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Question Number 18455 Answers: 1 Comments: 0

The equation cosec (x/2) + cosec (y/2) + cosec (z/2) = 6, where 0 < x, y, z < (π/2) and x + y + z = π, have (1) Three ordered triplet (x, y, z) solutions (2) Two ordered triplet (x, y, z) solutions (3) Just one ordered triplet (x, y, z) solution (4) No ordered triplet (x, y, z) solution

$$\mathrm{The}\:\mathrm{equation}\:\mathrm{cosec}\:\frac{{x}}{\mathrm{2}}\:+\:\mathrm{cosec}\:\frac{{y}}{\mathrm{2}}\:+ \\ $$$$\mathrm{cosec}\:\frac{{z}}{\mathrm{2}}\:=\:\mathrm{6},\:\mathrm{where}\:\mathrm{0}\:<\:{x},\:{y},\:{z}\:<\:\frac{\pi}{\mathrm{2}}\:\mathrm{and} \\ $$$${x}\:+\:{y}\:+\:{z}\:=\:\pi,\:\mathrm{have} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Three}\:\mathrm{ordered}\:\mathrm{triplet}\:\left({x},\:{y},\:{z}\right) \\ $$$$\mathrm{solutions} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Two}\:\mathrm{ordered}\:\mathrm{triplet}\:\left({x},\:{y},\:{z}\right) \\ $$$$\mathrm{solutions} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Just}\:\mathrm{one}\:\mathrm{ordered}\:\mathrm{triplet}\:\left({x},\:{y},\:{z}\right) \\ $$$$\mathrm{solution} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{No}\:\mathrm{ordered}\:\mathrm{triplet}\:\left({x},\:{y},\:{z}\right)\:\mathrm{solution} \\ $$

Question Number 18450 Answers: 0 Comments: 0

A $4000 note is signed, for 30 days at a discount rate of 12%. Find the proceeds. I′m not sure whether the $4000 is bank discount, principal or maturity value. Please help me

$${A}\:\$\mathrm{4000}\:{note}\:{is}\:{signed},\:{for}\:\mathrm{30}\:{days} \\ $$$${at}\:{a}\:{discount}\:{rate}\:{of}\:\mathrm{12\%}.\:{Find}\:{the} \\ $$$${proceeds}. \\ $$$$ \\ $$$${I}'{m}\:{not}\:{sure}\:{whether}\:{the}\:\$\mathrm{4000}\:{is} \\ $$$${bank}\:{discount},\:{principal}\:{or}\:{maturity}\:{value}. \\ $$$$ \\ $$$${Please}\:{help}\:{me} \\ $$

Question Number 18448 Answers: 1 Comments: 0

If a, b, c, d are in GP and a^x = b^y = c^z = d^u , then x, y, z, u are in

$$\mathrm{If}\:\:{a},\:{b},\:{c},\:{d}\:\mathrm{are}\:\mathrm{in}\:\mathrm{GP}\:\mathrm{and}\:\:{a}^{{x}} =\:{b}^{{y}} =\:{c}^{{z}} =\:{d}^{{u}} , \\ $$$$\mathrm{then}\:{x},\:{y},\:{z},\:{u}\:\mathrm{are}\:\mathrm{in} \\ $$

Question Number 18446 Answers: 0 Comments: 0

Question Number 18440 Answers: 0 Comments: 0

Question Number 18439 Answers: 0 Comments: 3

Question Number 18432 Answers: 1 Comments: 0

Find interval p so (p − 2)x^2 + 2px + p − 1 = 0 have negative roots

$$\mathrm{Find}\:\mathrm{interval}\:{p}\:\mathrm{so} \\ $$$$\left({p}\:−\:\mathrm{2}\right){x}^{\mathrm{2}} \:+\:\mathrm{2}{px}\:+\:{p}\:−\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\mathrm{have}\:\mathrm{negative}\:\mathrm{roots} \\ $$

Question Number 18431 Answers: 1 Comments: 0

x^2 =16^x find x

$$\mathrm{x}^{\mathrm{2}} =\mathrm{16}^{\mathrm{x}} \\ $$$$\mathrm{find}\:\mathrm{x} \\ $$

Question Number 18430 Answers: 1 Comments: 0

x^2 =3^x find x

$$\mathrm{x}^{\mathrm{2}} =\mathrm{3}^{\mathrm{x}} \\ $$$$\mathrm{find}\:\mathrm{x} \\ $$

