Check whether the following are quadratic equations: (i) (x+ 1)2=2(x-3) (ii) x – 2x = (- 2) (3-x) (iii) (x – 2) (x + 1) = (x – 1) (x + 3) (iv) (x – 3) (2x + 1) = x (x + 5) (v) (2x – 1) (x – 3) = (x + 5) (x – 1) (vi) x2 + 3x + 1 = (x – 2)2 (vii) (x + 2)3 = 2x(x2 – 1) (viii) x3 -4x2 -x + 1 = (x-2)3 -Maths 10th

Check whether the following are quadratic equations: (i) (x+ 1)2=2(x-3) (ii) x – 2x = (- 2) (3-x) (iii) (x – 2) (x + 1) = (x – 1) (x + 3) (iv) (x – 3) (2x + 1) = x (x + 5) (v) (2x – 1) (x – 3) = (x + 5) (x – 1) (vi) x2 + 3x + 1 = (x – 2)2 (vii) (x + 2)3 = 2x(x2 – 1) (viii) x3 -4x2 -x + 1 = (x-2)3 -Maths 10th
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Find the polynomial of least degree which should be subtracted from the polynomial x4 + 2x3 – 4x2 + 6x – 3 so that it is exactly divisible by x2 – x + 1. -Maths 10th

Description : Find the polynomial of least degree which should be subtracted from the polynomial x4 + 2x3 – 4x2 + 6x – 3 so that it is exactly divisible by x2 – x + 1. -Maths 10th

Last Answer : Here, p(x) = x4 + 2x3 - 4x2 + 6x - 3, g(x) = x2 - x +1 On dividing p(x) by g(x) Therefore (x-1) must be subtracted from the polynomial p(x) to make it divisible by g(x).

Convert each of the following fraction into a percentage: (i) 47 100 (ii) 9 20 (iii) 3 8 (iv) 8 125 (v) 19 500 (vi) 4 15 (vii) 2 3 (viii) 1 3 5 -Maths

Description : Convert each of the following fraction into a percentage: (i) 47 100 (ii) 9 20 (iii) 3 8 (iv) 8 125 (v) 19 500 (vi) 4 15 (vii) 2 3 (viii) 1 3 5 -Maths

Last Answer : We have the following: (i) 47100=47100×100% = 47% (ii) 920=920×100% =9×5%= 45% (iii) 38=38×100% =3×252%=752%=3712% (iv) 8125=8125×100%=8×45%=325%=6.4% (v) ... ;203%=803%=2623% (vii) 23=23×100%=2003%=6623% (viii) 135=85=85×100%=(8×20)%=160%

Convert each of the following into a fraction: (i) 32% (ii) 614% (iii) 2623% (iv) 120% (v) 6.25% (vi) 0.8% (vii) 0.06% (viii) 22.75% -Maths

Description : Convert each of the following into a fraction: (i) 32% (ii) 614% (iii) 2623% (iv) 120% (v) 6.25% (vi) 0.8% (vii) 0.06% (viii) 22.75% -Maths

Last Answer : answer:

I have some matchsticks set up like this: I + XI = X. If you do not know Roman numerals, here is a quick starter: I=1 II=2 III=3 IV=4 V=5 VI=6 VII=7 VIII=8 IX=9 X=10 XI=11 XII=12 XIII=13 XIV=14 XV=15...... Of course, I + XI = X is wrong. You are told to move the minimum number of matchsticks to make the question correct. So, how many do you have to move? -Riddles

Description : I have some matchsticks set up like this: I + XI = X. If you do not know Roman numerals, here is a quick starter: I=1 II=2 III=3 IV=4 V=5 VI=6 VII=7 VIII=8 IX=9 X=10 XI=11 ... are told to move the minimum number of matchsticks to make the question correct. So, how many do you have to move? -Riddles

Last Answer : Most people would get to one move at the least, but the real answer is zero moves. If you rotate the whole question around 180 degrees, it will display this: X = IX + I

