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Question Number 76910 Answers: 1 Comments: 1

Question Number 76904 Answers: 1 Comments: 2

montrer que: ∀x,y,z>0 x^3 +2y^2 +4z≥6xy^(2/3) z^(1/3)

$$\mathrm{montrer}\:\mathrm{que}: \\ $$$$\forall{x},{y},{z}>\mathrm{0}\:\:{x}^{\mathrm{3}} +\mathrm{2}{y}^{\mathrm{2}} +\mathrm{4}{z}\geqslant\mathrm{6}{xy}^{\mathrm{2}/\mathrm{3}} {z}^{\mathrm{1}/\mathrm{3}} \\ $$

Question Number 76898 Answers: 1 Comments: 0

Question Number 76892 Answers: 3 Comments: 0

Calculate ∫ ((√(9−x^2 ))/x^6 ) dx .

$$\mathcal{C}{alculate}\:\int\:\frac{\sqrt{\mathrm{9}−{x}^{\mathrm{2}} }}{{x}^{\mathrm{6}} }\:{dx}\:. \\ $$

Question Number 76888 Answers: 0 Comments: 3

happy new year sir to all member this forum

$${happy}\:{new}\:{year}\:{sir}\:{to}\:{all}\:{member}\: \\ $$$${this}\:{forum} \\ $$

Question Number 76883 Answers: 0 Comments: 1

what is the perimeter of the loop? 3ay^2 = x(x−3a)^(2 ) ?

$${what}\:{is}\:{the}\:{perimeter}\:{of}\:{the}\:{loop}? \\ $$$$\mathrm{3}{ay}^{\mathrm{2}} \:=\:{x}\left({x}−\mathrm{3}{a}\right)^{\mathrm{2}\:} ? \\ $$

Question Number 76880 Answers: 0 Comments: 2

(((c/5)−7)/5) −2= 5 Intead of 5 what else could be there?

$$\frac{\frac{{c}}{\mathrm{5}}−\mathrm{7}}{\mathrm{5}}\:−\mathrm{2}=\:\mathrm{5} \\ $$$${Intead}\:{of}\:\mathrm{5}\:{what}\:{else}\:{could}\:{be} \\ $$$${there}? \\ $$

Question Number 76872 Answers: 0 Comments: 4

the product of 3 integer x,y,z is 192 . z = 4 and t is equal to average? of x and y . what is the minimum posible value of t?

$${the}\:{product}\:{of}\:\mathrm{3}\:{integer}\:{x},{y},{z}\:{is}\:\mathrm{192} \\ $$$$.\:{z}\:=\:\mathrm{4}\:{and}\:{t}\:{is}\:{equal}\:{to}\:{average}? \\ $$$${of}\:{x}\:{and}\:{y}\:.\:{what}\:{is}\:{the}\:{minimum}\: \\ $$$${posible}\:{value}\:{of}\:{t}? \\ $$

Question Number 76862 Answers: 1 Comments: 0

Find the equation of parabola whose focus (−1,−2) and directrix x−2y+3=0 ??

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{parabola}\: \\ $$$$\mathrm{whose}\:\mathrm{focus}\:\left(−\mathrm{1},−\mathrm{2}\right)\:\mathrm{and}\:\mathrm{directrix} \\ $$$$\mathrm{x}−\mathrm{2y}+\mathrm{3}=\mathrm{0}\:?? \\ $$

Question Number 76858 Answers: 0 Comments: 0

Question Number 76855 Answers: 1 Comments: 2

Question Number 76842 Answers: 2 Comments: 1

what is solution y^(′′ ) + y = 0 .

$$ \\ $$$${what}\:{is}\:{solution}\:{y}^{''\:} +\:\:{y}\:=\:\mathrm{0}\:. \\ $$

Question Number 76860 Answers: 2 Comments: 0

Find the equation of the circle having (2,−2) as its centre and passing through 3x+y=14, 2x+5y=18 ??