Question Number 18426 Answers: 0 Comments: 0

The equation 2 cot 2x − 3 cot 3x = tan 2x has (1) Two solutions in (0, (π/3)) (2) One solution in (0, (π/3)) (3) No solution in (−∞, ∞) (4) Three solution in (0, π)

$$\mathrm{The}\:\mathrm{equation}\:\mathrm{2}\:\mathrm{cot}\:\mathrm{2}{x}\:−\:\mathrm{3}\:\mathrm{cot}\:\mathrm{3}{x}\:=\:\mathrm{tan}\:\mathrm{2}{x} \\ $$$$\mathrm{has} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Two}\:\mathrm{solutions}\:\mathrm{in}\:\left(\mathrm{0},\:\frac{\pi}{\mathrm{3}}\right) \\ $$$$\left(\mathrm{2}\right)\:\mathrm{One}\:\mathrm{solution}\:\mathrm{in}\:\left(\mathrm{0},\:\frac{\pi}{\mathrm{3}}\right) \\ $$$$\left(\mathrm{3}\right)\:\mathrm{No}\:\mathrm{solution}\:\mathrm{in}\:\left(−\infty,\:\infty\right) \\ $$$$\left(\mathrm{4}\right)\:\mathrm{Three}\:\mathrm{solution}\:\mathrm{in}\:\left(\mathrm{0},\:\pi\right) \\ $$

Question Number 18428 Answers: 0 Comments: 0

Question Number 18416 Answers: 0 Comments: 0

The number of integral values of x which satisfies (((x − 5)^(10) (x − 13)^(20) (x − 19)^(13) )/((x − 10)^(18) (x − 25)^(19) )) ≥ 0 and 2 ≤ x ≤ 30 are (1) 23 (2) 24 (3) 25 (4) 26

$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{integral}\:\mathrm{values}\:\mathrm{of}\:{x} \\ $$$$\mathrm{which}\:\mathrm{satisfies} \\ $$$$\frac{\left({x}\:−\:\mathrm{5}\right)^{\mathrm{10}} \left({x}\:−\:\mathrm{13}\right)^{\mathrm{20}} \left({x}\:−\:\mathrm{19}\right)^{\mathrm{13}} }{\left({x}\:−\:\mathrm{10}\right)^{\mathrm{18}} \left({x}\:−\:\mathrm{25}\right)^{\mathrm{19}} }\:\geqslant\:\mathrm{0}\:\mathrm{and} \\ $$$$\mathrm{2}\:\leqslant\:{x}\:\leqslant\:\mathrm{30}\:\mathrm{are} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{23} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{24} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{25} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{26} \\ $$

Question Number 18415 Answers: 1 Comments: 1

Calculate the magnetic field produced at ground level by a 15A current flowing in a long horizontal wire suspended at a height of 7.5m

$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{magnetic}\:\mathrm{field}\:\mathrm{produced}\:\mathrm{at}\:\mathrm{ground}\:\mathrm{level}\:\mathrm{by}\:\mathrm{a}\:\mathrm{15A}\:\mathrm{current} \\ $$$$\mathrm{flowing}\:\mathrm{in}\:\mathrm{a}\:\mathrm{long}\:\mathrm{horizontal}\:\mathrm{wire}\:\mathrm{suspended}\:\mathrm{at}\:\mathrm{a}\:\mathrm{height}\:\mathrm{of}\:\mathrm{7}.\mathrm{5m} \\ $$

Question Number 18411 Answers: 1 Comments: 0

A glass bulb contains 2.24 L of H_2 and 1.12 L of D_2 at S.T.P. It is connected to a fully evacuated bulb by a stopcock with a small opening. The stopcock is opened for sometime and then closed. The first bulb now contains 0.1 g of D_2 . Calculate the percentage composition by weight of the gases in the second bulb.

$$\mathrm{A}\:\mathrm{glass}\:\mathrm{bulb}\:\mathrm{contains}\:\mathrm{2}.\mathrm{24}\:\mathrm{L}\:\mathrm{of}\:\mathrm{H}_{\mathrm{2}} \:\mathrm{and} \\ $$$$\mathrm{1}.\mathrm{12}\:\mathrm{L}\:\mathrm{of}\:\mathrm{D}_{\mathrm{2}} \:\mathrm{at}\:\mathrm{S}.\mathrm{T}.\mathrm{P}.\:\mathrm{It}\:\mathrm{is}\:\mathrm{connected}\:\mathrm{to} \\ $$$$\mathrm{a}\:\mathrm{fully}\:\mathrm{evacuated}\:\mathrm{bulb}\:\mathrm{by}\:\mathrm{a}\:\mathrm{stopcock} \\ $$$$\mathrm{with}\:\mathrm{a}\:\mathrm{small}\:\mathrm{opening}.\:\mathrm{The}\:\mathrm{stopcock}\:\mathrm{is} \\ $$$$\mathrm{opened}\:\mathrm{for}\:\mathrm{sometime}\:\mathrm{and}\:\mathrm{then}\:\mathrm{closed}. \\ $$$$\mathrm{The}\:\mathrm{first}\:\mathrm{bulb}\:\mathrm{now}\:\mathrm{contains}\:\mathrm{0}.\mathrm{1}\:\mathrm{g}\:\mathrm{of}\:\mathrm{D}_{\mathrm{2}} . \\ $$$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{percentage}\:\mathrm{composition} \\ $$$$\mathrm{by}\:\mathrm{weight}\:\mathrm{of}\:\mathrm{the}\:\mathrm{gases}\:\mathrm{in}\:\mathrm{the}\:\mathrm{second} \\ $$$$\mathrm{bulb}. \\ $$