A company has to choose between two projects whose cash flows are as indicated below; Project 1: i. Investment – Rs. 15 Lakhs ii. Annual cost savings – Rs. 4 lakhs. iii. Bi-annual maintenance cost – Rs. 50,000/- iv. Reconditioning and overhaul during 5th year: 6 lakhs v. Life of the project – 8 years vi. Salvage value – Rs. 5 lakhs Project 2: vii. Investment – Rs. 14 Lakhs viii. Annual cost savings – Rs. 3.5 lakhs. ix. Annual Maintenance cost – Rs. 20,000/- x. Reconditioning and overhaul during 4th year: 5 lakhs xi. Life of the project – 8 years xii. Salvage Value- 2 lakhs Which project should the company choose? The annual discount rate is 12

Description :  A company has to choose between two projects whose cash flows are as indicated  below; Project 1:  i. Investment - Rs. 15 Lakhs ii. Annual cost savings - Rs. 4 lakhs. iii. Bi-annual ... years xii. Salvage Value- 2 lakhs Which project should the company choose? The annual discount rate is 12

Last Answer : Year Project 1 Project 2 Outgo Saving NPV Outgo Saving NPV 0 15.0 0 =-15.0 14.0 0 = -14 1 0 4.0 = (4 / (1+.12)1 = 3.571 0.2 3.5 = (3.3 / (1+.12)1 = 2.95 2 0.5 4.0 = (3.5 / (1+.12)2 = 2.79 0.2 3.5 ... (1+.12)8 = 3.43 0.2 5.5  (3.5+2) = (5.3 / (1+.12)8 = 2.14 NPV = + 2.301 @12% = + 0.15

Cold conc. `HNO_(3)` will completely dissolve: (i) Pb (ii) `Pb_(3)O_(4)` (iii) Fe (iv) Sn (v) Mg (vi) MgO (vii) Hg (viii) Au (ix) Ag (x) Pt

Description : Cold conc. `HNO_(3)` will completely dissolve: (i) Pb (ii) `Pb_(3)O_(4)` (iii) Fe (iv) Sn (v) Mg (vi) MgO (vii) Hg (viii) Au (ix) Ag (x) Pt

Last Answer : Cold conc. `HNO_(3)` will completely dissolve: (i) Pb (ii) `Pb_(3)O_(4)` (iii) Fe (iv) Sn (v) Mg (vi) MgO (vii) Hg (viii) Au (ix) Ag (x) Pt

Cold conc. `HNO_(3)` will completely dissolve: (i) Pb (ii) `Pb_(3)O_(4)` (iii) Fe (iv) Sn (v) Mg (vi) MgO (vii) Hg (viii) Au (ix) Ag (x) Pt

Description : Cold conc. `HNO_(3)` will completely dissolve: (i) Pb (ii) `Pb_(3)O_(4)` (iii) Fe (iv) Sn (v) Mg (vi) MgO (vii) Hg (viii) Au (ix) Ag (x) Pt

Last Answer : Cold conc. `HNO_(3)` will completely dissolve: (i) Pb (ii) `Pb_(3)O_(4)` (iii) Fe (iv) Sn (v) Mg (vi) MgO (vii) Hg (viii) Au (ix) Ag (x) Pt

Differentiate between any five of the following. (i) Artery and vein (ii) Hard water and soft water (iii) E-mail and Snail mail (iv) Apes and monkey (v) Hydrostatics and hydrodynamics (vi) Comet and meteor (vii) Barrage and dam (viii) Electron and hole (ix) Isobars and isotopes (x) Autopsy and biopsy

Description : Differentiate between any five of the following. (i) Artery and vein (ii) Hard water and soft water (iii) E-mail and Snail mail (iv) Apes and monkey (v) Hydrostatics and hydrodynamics (vi) Comet and ... (vii) Barrage and dam (viii) Electron and hole (ix) Isobars and isotopes (x) Autopsy and biopsy

Last Answer : Answers: (i) Artery and vein: Arteries: Muscular blood vessels that carry blood away from the heart. All arteries, with the exception of the pulmonary and umbilical arteries, carry oxygenated blood ... to determine cause of death Biopsy: Removal and study of a tissue sample for diagnostic purposes

