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Question Number 203638    Answers: 2   Comments: 2

Question Number 203636    Answers: 1   Comments: 1

Question Number 203634    Answers: 2   Comments: 0

$$\:\:\: \\ $$

Question Number 203632    Answers: 0   Comments: 0

Question Number 203625    Answers: 2   Comments: 0

Question Number 203615    Answers: 1   Comments: 0

I = ∫_0 ^(π/2) (x^2 /(1+sin^2 (x)))dx

$${I}\:=\:\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:\left({x}\right)}{dx} \\ $$

Question Number 203613    Answers: 1   Comments: 2

Question Number 203605    Answers: 0   Comments: 3

Value of x?

$$\mathrm{Value}\:\mathrm{of}\:\boldsymbol{\mathrm{x}}? \\ $$

Question Number 203602    Answers: 2   Comments: 0

Question Number 203577    Answers: 0   Comments: 0

Question Number 203576    Answers: 2   Comments: 0

Question Number 203573    Answers: 1   Comments: 0

Question Number 203570    Answers: 0   Comments: 2

Suggested solution method to question 203502 ze^z =1 Obviously the only real solution is z=W(1)≈.567143290 z=a+bi∧b≠0 (a+bi)e^(a+bi) =1 (a+bi)(cos b +isin b)e^a =1 e^a (acos b −bsin b)+e^a (asin b +bcos b)i=1 { ((e^a (acos b −bsin b)=1)),((e^a (asin b +bcos b)=0 ⇒ a=−bcot b)) :} Inserting & transforming leaves us with: { ((be^(−bcot b) +sin b =0)),((a=−bcot b)) :} We can only approximate. The first solutions are: b≈±4.37518515 ⇒ a≈−1.53391332 ⇒ z≈−1.53391332±4.37518515i b≈±10.7762995 ⇒ a≈−2.40158510 ⇒ z≈−2.40158510±10.7762995i These are the values of the complex LambertW−function W_n (1); n∈Z Following this path we can solve ze^z =w with z, w∈Z I hope this is helpful!

$$\mathrm{Suggested}\:\mathrm{solution}\:\mathrm{method}\:\mathrm{to} \\ $$$$\mathrm{question}\:\mathrm{203502} \\ $$$$ \\ $$$${z}\mathrm{e}^{{z}} =\mathrm{1} \\ $$$$\mathrm{Obviously}\:\mathrm{the}\:\mathrm{only}\:\mathrm{real}\:\mathrm{solution}\:\mathrm{is} \\ $$$${z}={W}\left(\mathrm{1}\right)\approx.\mathrm{567143290} \\ $$$$ \\ $$$${z}={a}+{b}\mathrm{i}\wedge{b}\neq\mathrm{0} \\ $$$$\left({a}+{b}\mathrm{i}\right)\mathrm{e}^{{a}+{b}\mathrm{i}} =\mathrm{1} \\ $$$$\left({a}+{b}\mathrm{i}\right)\left(\mathrm{cos}\:{b}\:+\mathrm{isin}\:{b}\right)\mathrm{e}^{{a}} =\mathrm{1} \\ $$$$\mathrm{e}^{{a}} \left({a}\mathrm{cos}\:{b}\:−{b}\mathrm{sin}\:{b}\right)+\mathrm{e}^{{a}} \left({a}\mathrm{sin}\:{b}\:+{b}\mathrm{cos}\:{b}\right)\mathrm{i}=\mathrm{1} \\ $$$$\begin{cases}{\mathrm{e}^{{a}} \left({a}\mathrm{cos}\:{b}\:−{b}\mathrm{sin}\:{b}\right)=\mathrm{1}}\\{\mathrm{e}^{{a}} \left({a}\mathrm{sin}\:{b}\:+{b}\mathrm{cos}\:{b}\right)=\mathrm{0}\:\Rightarrow\:{a}=−{b}\mathrm{cot}\:{b}}\end{cases} \\ $$$$\mathrm{Inserting}\:\&\:\mathrm{transforming}\:\mathrm{leaves}\:\mathrm{us}\:\mathrm{with}: \\ $$$$\begin{cases}{{b}\mathrm{e}^{−{b}\mathrm{cot}\:{b}} +\mathrm{sin}\:{b}\:=\mathrm{0}}\\{{a}=−{b}\mathrm{cot}\:{b}}\end{cases} \\ $$$$\mathrm{We}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate}. \\ $$$$\mathrm{The}\:\mathrm{first}\:\mathrm{solutions}\:\mathrm{are}: \\ $$$${b}\approx\pm\mathrm{4}.\mathrm{37518515}\:\Rightarrow\:{a}\approx−\mathrm{1}.\mathrm{53391332} \\ $$$$\Rightarrow\:{z}\approx−\mathrm{1}.\mathrm{53391332}\pm\mathrm{4}.\mathrm{37518515i} \\ $$$${b}\approx\pm\mathrm{10}.\mathrm{7762995}\:\Rightarrow\:{a}\approx−\mathrm{2}.\mathrm{40158510} \\ $$$$\Rightarrow\:{z}\approx−\mathrm{2}.\mathrm{40158510}\pm\mathrm{10}.\mathrm{7762995i} \\ $$$$\mathrm{These}\:\mathrm{are}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:\mathrm{the}\:\mathrm{complex} \\ $$$$\mathrm{LambertW}−\mathrm{function}\:{W}_{{n}} \left(\mathrm{1}\right);\:{n}\in\mathbb{Z} \\ $$$$\mathrm{Following}\:\mathrm{this}\:\mathrm{path}\:\mathrm{we}\:\mathrm{can}\:\mathrm{solve} \\ $$$${z}\mathrm{e}^{{z}} ={w}\:\mathrm{with}\:{z},\:{w}\in\mathbb{Z} \\ $$$$\mathrm{I}\:\mathrm{hope}\:\mathrm{this}\:\mathrm{is}\:\mathrm{helpful}! \\ $$