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{having} \\ $$$$\left(\mathrm{2},−\mathrm{2}\right)\:\mathrm{as}\:\mathrm{its}\:\mathrm{centre}\:\mathrm{and}\:\mathrm{passing} \\ $$$$\mathrm{through}\:\mathrm{3x}+\mathrm{y}=\mathrm{14},\:\mathrm{2x}+\mathrm{5y}=\mathrm{18}\:?? \\ $$

Question Number 76830 Answers: 1 Comments: 2

Question Number 76826 Answers: 1 Comments: 0

Question Number 76819 Answers: 0 Comments: 4

one of the foci of the ellipse (x^2 /9) + (y^2 /4) = 1 is A. (4,0) B. (9,0) C. (5,0) D. ((√5) , 0)

$$\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{foci}\:\mathrm{of}\:\mathrm{the}\:\mathrm{ellipse} \\ $$$$\:\:\:\:\:\frac{{x}^{\mathrm{2}} }{\mathrm{9}}\:+\:\frac{{y}^{\mathrm{2}} }{\mathrm{4}}\:=\:\mathrm{1}\:\mathrm{is} \\ $$$$\mathrm{A}.\:\left(\mathrm{4},\mathrm{0}\right) \\ $$$$\mathrm{B}.\:\left(\mathrm{9},\mathrm{0}\right) \\ $$$$\mathrm{C}.\:\left(\mathrm{5},\mathrm{0}\right) \\ $$$$\mathrm{D}.\:\left(\sqrt{\mathrm{5}}\:,\:\mathrm{0}\right) \\ $$

Question Number 76817 Answers: 0 Comments: 2

A compound pendulum ocsillates through angles θ about its equilibrium position such that 8aθ^2 = 9g cosθ, a>0. its period is A. 2π(√((8a)/(9g))) B. ((3π)/8)(√(a/g)) C. 2π(√((9g)/(8a))) D. ((8π)/3)(√(a/g))

$$\mathrm{A}\:\mathrm{compound}\:\mathrm{pendulum}\:\mathrm{ocsillates} \\ $$$$\mathrm{through}\:\mathrm{angles}\:\theta\:\mathrm{about}\:\mathrm{its}\:\mathrm{equilibrium} \\ $$$$\mathrm{position}\:\mathrm{such}\:\mathrm{that}\: \\ $$$$\mathrm{8}{a}\theta^{\mathrm{2}} \:=\:\mathrm{9}{g}\:{cos}\theta,\:{a}>\mathrm{0}.\:\mathrm{its}\:\mathrm{period}\:\mathrm{is}\: \\ $$$$\mathrm{A}.\:\mathrm{2}\pi\sqrt{\frac{\mathrm{8}{a}}{\mathrm{9}{g}}} \\ $$$$\mathrm{B}.\:\frac{\mathrm{3}\pi}{\mathrm{8}}\sqrt{\frac{{a}}{{g}}} \\ $$$$\mathrm{C}.\:\mathrm{2}\pi\sqrt{\frac{\mathrm{9}{g}}{\mathrm{8}{a}}} \\ $$$$\mathrm{D}.\:\frac{\mathrm{8}\pi}{\mathrm{3}}\sqrt{\frac{{a}}{{g}}} \\ $$

Question Number 76815 Answers: 1 Comments: 2

Which one of the following series is Not convergent? A. Σ_(r=1) ^∞ (1/r^3 ) B. Σ_(r=1) ^∞ (1/r^2 ) C. Σ_(r=1) ^∞ (1/(r^2 −4)) D. Σ_(r=1) ^∞ (1/(5r))

$$\mathrm{Which}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{series}\:\mathrm{is}\:\boldsymbol{\mathrm{N}}\mathrm{o}\boldsymbol{\mathrm{t}}\:\mathrm{convergent}? \\ $$$$\mathrm{A}.\:\underset{\mathrm{r}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{r}^{\mathrm{3}} } \\ $$$$\mathrm{B}.\:\underset{\mathrm{r}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{r}^{\mathrm{2}} } \\ $$$$\mathrm{C}.\:\underset{\mathrm{r}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{r}^{\mathrm{2}} −\mathrm{4}} \\ $$$$\mathrm{D}.\:\underset{\mathrm{r}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{5r}} \\ $$