Name the quantities which are measured by the following units. (i) Newton (ii) Joule (iii) Watt (iv) Volt (v) Light year (vi) Angstrom (vii) Acre-foot (viii) Becquerel (ix) Hertz (x) Cusec

Description : Name the quantities which are measured by the following units. (i) Newton (ii) Joule (iii) Watt (iv) Volt (v) Light year (vi) Angstrom (vii) Acre-foot (viii) Becquerel (ix) Hertz (x) Cusec

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Which quantity do the following units measure: (i) Volt (ii) Coulomb (iii) Walt (iv) Ohm (v) Mho (vi) Ampere (vii) Dyne (viii) Celsius (ix) Joule (x) Calorie

Description : Which quantity do the following units measure: (i) Volt (ii) Coulomb (iii) Walt (iv) Ohm (v) Mho (vi) Ampere (vii) Dyne (viii) Celsius (ix) Joule (x) Calorie

Last Answer : Answers: (i) Voltage (ii) Charge of Electricity (iii) Power (iv) Resistance (v) Conductivity (vi) Current (vii) Force (viii) Temperature (ix) Energy (x) Heat

Cold dil. `H_(2)SO_(4)` will completely dissolve : (i) Pb (ii) `Fe_(3)O_(4)` (iii) Fe (iv) Cu (v) Mg (vi) MgO (vii) `CoCO_(3)` (viii) `CuCO_(2)` (ix)

Description : Cold dil. `H_(2)SO_(4)` will completely dissolve : (i) Pb (ii) `Fe_(3)O_(4)` (iii) Fe (iv) Cu (v) Mg (vi) MgO (vii) `CoCO_(3)` (viii) `CuCO_(2)` (ix)

Last Answer : Cold dil. `H_(2)SO_(4)` will completely dissolve : (i) Pb (ii) `Fe_(3)O_(4)` (iii) Fe (iv) Cu (v) ... ) `CoCO_(3)` (viii) `CuCO_(2)` (ix) `SrCO_(3)`

(i) `NH_(3)` (ii) `N_(2)H_(4)` (iii) `HN_(3)` (iv) `PH_(3)` (v) `H_(2)S` (vi) `AsH_(3)` (vii) `SbH_(3)` (viii) `H_(2)Se` (ix) `H_(2)Te` Number of mole

Description : (i) `NH_(3)` (ii) `N_(2)H_(4)` (iii) `HN_(3)` (iv) `PH_(3)` (v) `H_(2)S` (vi) `AsH_(3)` (vii) `SbH_(3)` (viii) `H_(2)Se` (ix) `H_(2)Te` Number of mole

Last Answer : (i) `NH_(3)` (ii) `N_(2)H_(4)` (iii) `HN_(3)` (iv) `PH_(3)` (v) `H_(2) ... lone pair of electrons on the central atom is present in pure s-orbital

Cold dil. `H_(2)SO_(4)` will completely dissolve : (i) Pb (ii) `Fe_(3)O_(4)` (iii) Fe (iv) Cu (v) Mg (vi) MgO (vii) `CoCO_(3)` (viii) `CuCO_(2)` (ix)

Description : Cold dil. `H_(2)SO_(4)` will completely dissolve : (i) Pb (ii) `Fe_(3)O_(4)` (iii) Fe (iv) Cu (v) Mg (vi) MgO (vii) `CoCO_(3)` (viii) `CuCO_(2)` (ix)

Last Answer : Cold dil. `H_(2)SO_(4)` will completely dissolve : (i) Pb (ii) `Fe_(3)O_(4)` (iii) Fe (iv) Cu (v) ... ) `CoCO_(3)` (viii) `CuCO_(2)` (ix) `SrCO_(3)`

(i) `NH_(3)` (ii) `N_(2)H_(4)` (iii) `HN_(3)` (iv) `PH_(3)` (v) `H_(2)S` (vi) `AsH_(3)` (vii) `SbH_(3)` (viii) `H_(2)Se` (ix) `H_(2)Te` Number of mole