Question Number 203569    Answers: 1   Comments: 3

Question Number 203567    Answers: 0   Comments: 0

Question Number 203566    Answers: 1   Comments: 0

′P′ is a prime number (P>1000). If P ≡ r (mod 1000). how many value of ′r′.

$$'\mathrm{P}'\:\mathrm{is}\:\mathrm{a}\:\mathrm{prime}\:\mathrm{number}\:\left(\mathrm{P}>\mathrm{1000}\right).\: \\ $$$$\mathrm{If}\:\:\:\mathrm{P}\:\equiv\:\mathrm{r}\:\left(\mathrm{mod}\:\mathrm{1000}\right).\:\mathrm{how}\:\mathrm{many}\:\mathrm{value}\:\mathrm{of}\:'\mathrm{r}'. \\ $$

Question Number 203565    Answers: 1   Comments: 0

Solve the following equation simultaneously and find the stationary points: 2xy^2 c^2 − 4x^3 y^2 − 2xy^4 = 0 -----(1) 2x^2 yc^2 − 2x^4 y − 4x^2 y^3 = 0 -----(2) Please, I need a well detail calculation Thank you

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{following}\:\mathrm{equation}\:\mathrm{simultaneously} \\ $$$$\mathrm{and}\:\mathrm{find}\:\mathrm{the}\:\mathrm{stationary}\:\mathrm{points}: \\ $$$$\mathrm{2xy}^{\mathrm{2}} \mathrm{c}^{\mathrm{2}} \:−\:\mathrm{4x}^{\mathrm{3}} \mathrm{y}^{\mathrm{2}} \:−\:\mathrm{2xy}^{\mathrm{4}} \:=\:\mathrm{0}\:-----\left(\mathrm{1}\right) \\ $$$$\mathrm{2x}^{\mathrm{2}} \mathrm{yc}^{\mathrm{2}} \:−\:\mathrm{2x}^{\mathrm{4}} \mathrm{y}\:−\:\mathrm{4x}^{\mathrm{2}} \mathrm{y}^{\mathrm{3}} \:=\:\mathrm{0}\:-----\left(\mathrm{2}\right) \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Please},\:\mathrm{I}\:\mathrm{need}\:\mathrm{a}\:\mathrm{well}\:\mathrm{detail}\:\mathrm{calculation} \\ $$$$\mathrm{Thank}\:\mathrm{you} \\ $$

Question Number 203564    Answers: 1   Comments: 0

∫_0 ^( ∞) ((sin^( 3) (x))/x^( 2) ) dx= ?