Question Number 20955 Answers: 1 Comments: 0

lemme join miss Tawa Tawa here. It takes 8 painters working at the same rate ,5 hours to paint a house.If 6 painters are working at 2/3 the rate of the 8 painters,how long would it take them to paint the same house?

$$\mathrm{lemme}\:\mathrm{join}\:\mathrm{miss}\:\mathrm{Tawa}\:\mathrm{Tawa}\:\mathrm{here}. \\ $$$$ \\ $$$$\mathrm{It}\:\mathrm{takes}\:\mathrm{8}\:\mathrm{painters}\:\mathrm{working}\:\mathrm{at}\:\mathrm{the} \\ $$$$\mathrm{same}\:\mathrm{rate}\:,\mathrm{5}\:\mathrm{hours}\:\mathrm{to}\:\mathrm{paint}\:\mathrm{a} \\ $$$$\mathrm{house}.\mathrm{If}\:\mathrm{6}\:\mathrm{painters}\:\mathrm{are}\:\mathrm{working}\:\mathrm{at} \\ $$$$\mathrm{2}/\mathrm{3}\:\mathrm{the}\:\mathrm{rate}\:\mathrm{of}\:\mathrm{the}\:\mathrm{8}\:\mathrm{painters},\mathrm{how} \\ $$$$\mathrm{long}\:\mathrm{would}\:\mathrm{it}\:\mathrm{take}\:\mathrm{them}\:\mathrm{to}\:\mathrm{paint} \\ $$$$\mathrm{the}\:\mathrm{same}\:\mathrm{house}? \\ $$

Question Number 18429 Answers: 0 Comments: 6

30cm^3 of hydrogen at s.t.p combines with 20cm^3 of oxygen to form steam according to the following equation, 2H_2 (g) + O_2 (g) → 2H_2 O (g). Calculate the total volume of gaseous mixture at the end of the reaction.

$$\mathrm{30cm}^{\mathrm{3}} \:\mathrm{of}\:\mathrm{hydrogen}\:\mathrm{at}\:\mathrm{s}.\mathrm{t}.\mathrm{p}\:\mathrm{combines}\:\mathrm{with}\:\mathrm{20cm}^{\mathrm{3}} \:\mathrm{of}\:\mathrm{oxygen}\:\mathrm{to}\:\mathrm{form}\:\mathrm{steam}\: \\ $$$$\mathrm{according}\:\mathrm{to}\:\mathrm{the}\:\mathrm{following}\:\mathrm{equation},\:\:\mathrm{2H}_{\mathrm{2}} \:\left(\mathrm{g}\right)\:+\:\mathrm{O}_{\mathrm{2}} \:\left(\mathrm{g}\right)\:\rightarrow\:\mathrm{2H}_{\mathrm{2}} \mathrm{O}\:\left(\mathrm{g}\right). \\ $$$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{total}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{gaseous}\:\mathrm{mixture}\:\mathrm{at}\:\mathrm{the}\:\mathrm{end}\:\mathrm{of}\:\mathrm{the}\:\mathrm{reaction}. \\ $$

Question Number 18402 Answers: 1 Comments: 0

If P_n = cos^n θ + sin^n θ, θ ∈ [0, (π/2)], n ∈ (−∞, 2), then minimum of P_n will be (1) 1 (2) (1/2) (3) (√2) (4) (1/(√2))

$$\mathrm{If}\:{P}_{{n}} \:=\:\mathrm{cos}^{{n}} \:\theta\:+\:\mathrm{sin}^{{n}} \:\theta,\:\theta\:\in\:\left[\mathrm{0},\:\frac{\pi}{\mathrm{2}}\right],\:{n}\:\in \\ $$$$\left(−\infty,\:\mathrm{2}\right),\:\mathrm{then}\:\mathrm{minimum}\:\mathrm{of}\:{P}_{{n}} \:\mathrm{will}\:\mathrm{be} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{1} \\ $$$$\left(\mathrm{2}\right)\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left(\mathrm{3}\right)\:\sqrt{\mathrm{2}} \\ $$$$\left(\mathrm{4}\right)\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}} \\ $$

