectively of an equilateral triangle ABC.Show that △DEF is also an euilateral triangle -Maths 9th

Description : D,E and F are the mid-points of the sides BC,CA and AB,respectively of an equilateral triangle ABC.Show that △DEF is also an euilateral triangle -Maths 9th

Last Answer : Solution :-

E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. Prove that EF||AB and EF = 1/2 (AB +CD). -Maths 9th

Description : E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. Prove that EF||AB and EF = 1/2 (AB +CD). -Maths 9th

Last Answer : Solution :-

In Fig. 8.53,ABCD is a parallelogram and E is the mid - point of AD. A line through D, drawn parallel to EB, meets AB produced at F and BC at L.Prove that (i) AF = 2DC (ii) DF = 2DL -Maths 9th

Description : In Fig. 8.53,ABCD is a parallelogram and E is the mid - point of AD. A line through D, drawn parallel to EB, meets AB produced at F and BC at L.Prove that (i) AF = 2DC (ii) DF = 2DL -Maths 9th

Last Answer : Given, E is mid point of AD Also EB∥DF ⇒ B is mid point of AF [mid--point theorem] so, AF=2AB (1) Since, ABCD is a parallelogram, CD=AB ⇒AF=2CD AD∥BC⇒LB∥AD In ΔFDA ⇒LB∥AD ⇒LDLF​=ABFB​=1 from (1) ⇒LF=LD so, DF=2DL

In Fig. 9.23, ABCD is a parallelogram in which BC is produced to E such A B that CE = BC. AE intersects CD at F. If area of △BDF = 3 cm2, find the area of parallelogram ABCD. -Maths 9th

Description : In Fig. 9.23, ABCD is a parallelogram in which BC is produced to E such A B that CE = BC. AE intersects CD at F. If area of △BDF = 3 cm2, find the area of parallelogram ABCD. -Maths 9th

Last Answer : Solution :-

Bisectors of angles A, B and C of a triangle ABC intersects its circumcircle at D, E and F respectively. Prove that angles of triangle DEF are 90° - A/2, 90° - B/2 and 90° - C/2. -Maths 9th

Description : Bisectors of angles A, B and C of a triangle ABC intersects its circumcircle at D, E and F respectively. Prove that angles of triangle DEF are 90° - A/2, 90° - B/2 and 90° - C/2. -Maths 9th

Last Answer : We have ∠BED = ∠BAD (Angles in the same segment) ⇒ ∠BED = 1/2∠A ...(i) Also, ∠BEF = ∠BCF (Angles in the same segment) ⇒ ∠BEF = 1/2∠C ...(ii) From (i) and (ii) ∠BED + ∠BEF = 1/2∠A + 1/2∠C ∠DEF ... ∠A + ∠C) ⇒ ∠DEF = 1/2(180° - ∠B) (Since, ∠A + ∠B + ∠C = 180°) ⇒ ∠DEF = 90° - 1/2∠B

ABCD is a rhombus and AB is produved to E and F such that AE=AB=BF prove that ED and FC are perpendicular to each other -Maths 9th

Description : ABCD is a rhombus and AB is produved to E and F such that AE=AB=BF prove that ED and FC are perpendicular to each other -Maths 9th

Last Answer : This answer was deleted by our moderators...

Let ABC be a triangle. Let D, E, F be points respectively on segments BC, CA, AB such that AD, BE and CF concur at point K. -Maths 9th

Description : Let ABC be a triangle. Let D, E, F be points respectively on segments BC, CA, AB such that AD, BE and CF concur at point K. -Maths 9th

Last Answer : answer:

f(x) = x^4 – 2x^3 + 3x^2 – ax + b is a polynomial such that when it is divided by (x – 1) and (x + 1), the remainders are respectively 5 and 19. -Maths 9th

Description : f(x) = x^4 – 2x^3 + 3x^2 – ax + b is a polynomial such that when it is divided by (x – 1) and (x + 1), the remainders are respectively 5 and 19. -Maths 9th

Last Answer : answer:

A cubic polynomial f(x) is such that f(1) = 1, f(2) = 2, f(3) = 3 and f(4) = 5, then f(6) equals : -Maths 9th

Description : A cubic polynomial f(x) is such that f(1) = 1, f(2) = 2, f(3) = 3 and f(4) = 5, then f(6) equals : -Maths 9th

Last Answer : answer:

When a polynomial f(x) is divided by (x – 3) and (x + 6), the respective remainders are 7 and 22. What is the remainder when f(x) is -Maths 9th

Description : When a polynomial f(x) is divided by (x – 3) and (x + 6), the respective remainders are 7 and 22. What is the remainder when f(x) is -Maths 9th

Last Answer : answer:

If p(x) is a common multiple of degree 6 of the polynomials f(x) = x^3 + x^2 – x – 1 and g(x) = x^3 – x^2 + x – 1, then which -Maths 9th

Description : If p(x) is a common multiple of degree 6 of the polynomials f(x) = x^3 + x^2 – x – 1 and g(x) = x^3 – x^2 + x – 1, then which -Maths 9th

Last Answer : answer:

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.