Question Number 76813 Answers: 1 Comments: 7

Σ_(k=1) ^(2n) (−1)^k = A. ∞ B. 1 C. −1 D. 0

$$\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{2n}} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{k}} \:=\: \\ $$$$\mathrm{A}.\:\infty \\ $$$$\mathrm{B}.\:\mathrm{1} \\ $$$$\mathrm{C}.\:−\mathrm{1} \\ $$$$\mathrm{D}.\:\mathrm{0} \\ $$

Question Number 76812 Answers: 1 Comments: 0

Given that A and B are 3 × 3 invertible matrices, then (A^(−1) B)^(−1) = A. AB^(−1) B.B^(−1) A C. B^(−1) A^(−1) D. BA^(−1)

$$\mathrm{Given}\:\mathrm{that}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{are}\:\mathrm{3}\:×\:\mathrm{3}\:\mathrm{invertible}\:\mathrm{matrices}, \\ $$$$\:\mathrm{then}\:\left(\mathrm{A}^{−\mathrm{1}} \mathrm{B}\right)^{−\mathrm{1}} \:= \\ $$$$\mathrm{A}.\:\mathrm{AB}^{−\mathrm{1}} \\ $$$$\mathrm{B}.\mathrm{B}^{−\mathrm{1}} \mathrm{A} \\ $$$$\mathrm{C}.\:\mathrm{B}^{−\mathrm{1}} \mathrm{A}^{−\mathrm{1}} \\ $$$$\mathrm{D}.\:\mathrm{BA}^{−\mathrm{1}} \\ $$

Question Number 76811 Answers: 0 Comments: 2

The eccentricity of the hyperbola (x^2 /(64)) − (y^2 /(36)) = 1 is A. (5/4) B. (3/4) C. (4/5) D. (4/3)

$$\mathrm{The}\:\mathrm{eccentricity}\:\mathrm{of}\:\mathrm{the}\:\mathrm{hyperbola} \\ $$$$\:\:\:\frac{{x}^{\mathrm{2}} }{\mathrm{64}}\:−\:\frac{{y}^{\mathrm{2}} }{\mathrm{36}}\:=\:\mathrm{1}\:\mathrm{is}\: \\ $$$$\mathrm{A}.\:\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$\mathrm{B}.\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\mathrm{C}.\:\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\mathrm{D}.\:\frac{\mathrm{4}}{\mathrm{3}} \\ $$

Question Number 76809 Answers: 0 Comments: 3

Question Number 76808 Answers: 1 Comments: 3

∫(1/(√(x^3 +1)))dx

$$\int\frac{\mathrm{1}}{\sqrt{{x}^{\mathrm{3}} +\mathrm{1}}}{dx} \\ $$

Question Number 76807 Answers: 0 Comments: 0

Question Number 76802 Answers: 0 Comments: 0

3)αparticle of energy 5MeV pass through an ionisation chamber at the rate of 10 pwe second . Assum that all the energy is used in producing ion pairs,calculate the current produced. (35MeV is required for producing an ion pair and e=1.6×10^(-19) C) solution: Energy of α particles=5×10^6 eV Energy required for producing one ion pair=35eV No.of ion pairs produced by one α particle =((5×10^6 )/(35))=1.426×10^5 since 10 particle enter the chamber in one second =1.426×10^5 ×10=1.426×10^6 charge on dach ion=1.6×10^(-19) C Current=(1.426×10^6 )×(1.6×10^(-19) )C/s =2.287×10^(-13) A.

