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Question Number 44610 Answers: 0 Comments: 1

Let A be 2×3 matrix , whereas B be 3×2 matrix. If determinant(AB)=4, then the value of determinant (BA) ?

$${Let}\:{A}\:{be}\:\mathrm{2}×\mathrm{3}\:{matrix}\:,\:{whereas}\:{B}\:{be} \\ $$$$\mathrm{3}×\mathrm{2}\:{matrix}.\:{If}\:{determinant}\left({AB}\right)=\mathrm{4}, \\ $$$${then}\:{the}\:{value}\:{of}\:{determinant}\:\left({BA}\right)\:? \\ $$

Question Number 41691 Answers: 2 Comments: 0

tan^2 20+tan^2 40+tan^2 80=33

$${tan}^{\mathrm{2}} \mathrm{20}+{tan}^{\mathrm{2}} \mathrm{40}+{tan}^{\mathrm{2}} \mathrm{80}=\mathrm{33} \\ $$

Question Number 41686 Answers: 1 Comments: 2

Question Number 41682 Answers: 2 Comments: 0

find radius of curvature to y=sin x at x=π/6 .

$${find}\:{radius}\:{of}\:{curvature}\:{to} \\ $$$${y}=\mathrm{sin}\:{x}\:\:{at}\:\:{x}=\pi/\mathrm{6}\:. \\ $$

Question Number 41679 Answers: 1 Comments: 5

let f(x) = ∫_0 ^1 ln(1+t +xt^2 )dt 1) calculate f^′ (x) then find a simple form of f(x) 2) calculate ∫_0 ^1 ln(1+t +t^2 )dt 3) calculate ∫_0 ^1 ln(1−t^3 )dt .

$${let}\:{f}\left({x}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{t}\:+{xt}^{\mathrm{2}} \right){dt} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{'} \left({x}\right)\:{then}\:{find}\:{a}\:{simple}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{t}\:+{t}^{\mathrm{2}} \right){dt} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{ln}\left(\mathrm{1}−{t}^{\mathrm{3}} \right){dt}\:. \\ $$

Question Number 41678 Answers: 1 Comments: 0

prove that ∫_0 ^∞ cos(x^2 )dx=∫_0 ^∞ sin(x^2 )dx by using only series.

$${prove}\:{that}\:\:\int_{\mathrm{0}} ^{\infty} \:{cos}\left({x}^{\mathrm{2}} \right){dx}=\int_{\mathrm{0}} ^{\infty} \:{sin}\left({x}^{\mathrm{2}} \right){dx}\:{by}\:{using} \\ $$$${only}\:{series}. \\ $$

Question Number 41677 Answers: 2 Comments: 2

calculate A = ∫_0 ^(π/4) cos^8 xdx and B= ∫_0 ^(π/4) sin^8 xdx 2) calculate A +B and A−B 3) calculate A^2 −B^2

$${calculate}\:{A}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cos}^{\mathrm{8}} {xdx}\:{and}\: \\ $$$${B}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{sin}^{\mathrm{8}} {xdx} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{A}\:+{B}\:{and}\:{A}−{B} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:{A}^{\mathrm{2}} \:−{B}^{\mathrm{2}} \\ $$

Question Number 41675 Answers: 1 Comments: 1

Question Number 41672 Answers: 0 Comments: 1

find Σ_(k=0) ^n (1/(3k+1)) interms of H_n

$${find}\:\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\frac{\mathrm{1}}{\mathrm{3}{k}+\mathrm{1}}\:{interms}\:{of}\:{H}_{{n}} \\ $$

Question Number 41671 Answers: 0 Comments: 0

if p=6.4×10^4 and q=3.2×10^5 find the values of (i)p×q (ii)p+q write the answers in standard form

Question Number 41651 Answers: 2 Comments: 1

∫( 1+2x+3x^2 +4x^3 +.........) dx , (0<∣x∣<1)

$$\int\left(\:\mathrm{1}+\mathrm{2}{x}+\mathrm{3}{x}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{3}} +.........\right)\:{dx}\:,\:\:\: \\ $$$$\left(\mathrm{0}<\mid{x}\mid<\mathrm{1}\right) \\ $$

Question Number 41642 Answers: 1 Comments: 0

n(n − 1)(n − 2)(n − 3) .... (n − r + 1) = ??

$$\mathrm{n}\left(\mathrm{n}\:−\:\mathrm{1}\right)\left(\mathrm{n}\:−\:\mathrm{2}\right)\left(\mathrm{n}\:−\:\mathrm{3}\right)\:....\:\left(\mathrm{n}\:−\:\mathrm{r}\:+\:\mathrm{1}\right)\:=\:?? \\ $$

Question Number 41634 Answers: 2 Comments: 1

Question Number 41622 Answers: 4 Comments: 9

let z_1 and z_2 the roots of x^2 −2x+2=0 1) calculate z_1 ^3 +z_2 ^3 then (1/z_1 ^3 ) +(1/z_2 ^3 ) 2) calculate z_1 ^4 +z_2 ^4 then (1/z_1 ^4 ) +(1/z_2 ^4 ) 3) let n from N simplify A_n = z_1 ^n +z_2 ^n and B_n = z_1 ^n −z_2 ^n 4) simplify S_n =Σ_(k=0) ^(n−1) (z_1 ^k +z_2 ^k )

$${let}\:{z}_{\mathrm{1}} \:{and}\:{z}_{\mathrm{2}} \:{the}\:{roots}\:{of}\:{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}=\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{z}_{\mathrm{1}} ^{\mathrm{3}} \:+{z}_{\mathrm{2}} ^{\mathrm{3}} \:\:\:{then}\:\:\frac{\mathrm{1}}{{z}_{\mathrm{1}} ^{\mathrm{3}} }\:+\frac{\mathrm{1}}{{z}_{\mathrm{2}} ^{\mathrm{3}} } \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{z}_{\mathrm{1}} ^{\mathrm{4}} \:+{z}_{\mathrm{2}} ^{\mathrm{4}} \:\:{then}\:\:\frac{\mathrm{1}}{{z}_{\mathrm{1}} ^{\mathrm{4}} }\:+\frac{\mathrm{1}}{{z}_{\mathrm{2}} ^{\mathrm{4}} } \\ $$$$\left.\mathrm{3}\right)\:{let}\:{n}\:{from}\:{N}\:\:{simplify} \\ $$$${A}_{{n}} =\:{z}_{\mathrm{1}} ^{{n}} \:+{z}_{\mathrm{2}} ^{{n}} \:\:\:\:\:\:\:{and}\:\:{B}_{{n}} =\:{z}_{\mathrm{1}} ^{{n}} \:−{z}_{\mathrm{2}} ^{{n}} \\ $$$$\left.\mathrm{4}\right)\:{simplify}\:\:{S}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\left({z}_{\mathrm{1}} ^{{k}} \:\:\:+{z}_{\mathrm{2}} ^{{k}} \right) \\ $$

Question Number 41620 Answers: 2 Comments: 1