$$ \\ $$$$\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}^{\:\mathrm{3}} \left({x}\right)}{{x}^{\:\mathrm{2}} }\:{dx}=\:?\:\:\:\:\: \\ $$

Question Number 203560    Answers: 0   Comments: 1

Question Number 203561    Answers: 0   Comments: 1

(√(2x−5+3=?))

$$\sqrt{\mathrm{2}{x}−\mathrm{5}+\mathrm{3}=?} \\ $$

Question Number 203557    Answers: 1   Comments: 1

Question Number 203554    Answers: 0   Comments: 0

Question Number 203547    Answers: 0   Comments: 0

Question Number 203545    Answers: 1   Comments: 0

If A,B,C are finite sets whose elements are from the same universal set U and n(A) denotes the number of element in the set A (a) Show by means of venn diagram that n(A ∪ B) = n(A) + n(B) − n(A ∩ B) (b) Using the fact that (A ∪ B ∪ C) = (A∪B)∪C =A∪(B∪C) deduce an expression for (A∪B∪C) (c) If n(A∪B)= n(A∩B), what can be said about A and B? How did you reach your conclusion. Thank you in advance

$$\mathrm{If}\:\mathrm{A},\mathrm{B},\mathrm{C}\:\mathrm{are}\:\mathrm{finite}\:\mathrm{sets}\:\mathrm{whose}\:\mathrm{elements}\:\mathrm{are}\: \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{same}\:\mathrm{universal}\:\mathrm{set}\:\mathrm{U}\:\mathrm{and}\:\mathrm{n}\left(\mathrm{A}\right)\: \\ $$$$\mathrm{denotes}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{element}\:\mathrm{in}\:\mathrm{the}\:\mathrm{set}\:\mathrm{A} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{Show}\:\mathrm{by}\:\mathrm{means}\:\mathrm{of}\:\mathrm{venn}\:\mathrm{diagram}\:\mathrm{that} \\ $$$$\mathrm{n}\left(\mathrm{A}\:\cup\:\mathrm{B}\right)\:=\:\mathrm{n}\left(\mathrm{A}\right)\:+\:\mathrm{n}\left(\mathrm{B}\right)\:−\:\mathrm{n}\left(\mathrm{A}\:\cap\:\mathrm{B}\right) \\ $$$$\left(\mathrm{b}\right)\:\mathrm{Using}\:\mathrm{the}\:\mathrm{fact}\:\mathrm{that}\:\left(\mathrm{A}\:\cup\:\mathrm{B}\:\cup\:\mathrm{C}\right)\:=\:\left(\mathrm{A}\cup\mathrm{B}\right)\cup\mathrm{C} \\ $$$$=\mathrm{A}\cup\left(\mathrm{B}\cup\mathrm{C}\right)\:\mathrm{deduce}\:\mathrm{an}\:\mathrm{expression}\:\mathrm{for} \\ $$$$\left(\mathrm{A}\cup\mathrm{B}\cup\mathrm{C}\right) \\ $$$$\left(\mathrm{c}\right)\:\mathrm{If}\:\mathrm{n}\left(\mathrm{A}\cup\mathrm{B}\right)=\:\mathrm{n}\left(\mathrm{A}\cap\mathrm{B}\right),\:\:\mathrm{what}\:\mathrm{can}\:\mathrm{be}\:\mathrm{said} \\ $$$$\mathrm{about}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}?\:\mathrm{How}\:\mathrm{did}\:\mathrm{you}\:\mathrm{reach}\:\mathrm{your}\:\mathrm{conclusion}. \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{in}\:\mathrm{advance} \\ $$

Question Number 203544    Answers: 1   Comments: 0

∫_0 ^1_ (dx/((1−x^6 )^(1/6) )) =(π/3)

$$\int_{\mathrm{0}} ^{\mathrm{1}_{} } \frac{{dx}}{\left(\mathrm{1}−{x}^{\mathrm{6}} \right)^{\frac{\mathrm{1}}{\mathrm{6}}} }\:=\frac{\pi}{\mathrm{3}} \\ $$

Question Number 203533    Answers: 2   Comments: 0

Solve for x : 3^(x+2) =15^(x−1)

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x}\:: \\ $$$$\mathrm{3}^{\mathrm{x}+\mathrm{2}} =\mathrm{15}^{\mathrm{x}−\mathrm{1}} \\ $$

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