Description : (i) `NH_(3)` (ii) `N_(2)H_(4)` (iii) `HN_(3)` (iv) `PH_(3)` (v) `H_(2)S` (vi) `AsH_(3)` (vii) `SbH_(3)` (viii) `H_(2)Se` (ix) `H_(2)Te` Number of mole

Last Answer : (i) `NH_(3)` (ii) `N_(2)H_(4)` (iii) `HN_(3)` (iv) `PH_(3)` (v) `H_(2) ... lone pair of electrons on the central atom is present in pure s-orbital

How many of the following are oxides ores (i) Carnalite (ii) Cuprite (iii) Cassiterite (iv) Chromite (v) Cinnabar (vi) Calamine (vii) Cerussite (Viii)

Description : How many of the following are oxides ores (i) Carnalite (ii) Cuprite (iii) Cassiterite (iv) Chromite (v) Cinnabar (vi) Calamine (vii) Cerussite (Viii)

Last Answer : How many of the following are oxides ores (i) Carnalite (ii) Cuprite (iii) Cassiterite (iv) ... (vii) Cerussite (Viii) Chalocopyrite (ix) Chalcocite

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Description : Identify the components labelled A, B, C and D in the diagram below from the list (i) to (viii) given along with Components : (i) Cristae of mitochondria (ii) Inner membrane of mitochondria (iii) Cytoplasm (iv) Smooth endoplasmic ... viii) (vi) (c) (vi) (v) (iv) (vii) (d) (v) (i) (iii) (ii)

Last Answer : (a) (v) (iv) (viii) (iii)

Compare pin type and suspension insulators on given points: (i) Position of insulator on cross arm. (ii) Position of conductor on insulator. (iii) Reaction on cross arm. (iv) Possibility of flash over due to large birds (v) Maintenance/Replacement cost (vi) Maximum voltage level (vii) Effect on height of supporting structure. (viii) Life

Description : Compare pin type and suspension insulators on given points: (i) Position of insulator on cross arm. (ii) Position of conductor on insulator. (iii) Reaction on cross arm. (iv) Possibility of ... Replacement cost (vi) Maximum voltage level (vii) Effect on height of supporting structure. (viii) Life 

Last Answer : S.No  Points Pin Type insulator Suspension or Disc Type insulator 1 Position of insulator on cross arm It is fixed on top of cross arm by using galvanized steel pin. So it is ... so to maintain minimum ground clearance height of pole increase.  8 Life Less More

Find the zeroes of the of the polynomials p(x) = 4x2 – 12x + 9 -Maths 10th

Description : Find the zeroes of the of the polynomials p(x) = 4x2 – 12x + 9 -Maths 10th

Last Answer : P(x)= 4x² - 12x + 9 4x²- 12x + 9 = 0 4x²- 6x - 6x +9 = 0 2x(2x - 3) - 3(2x - 3) = 0 (2x - 3) (2x - 3) x = 3/2 x = 3/2 x = 3/2 ; 3/2 are the zeros of the given polynomial.

If the roots of the equation (b – c)x2 + (c – a)x + (a – b) = 0 are equal, then prove that 2b = a + c. -Maths 10th

Description : If the roots of the equation (b – c)x2 + (c – a)x + (a – b) = 0 are equal, then prove that 2b = a + c. -Maths 10th

Last Answer : following is the equation of 2b = a+c =

If the roots ff the equation (c2 – ab)x2 – 2(a2 – bc)x + b2 – ac = 0 are equal, then prove that either a = 0 or a3 + b3 + c3 = 3abc -Maths 10th

Description : If the roots ff the equation (c2 – ab)x2 – 2(a2 – bc)x + b2 – ac = 0 are equal, then prove that either a = 0 or a3 + b3 + c3 = 3abc -Maths 10th

Last Answer : (c2 – ab) x2 + 2(bc - a2 ) x+ (b2 – ac) = 0 Comparing with Ax2 + Bx + C = 0 A = (c2 – ab), B = 2(bc - a2 ) and C = b2 – ac According to the question, B2 - 4AC = 0 Put the values in the above equation we get 4a(a3 + b3 + c3 -3abc) = 0 hence, a = 0 or a3 + b3 + c3 = 3ab