$$\left.\mathrm{3}\right)\alpha\mathrm{particle}\:\mathrm{of}\:\mathrm{energy}\:\mathrm{5MeV}\:\mathrm{pass} \\ $$$$\mathrm{through}\:\mathrm{an}\:\mathrm{ionisation}\:\mathrm{chamber}\:\mathrm{at}\:\mathrm{the} \\ $$$$\mathrm{rate}\:\mathrm{of}\:\mathrm{10}\:\mathrm{pwe}\:\mathrm{second}\:.\:\mathrm{Assum}\:\mathrm{that}\:\mathrm{all} \\ $$$$\mathrm{the}\:\mathrm{energy}\:\mathrm{is}\:\mathrm{used}\:\mathrm{in}\:\mathrm{producing}\:\mathrm{ion}\: \\ $$$$\mathrm{pairs},\mathrm{calculate}\:\mathrm{the}\:\mathrm{current}\:\mathrm{produced}. \\ $$$$\left(\mathrm{35MeV}\:\mathrm{is}\:\mathrm{required}\:\mathrm{for}\:\mathrm{producing}\:\:\mathrm{an}\:\right. \\ $$$$\left.\mathrm{ion}\:\mathrm{pair}\:\mathrm{and}\:\mathrm{e}=\mathrm{1}.\mathrm{6}×\mathrm{10}^{-\mathrm{19}} \:\mathrm{C}\right) \\ $$$$\boldsymbol{\mathrm{solution}}: \\ $$$$\:\:\:\:\mathrm{Energy}\:\mathrm{of}\:\alpha\:\mathrm{particles}=\mathrm{5}×\mathrm{10}^{\mathrm{6}} \mathrm{eV} \\ $$$$\:\:\:\:\mathrm{Energy}\:\mathrm{required}\:\mathrm{for}\:\mathrm{producing}\:\mathrm{one} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{ion}\:\mathrm{pair}=\mathrm{35eV} \\ $$$$\:\:\:\:\mathrm{No}.\mathrm{of}\:\mathrm{ion}\:\mathrm{pairs}\:\mathrm{produced}\:\mathrm{by}\:\mathrm{one}\:\alpha \\ $$$$\:\:\:\:\:\:\:\:\mathrm{particle}\:=\frac{\mathrm{5}×\mathrm{10}^{\mathrm{6}} }{\mathrm{35}}=\mathrm{1}.\mathrm{426}×\mathrm{10}^{\mathrm{5}} \\ $$$$\mathrm{since}\:\mathrm{10}\:\mathrm{particle}\:\mathrm{enter}\:\mathrm{the}\:\mathrm{chamber}\:\mathrm{in} \\ $$$$\mathrm{one}\:\mathrm{second} \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{1}.\mathrm{426}×\mathrm{10}^{\mathrm{5}} ×\mathrm{10}=\mathrm{1}.\mathrm{426}×\mathrm{10}^{\mathrm{6}} \\ $$$$\mathrm{charge}\:\mathrm{on}\:\mathrm{dach}\:\mathrm{ion}=\mathrm{1}.\mathrm{6}×\mathrm{10}^{-\mathrm{19}} \mathrm{C} \\ $$$$\mathrm{Current}=\left(\mathrm{1}.\mathrm{426}×\mathrm{10}^{\mathrm{6}} \right)×\left(\mathrm{1}.\mathrm{6}×\mathrm{10}^{-\mathrm{19}} \right)\mathrm{C}/\mathrm{s} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}.\mathrm{287}×\mathrm{10}^{-\mathrm{13}} \mathrm{A}. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Question Number 76794 Answers: 1 Comments: 1

Find the sum of x + (x/(1+x)) + (x/((1+x)^2 )) + (x/((1+x)^3 ))+.... for ∣(1/(1+x))∣<1

$${Find}\:{the}\:{sum}\:{of} \\ $$$${x}\:+\:\frac{{x}}{\mathrm{1}+{x}}\:+\:\frac{{x}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }\:+\:\frac{{x}}{\left(\mathrm{1}+{x}\right)^{\mathrm{3}} }+....\:{for} \\ $$$$\mid\frac{\mathrm{1}}{\mathrm{1}+{x}}\mid<\mathrm{1} \\ $$

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