Question Number 18400 Answers: 0 Comments: 1

Question Number 18397 Answers: 0 Comments: 0

Prove : ∀n ∈ N, n ≥ 2, so ∃ x,y,z ∣ x,y,z ∈ N such that (4/n) = (1/x) + (1/y) + (1/z) Example: choose n = 2 (4/2) = (1/x) + (1/y) + (1/z) If x = 1 , y = 2 and z = 2, the equa- tion is correct!

$${Prove}\::\:\forall{n}\:\in\:\mathbb{N},\:{n}\:\geqslant\:\mathrm{2},\:{so}\:\exists\:{x},{y},{z}\:\mid\: \\ $$$${x},{y},{z}\:\in\:\mathbb{N}\:{such}\:{that}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{4}}{{n}}\:=\:\frac{\mathrm{1}}{{x}}\:+\:\frac{\mathrm{1}}{{y}}\:+\:\frac{\mathrm{1}}{{z}} \\ $$$${Example}: \\ $$$${choose}\:{n}\:=\:\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{4}}{\mathrm{2}}\:=\:\frac{\mathrm{1}}{{x}}\:+\:\frac{\mathrm{1}}{{y}}\:+\:\frac{\mathrm{1}}{{z}} \\ $$$${If}\:{x}\:=\:\mathrm{1}\:\:,\:\:\:{y}\:=\:\mathrm{2}\:\:{and}\:\:\:{z}\:=\:\mathrm{2},\:{the}\:{equa}- \\ $$$${tion}\:{is}\:{correct}! \\ $$

Question Number 18396 Answers: 1 Comments: 0

∫ (√(1 + (1/x^2 ) + (1/((x + 1)^2 )))) dx

$$\int\:\sqrt{\mathrm{1}\:+\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\left({x}\:+\:\mathrm{1}\right)^{\mathrm{2}} }}\:\:{dx} \\ $$

Question Number 18394 Answers: 0 Comments: 0

Question Number 18392 Answers: 1 Comments: 1

Question Number 18388 Answers: 1 Comments: 0

Solve simultaneously. x + y = 5 ....... (i) 5^x + y = 15 ...... (ii)

$$\mathrm{Solve}\:\mathrm{simultaneously}.\: \\ $$$$\mathrm{x}\:+\:\mathrm{y}\:=\:\mathrm{5}\:\:\:\:\:.......\:\left(\mathrm{i}\right) \\ $$$$\mathrm{5}^{\mathrm{x}} \:+\:\mathrm{y}\:=\:\mathrm{15}\:\:\:\:......\:\left(\mathrm{ii}\right) \\ $$

Question Number 18386 Answers: 1 Comments: 1

Prove that ((2 + (√5)))^(1/3) + ((2 − (√5)))^(1/3) is a rational number.

$$\mathrm{Prove}\:\mathrm{that}\:\sqrt[{\mathrm{3}}]{\mathrm{2}\:+\:\sqrt{\mathrm{5}}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{2}\:−\:\sqrt{\mathrm{5}}}\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{rational}\:\mathrm{number}. \\ $$

Question Number 18385 Answers: 1 Comments: 0

Find all the integers which are equal to 11 times the sum of their digits.

$$\mathrm{Find}\:\mathrm{all}\:\mathrm{the}\:\mathrm{integers}\:\mathrm{which}\:\mathrm{are}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\mathrm{11}\:\mathrm{times}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{their}\:\mathrm{digits}. \\ $$

Question Number 18384 Answers: 1 Comments: 0

Let a, b, c ∈ R, a ≠ 0, such that a and 4a + 3b + 2c have the same sign. Show that the equation ax^2 + bx + c = 0 can not have both roots in the interval (1, 2).

$$\mathrm{Let}\:{a},\:{b},\:{c}\:\in\:{R},\:{a}\:\neq\:\mathrm{0},\:\mathrm{such}\:\mathrm{that}\:{a}\:\mathrm{and} \\ $$$$\mathrm{4}{a}\:+\:\mathrm{3}{b}\:+\:\mathrm{2}{c}\:\mathrm{have}\:\mathrm{the}\:\mathrm{same}\:\mathrm{sign}.\:\mathrm{Show} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{equation}\:{ax}^{\mathrm{2}} \:+\:{bx}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{can} \\ $$$$\mathrm{not}\:\mathrm{have}\:\mathrm{both}\:\mathrm{roots}\:\mathrm{in}\:\mathrm{the}\:\mathrm{interval} \\ $$$$\left(\mathrm{1},\:\mathrm{2}\right). \\ $$

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