Compare detonation and diesel knock on the basis of (i) Fuel Ignition Temperature (ii) Compression Ratio (iii) Speed (iv) inlet temperature (v) inlet pressure (vi) cylinder wall temperature (vii) cylinder size

Description : Compare detonation and diesel knock on the basis of (i) Fuel Ignition Temperature (ii) Compression Ratio (iii) Speed (iv) inlet temperature (v) inlet pressure (vi) cylinder wall temperature (vii) cylinder size 

Last Answer : Comparison between detonation and diesel knock: Parameter To avoid Detonation in SI Engine  To avoid Diesel knock i) Fuel Ignition Temperature High Low ii) Compression Ratio ... Cylinder wall temperature Low High viii) Cylinder size Small Large

Evaluate each of the following using identities: (i) (2x –1x)2 (ii) (2x + y) (2x – y) (iii) (a2b – b2a)2 (iv) (a – 0.1) (a + 0.1) (v) (1.5.x2 – 0.3y2) (1.5x2 + 0.3y2) -Maths 9th

Description : Evaluate each of the following using identities: (i) (2x –1x)2 (ii) (2x + y) (2x – y) (iii) (a2b – b2a)2 (iv) (a – 0.1) (a + 0.1) (v) (1.5.x2 – 0.3y2) (1.5x2 + 0.3y2) -Maths 9th

Last Answer : (i) (2x - 1/x)2 [Use identity: (a - b)2 = a2 + b2 - 2ab ] (2x - 1/x)2 = (2x) 2 + (1/x)2 - 2 (2x)(1/x) = 4x2 + 1/x2 - 4 (ii) (2x + y) (2x - y) [Use identity: (a - b)(a + b) = a2 - b 2 ] (2x + y) (2x - ... ) = a2 - b 2 ](1.5 x 2 - 0.3y2 ) (1.5x2 + 0.3y2 ) = (1.5 x 2 ) 2 - (0.3y2 ) 2 = 2.25 x4 - 0.09y4

Simplify: (i) (a + b + c)2 + (a – b + c)2 (ii) (a + b + c)2 – (a – b + c)2 (iii) (a + b + c)2 + (a – b + c)2 + (a + b – c)2 (iv) (2x + p – c)2 – (2x – p + c)2 (v) (x2 + y2 – z2)2 – (x2 – y2 + z2)2 -Maths 9th

Description : Simplify: (i) (a + b + c)2 + (a – b + c)2 (ii) (a + b + c)2 – (a – b + c)2 (iii) (a + b + c)2 + (a – b + c)2 + (a + b – c)2 (iv) (2x + p – c)2 – (2x – p + c)2 (v) (x2 + y2 – z2)2 – (x2 – y2 + z2)2 -Maths 9th

Last Answer : answer:

How many solutions does the pair of equations y = 0 and y = -5 have? -Maths 10th

Description : How many solutions does the pair of equations y = 0 and y = -5 have? -Maths 10th

Last Answer : y = 0 and y = -5 are Parallel lines, hence no solution.

If the roots of the equations ax2 + 2bx + c = 0 and bx2 – are simultanously real then prove that b2 = 4ac. -Maths 10th

Description : If the roots of the equations ax2 + 2bx + c = 0 and bx2 – are simultanously real then prove that b2 = 4ac. -Maths 10th

Last Answer : Let D1​ and D2​ be the discriminants of the two equations, then D1​≥0 and D2​≥0 ⇒4b2−4ac≥0 and 4ac−4b2≥0 ⇒b2≥ac and ac≥b2 ⇒b2=ac

Find the distance between the following pairs of points: (i) (2, 3), (4, 1) (ii) (-5, 7), (-1, 3) (iii) (a, b), (- a, – b) -Maths 10th

Description : Find the distance between the following pairs of points: (i) (2, 3), (4, 1) (ii) (-5, 7), (-1, 3) (iii) (a, b), (- a, – b) -Maths 10th

