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Question Number 94811    Answers: 3   Comments: 0

∫(dx/(x+(√x))) (2,3)=?

$$\int\frac{\mathrm{dx}}{\mathrm{x}+\sqrt{\mathrm{x}}}\:\:\:\left(\mathrm{2},\mathrm{3}\right)=? \\ $$

Question Number 94806    Answers: 0   Comments: 2

Question Number 94803    Answers: 1   Comments: 0

Question Number 94801    Answers: 0   Comments: 0

Question Number 94797    Answers: 1   Comments: 0

Question Number 94786    Answers: 0   Comments: 1

x, y ∈N\{0, 1} ∧ x≤y find z∈N with z!=x!y!

$${x},\:{y}\:\in\mathbb{N}\backslash\left\{\mathrm{0},\:\mathrm{1}\right\}\:\wedge\:{x}\leqslant{y} \\ $$$$\mathrm{find}\:{z}\in\mathbb{N}\:\mathrm{with}\:{z}!={x}!{y}! \\ $$

Question Number 94782    Answers: 1   Comments: 0

The velocity of physical quantities is given by v = (√((P + (1/n))/x)) , where P is the pressure. Find the dimention of n and x.

$$\mathrm{The}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{physical}\:\mathrm{quantities}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by} \\ $$$$\:\:\mathrm{v}\:\:=\:\:\sqrt{\frac{\mathrm{P}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{n}}}{\mathrm{x}}}\:,\:\:\mathrm{where}\:\:\mathrm{P}\:\mathrm{is}\:\mathrm{the}\:\mathrm{pressure}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{dimention}\:\mathrm{of}\:\:\:\mathrm{n}\:\:\mathrm{and}\:\:\mathrm{x}. \\ $$

Question Number 94780    Answers: 0   Comments: 1

x!(x−4)!=12(2x−7)! x=?

$$\mathrm{x}!\left(\mathrm{x}−\mathrm{4}\right)!=\mathrm{12}\left(\mathrm{2x}−\mathrm{7}\right)! \\ $$$$\mathrm{x}=? \\ $$

Question Number 94776    Answers: 0   Comments: 0

Question Number 94774    Answers: 2   Comments: 1

Question Number 94773    Answers: 1   Comments: 0

Question Number 94772    Answers: 0   Comments: 0

Question Number 94769    Answers: 2   Comments: 0

Question Number 94768    Answers: 1   Comments: 0

Question Number 94767    Answers: 1   Comments: 0

Question Number 94764    Answers: 2   Comments: 0

Question Number 94765    Answers: 0   Comments: 0

Question Number 94756    Answers: 0   Comments: 2

solution: Q1)a) n=((ln(m/m_0 ))/(ln0.5)) = ((ln(12/75))/(ln0.5)) =2.64 N=N_0 (0.5)^n = 6.02×10^(23) (0.5)^(2.64) = 9.66×10^(22) A=λN=1.5×10^(−4) × 9.66×10^(22) =1.45×10^(19) Bq

$$\left.{s}\left.{olution}:\:\mathrm{Q1}\right){a}\right)\:{n}=\frac{{ln}\left({m}/{m}_{\mathrm{0}} \right)}{{ln}\mathrm{0}.\mathrm{5}}\:=\:\frac{{ln}\left(\mathrm{12}/\mathrm{75}\right)}{{ln}\mathrm{0}.\mathrm{5}}\:=\mathrm{2}.\mathrm{64} \\ $$$${N}=\mathrm{N}_{\mathrm{0}} \left(\mathrm{0}.\mathrm{5}\right)^{{n}} \:=\:\mathrm{6}.\mathrm{02}×\mathrm{10}^{\mathrm{23}} \left(\mathrm{0}.\mathrm{5}\right)^{\mathrm{2}.\mathrm{64}} =\:\mathrm{9}.\mathrm{66}×\mathrm{10}^{\mathrm{22}} \\ $$$${A}=\lambda{N}=\mathrm{1}.\mathrm{5}×\mathrm{10}^{−\mathrm{4}} \:×\:\mathrm{9}.\mathrm{66}×\mathrm{10}^{\mathrm{22}} =\mathrm{1}.\mathrm{45}×\mathrm{10}^{\mathrm{19}} \:{Bq} \\ $$

Question Number 94752    Answers: 0   Comments: 1

Find the volume of the region bounded above by the surface z=x and below by the region x^2 +y^2 −2y=0 ? pleas sir can you help me becouse im very nedd ?