Last Answer : Distance between two points (x1 ,y1 ) and (x2 ,y2 ) is (x1 −x2 )2+(y1 −y2 )2 (i) Distance between(2,3) and (4,1) is =(2−4)2+(3−1)2 =(−2)2+(2)2 =4+4 =8 =22 (ii) Distance between (−5,7) and (−1,3) ... 42 (iii )Distance between (a,b) and(−a,−b) is =(a−(−a))2+(b−(−b))2 =(2a)2+(2b)2 =4a2+4b2 =2a2+b2

Use Euclid’s division algorithm to find the HCF of: i. 135 and 225 ii. 196 and 38220 iii. 867 and 225 -Maths 10th

Description : Use Euclid’s division algorithm to find the HCF of: i. 135 and 225 ii. 196 and 38220 iii. 867 and 225 -Maths 10th

Last Answer : 135 and 225 As you can see, from the question 225 is greater than 135. Therefore, by Euclid's division algorithm, we have, 225 = 135 1 + 90 Now, remainder 90 ≠ 0, thus again using division lemma for 90, we get, ... (867,225) = HCF(225,102) = HCF(102,51) = 51. Hence, the HCF of 867 and 225 is 51

Use Euclid’s division algorithm to find the HCF of : (i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255 -Maths 10th

Description : Use Euclid’s division algorithm to find the HCF of : (i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255 -Maths 10th

Last Answer : (i) Given numbers are 135 and 225. On applying Euclid's division algorithm, we have 225 = 135 x 1 + 90 Since the remainder 90 ≠ 0, so again we apply Euclid's division algorithm to 135 and 90, to get 135 = 90 x ... division lemma, to a =867 and b=255 to find q and r such that 867 = 255q + r, 0 ≤ r

Solve the following in equalities `(i) |x+7| gt 5` `(ii) |x+3| lt 10` `(iii) (x+2) lt |x^(2)+3x+5|``(iv) |(2x-1)/(x-1)| gt 2` `(v) |x-6| le x^(2)-5x+9

Description : Solve the following in equalities `(i) |x+7| gt 5` `(ii) |x+3| lt 10` `(iii) (x+2) lt |x^(2)+3x+5|``(iv) |(2x-1)/(x-1)| gt 2` `(v) |x-6| le x^(2)-5x+9

Last Answer : Solve the following in equalities `(i) |x+7| gt 5` `(ii) |x+3| lt 10` `(iii) (x+2) lt |x^(2)+3x+5|``(iv) ... -5) lt 0` `(viii) |x-1|+|x-2|+|x-3| le 6`

Solve for `x` `(i) |x+1|=4x+3` `(ii) |x+1|=|x+3|` `(iii) 7|x-2|-|x-7|=5` `(iv) ||x-1|-2|=6x+8` `(v) |2x^(2)-3x+1|=|x^(2)+x-3|`

Description : Solve for `x` `(i) |x+1|=4x+3` `(ii) |x+1|=|x+3|` `(iii) 7|x-2|-|x-7|=5` `(iv) ||x-1|-2|=6x+8` `(v) |2x^(2)-3x+1|=|x^(2)+x-3|`

Last Answer : Solve for `x` `(i) |x+1|=4x+3` `(ii) |x+1|=|x+3|` `(iii) 7|x-2|-|x-7|=5` `(iv) ||x-1|-2|=6x+8` `(v) |2x^(2)-3x+1|=|x^(2)+x-3|`

For the complex [Fe(en)2Cl2], Cl, (en = ethylene diamine), identify (i) the oxidation number of iron, (ii) the hybrid orbitals and the shape of the complex, (iii) the magnetic behaviour of the complex, (iv) the number of geometrical isomers, (v) whether there is an optical isomer also, and (vi) name of the complex. (At. no. of Fe = 26) \ -Chemistry

Description : For the complex [Fe(en)2Cl2], Cl, (en = ethylene diamine), identify (i) the oxidation number of iron, (ii) the hybrid orbitals and the shape of the complex, (iii) the magnetic behaviour of the complex, (iv) ... an optical isomer also, and (vi) name of the complex. (At. no. of Fe = 26) \ -Chemistry

Last Answer : i) [Fe(en)2Cl2] Cl or x + 0 + 2 (-1) + (-1) = 0 x + (- 3) = 0 or x = + 3 ∴ Oxidation number of iron, x = + 3 (ii) The complex has two bidentate ligands and two ... )2Cl2] Cl, only cis-isomer shows optical isomerism. (vi) Name of complex: Dichloridobis (ethane-1, 2- diamine) Iron (III) chloride

It was India that took the matter to the UN under Chapter _____ of the UN Charter dealing with Pacific Settlement of Disputes A. VI B. V C. VII D. III

Description : It was India that took the matter to the UN under Chapter _____ of the UN Charter dealing with Pacific Settlement of Disputes A. VI B. V C. VII D. III

Last Answer : ANSWER: A

It was India that took the matter to the UN under Chapter _____ of the UN Charter dealing with Pacific Settlement of Disputes A. VI B. V C. VII D. III

Description : It was India that took the matter to the UN under Chapter _____ of the UN Charter dealing with Pacific Settlement of Disputes A. VI B. V C. VII D. III

Last Answer : ANSWER: A

It was India that took the matter to the UN under Chapter _____ of the UN Charter dealing with Pacific Settlement of Disputes A. VI B. V C. VII D. III

Description : It was India that took the matter to the UN under Chapter _____ of the UN Charter dealing with Pacific Settlement of Disputes A. VI B. V C. VII D. III

Last Answer : ANSWER: A

Find: (i) 25% of Rs 76 (ii) 20% of Rs 132 (iii) 7.5% of 600 m (iv) 313% of 90 km (v) 8.5% of 5 kg (vi) 20% of 12 litres -Maths

Description : Find: (i) 25% of Rs 76 (ii) 20% of Rs 132 (iii) 7.5% of 600 m (iv) 313% of 90 km (v) 8.5% of 5 kg (vi) 20% of 12 litres -Maths

Last Answer : answer:

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]

What percentage is (i) Rs 15 of Rs 120? (ii) 36 minutes of 2 hours? (iii) 8 hours of 2 days (iv) 160 metres of 4 km? (v) 175 mL of 1 litre? (vi) 25 paise of Rs 4? -Maths

Description : What percentage is (i) Rs 15 of Rs 120? (ii) 36 minutes of 2 hours? (iii) 8 hours of 2 days (iv) 160 metres of 4 km? (v) 175 mL of 1 litre? (vi) 25 paise of Rs 4? -Maths

Last Answer : answer:

Find: (i) 32% of 425 (ii) 1623% of 16 (iii) 6.5% of 400 (iv) 136% of 70 (v) 2.8% of 35 (vi) 0.6% of 45 -Maths

Description : Find: (i) 32% of 425 (ii) 1623% of 16 (iii) 6.5% of 400 (iv) 136% of 70 (v) 2.8% of 35 (vi) 0.6% of 45 -Maths

Last Answer : answer:

Evaluate each of the following: (i) (103)3 (ii) (98)3 (iii) (9.9)3 (iv) (10.4)3 (v) (598)3 (vi) (99)3 -Maths 9th

Description : Evaluate each of the following: (i) (103)3 (ii) (98)3 (iii) (9.9)3 (iv) (10.4)3 (v) (598)3 (vi) (99)3 -Maths 9th

Last Answer : answer:

During the British rule the only British King to visit India and hold his Darbar was: A. Edward VII B. George V C. James II D. Edward VI

Description : During the British rule the only British King to visit India and hold his Darbar was: A. Edward VII B. George V C. James II D. Edward VI

Last Answer : ANSWER: B

During the British rule the only British King to visit India and hold his Darbar was: A. Edward VII B. George V C. James II D. Edward VI

Description : During the British rule the only British King to visit India and hold his Darbar was: A. Edward VII B. George V C. James II D. Edward VI

Last Answer : ANSWER: B

During the British rule the only British King to visit India and hold his Darbar was: A. Edward VII B. George V C. James II D. Edward VI

Description : During the British rule the only British King to visit India and hold his Darbar was: A. Edward VII B. George V C. James II D. Edward VI

Last Answer : ANSWER: B

During the British rule the only British King to visit India and hold his Darbar was: A. Edward VII B. George V C. James II D. Edward VI