$${Find}\:{the}\:{volume}\:{of}\:{the}\:{region}\:{bounded}\:{above}\:{by}\:{the}\:{surface}\:{z}={x} \\ $$$${and}\:{below}\:{by}\:{the}\:{region}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{y}=\mathrm{0}\:? \\ $$$${pleas}\:{sir}\:{can}\:{you}\:{help}\:{me}\:{becouse}\:{im}\:{very}\:{nedd}\:? \\ $$

Question Number 94742    Answers: 0   Comments: 1

Question Number 94739    Answers: 2   Comments: 2

Question Number 94735    Answers: 0   Comments: 9

∫((2t)/((1+t^4 )(1+t)))dt=?

$$\int\frac{\mathrm{2t}}{\left(\mathrm{1}+\mathrm{t}^{\mathrm{4}} \right)\left(\mathrm{1}+\mathrm{t}\right)}\mathrm{dt}=? \\ $$

Question Number 94753    Answers: 1   Comments: 0

Question Number 94730    Answers: 2   Comments: 0

Question Number 94723    Answers: 2   Comments: 0

lim_(x→0) ((∫_0 ^x^2 (√(4+t^3 )) dt)/x^2 ) ?

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\underset{\mathrm{0}} {\overset{{x}^{\mathrm{2}} } {\int}}\:\sqrt{\mathrm{4}+{t}^{\mathrm{3}} \:}\:{dt}}{{x}^{\mathrm{2}} }\:?\: \\ $$

Question Number 94718    Answers: 0   Comments: 4

∫ (√(tan x)) dx = ∫(((√(tan x))+(√(cot x)))/2) dx + ∫ (((√(tan x))−(√(cot x)))/2) dx =(1/((√2) ))∫ ((sin x+cos x)/(√(sin 2x))) dx + (1/((√2) ))∫ ((sin x−cos x)/(√(sin 2x))) dx = (1/((√2) ))∫ ((sin x+cos x)/(√(1−(sin x−cos x)^2 ))) dx + (1/(√2)) ∫ ((sin x−cos x)/(√((sin x+cos x)^2 −1))) dx = (1/((√2) ))∫ (dt/(√(1−t^2 ))) +(1/((√2) ))∫ ((−du)/(√(u^2 −1))) = (1/((√2) ))sin^(−1) (t) −(1/(√2)) ln(u+(√(u^2 −1))) +c = (1/(√2)) sin^(−1) (sin x−cos x)− (1/(√2)) ln (sin x+cos x+(√(sin 2x))) + c where t = sin x−cos x ; u = sin x+cos x

$$\int\:\sqrt{\mathrm{tan}\:\mathrm{x}}\:\mathrm{dx}\:= \\ $$$$\int\frac{\sqrt{\mathrm{tan}\:\mathrm{x}}+\sqrt{\mathrm{cot}\:\mathrm{x}}}{\mathrm{2}}\:\mathrm{dx}\:+\:\int\:\frac{\sqrt{\mathrm{tan}\:\mathrm{x}}−\sqrt{\mathrm{cot}\:\mathrm{x}}}{\mathrm{2}}\:\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}\:}\int\:\frac{\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}}{\sqrt{\mathrm{sin}\:\mathrm{2x}}}\:\mathrm{dx}\:+\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}\:}\int\:\frac{\mathrm{sin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}}{\sqrt{\mathrm{sin}\:\mathrm{2x}}}\:\mathrm{dx} \\ $$$$=\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}\:}\int\:\frac{\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}}{\sqrt{\mathrm{1}−\left(\mathrm{sin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{2}} }}\:\mathrm{dx}\:+\: \\ $$$$\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:\int\:\frac{\mathrm{sin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}}{\sqrt{\left(\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{2}} −\mathrm{1}}}\:\mathrm{dx}\: \\ $$$$=\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}\:}\int\:\frac{\mathrm{dt}}{\sqrt{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }}\:+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}\:}\int\:\frac{−\mathrm{du}}{\sqrt{\mathrm{u}^{\mathrm{2}} −\mathrm{1}}} \\ $$$$=\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}\:}\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{t}\right)\:−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:\mathrm{ln}\left(\mathrm{u}+\sqrt{\mathrm{u}^{\mathrm{2}} −\mathrm{1}}\right)\:+\mathrm{c} \\ $$$$=\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{sin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}\right)− \\ $$$$\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:\mathrm{ln}\:\left(\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}+\sqrt{\mathrm{sin}\:\mathrm{2x}}\right)\:+\:\mathrm{c}\: \\ $$$$\mathrm{where}\:\mathrm{t}\:=\:\mathrm{sin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}\:;\: \\ $$$$\mathrm{u}\:=\:\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}\: \\ $$

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