Description : During the British rule the only British King to visit India and hold his Darbar was: A. Edward VII B. George V C. James II D. Edward VI

Last Answer : ANSWER: B

Solve the following equations : `(i) log_(x)(4x-3)=2` `(ii) log_2)(x-1)+log_(2)(x-3)=3` `(iii) log_(2)(log_(8)(x^(2)-1))=0` `(iv) 4^(log_(2)x)-2x-3=0`

Description : Solve the following equations : `(i) log_(x)(4x-3)=2` `(ii) log_2)(x-1)+log_(2)(x-3)=3` `(iii) log_(2)(log_(8)(x^(2)-1))=0` `(iv) 4^(log_(2)x)-2x-3=0`

Last Answer : Solve the following equations : `(i) log_(x)(4x-3)=2` `(ii) log_2)(x-1)+log_(2)(x-3)=3` `(iii) log_(2)(log_(8)(x^(2)-1))=0` `(iv) 4^(log_(2)x)-2x-3=0`

If `f(x)=|(1,x,x+1),(2x,x(x-1),(x+1)x),(3x(x-1),x(x-1)(x-2),(x+1)x(x-1))|,` then f(100) is equal to - (i)`0` (ii)`1` (iii)`100` (iv)`-100`

Description : If `f(x)=|(1,x,x+1),(2x,x(x-1),(x+1)x),(3x(x-1),x(x-1)(x-2),(x+1)x(x-1))|,` then f(100) is equal to - (i)`0` (ii)`1` (iii)`100` (iv)`-100`

Last Answer : If `f(x)=|(1,x,x+1),(2x,x(x-1),(x+1)x),(3x(x-1),x(x-1)(x-2),(x+1)x(x-1))|,` then f(100) is equal to - (i)`0` (ii)`1` (iii)`100` (iv)`-100`

If `f(x)=|(1,x,x+1),(2x,x(x-1),(x+1)x),(3x(x-1),x(x-1)(x-2),(x+1)x(x-1))|,` then f(100) is equal to - (i)`0` (ii)`1` (iii)`100` (iv)`-100`

Description : If `f(x)=|(1,x,x+1),(2x,x(x-1),(x+1)x),(3x(x-1),x(x-1)(x-2),(x+1)x(x-1))|,` then f(100) is equal to - (i)`0` (ii)`1` (iii)`100` (iv)`-100`

Last Answer : If `f(x)=|(1,x,x+1),(2x,x(x-1),(x+1)x),(3x(x-1),x(x-1)(x-2),(x+1)x(x-1))|,` then f(100) is ... `1` (iii)`100` (iv)`-100` A. 1 B. 0 C. 100 D. `-100`

If ax + by = a2 – b2 and bx + ay = 0, find the value of (x + y). -Maths 10th

Description : If ax + by = a2 – b2 and bx + ay = 0, find the value of (x + y). -Maths 10th

Last Answer : Given, ax+by=a2−b2......(1) bx+ay=0.......(2). Now adding (1) and (2) we get, x(a+b)x+(a+b)y=a2−b2 or, (a+b)(x+y)=a2−b2 or, x+y=a−b.

The value of quadratic polynomial f (x) = 2x 2 – 3x- 2 at x = -2 is ...... (a) 12 (b) 15 (c) -12 (d) 16

Description : The value of quadratic polynomial f (x) = 2x 2 – 3x- 2 at x = -2 is ...... (a) 12 (b) 15 (c) -12 (d) 16

Last Answer : (a) 12

Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases: (i) p(x) = 2x3+x2–2x–1, g(x) = x+1 -Maths 9th

Description : Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases: (i) p(x) = 2x3+x2–2x–1, g(x) = x+1 -Maths 9th

Last Answer : Solution: p(x) = 2x3+x2–2x–1, g(x) = x+1 g(x) = 0 ⇒ x+1 = 0 ⇒ x = −1 ∴Zero of g(x) is -1. Now, p(−1) = 2(−1)3+(−1)2–2(−1)–1 = −2+1+2−1 = 0 ∴By factor theorem, g(x) is a factor of p(x