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Question Number 184656    Answers: 1   Comments: 0

prove that the area of a triangle whose two sides are A^− and B^− is given by (1/2)∣A×B∣. Also find the direction−cosine of normal to this area. Help!

$$\mathrm{prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle} \\ $$$$\mathrm{whose}\:\mathrm{two}\:\mathrm{sides}\:\mathrm{are}\:\overset{−} {\mathrm{A}}\:\mathrm{and}\:\overset{−} {\mathrm{B}}\:\mathrm{is} \\ $$$$\mathrm{given}\:\mathrm{by}\:\frac{\mathrm{1}}{\mathrm{2}}\mid\mathrm{A}×\mathrm{B}\mid. \\ $$$$\mathrm{Also}\:\mathrm{find}\:\mathrm{the}\:\mathrm{direction}−\mathrm{cosine} \\ $$$$\mathrm{of}\:\mathrm{normal}\:\mathrm{to}\:\mathrm{this}\:\mathrm{area}. \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Help}! \\ $$

Question Number 184655    Answers: 1   Comments: 0

prove that an angle inscribe in a semi−circle is a right angle. Help!

$$\mathrm{prove}\:\mathrm{that}\:\mathrm{an}\:\mathrm{angle}\:\mathrm{inscribe}\:\mathrm{in}\:\mathrm{a}\: \\ $$$$\mathrm{semi}−\mathrm{circle}\:\mathrm{is}\:\mathrm{a}\:\mathrm{right}\:\mathrm{angle}. \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Help}! \\ $$

Question Number 184574    Answers: 0   Comments: 5

Test whether this is Convergent or Divergent Σ_(n=0) ^∞ (−1)^n ((n!x^n )/(5n)) Help!

$$\mathrm{Test}\:\mathrm{whether}\:\mathrm{this}\:\mathrm{is}\:\mathrm{Convergent}\:\mathrm{or} \\ $$$$\mathrm{Divergent} \\ $$$$\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{n}} \frac{\mathrm{n}!\mathrm{x}^{\mathrm{n}} }{\mathrm{5n}} \\ $$$$ \\ $$$$\mathrm{Help}! \\ $$

Question Number 184497    Answers: 1   Comments: 0

Question Number 184487    Answers: 0   Comments: 1

Question Number 184454    Answers: 0   Comments: 1

Question Number 184320    Answers: 0   Comments: 4

All-time Universal Formula determinant (((OLD+1=NEW))) Year:-The above formula applies every year. Month:-It also applies every month. Day:-It also applies every day. .... Second:-It also applies every second. ... SO, along with Happy New Year! also: Happy New Month! Happy New Day! .... Happy New Second! ...

$$\boldsymbol{\mathrm{All}}-\boldsymbol{\mathrm{time}}\:\boldsymbol{\mathrm{Universal}}\:\boldsymbol{\mathrm{Formula}} \\ $$$$\:\begin{array}{|c|}{\boldsymbol{\mathrm{OLD}}+\mathrm{1}=\boldsymbol{\mathrm{NEW}}}\\\hline\end{array}\: \\ $$$$\boldsymbol{{Year}}:-\boldsymbol{\mathcal{T}{he}}\:\boldsymbol{{above}}\:\boldsymbol{{formula}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{applies}}\:\boldsymbol{{every}}\:\boldsymbol{{year}}. \\ $$$$\boldsymbol{{Month}}:-\boldsymbol{{It}}\:\boldsymbol{{also}}\:\boldsymbol{{applies}}\:\boldsymbol{{every}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{month}}. \\ $$$$\boldsymbol{{Day}}:-\boldsymbol{{It}}\:\boldsymbol{{also}}\:\boldsymbol{{applies}}\:\boldsymbol{{every}}\:\boldsymbol{{day}}. \\ $$$$.... \\ $$$$\boldsymbol{{Second}}:-\boldsymbol{{It}}\:\boldsymbol{{also}}\:\boldsymbol{{applies}}\:\boldsymbol{{every}}\:\:\boldsymbol{{second}}. \\ $$$$... \\ $$$$\boldsymbol{\mathrm{SO}}, \\ $$$$\boldsymbol{\mathrm{along}}\:\boldsymbol{\mathrm{with}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{Happy}}\:\boldsymbol{\mathrm{New}}\:\boldsymbol{\mathrm{Year}}! \\ $$$$\boldsymbol{\mathrm{also}}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{Happy}}\:\boldsymbol{\mathrm{New}}\:\boldsymbol{\mathrm{Month}}! \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{Happy}}\:\boldsymbol{\mathrm{New}}\:\boldsymbol{\mathrm{Day}}! \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:.... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{Happy}}\:\boldsymbol{\mathrm{New}}\:\boldsymbol{\mathrm{Second}}! \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:... \\ $$$$ \\ $$

Question Number 184311    Answers: 1   Comments: 0

Question Number 184307    Answers: 1   Comments: 0

Show that the boundary−value problem y′′+λy=0 y(0)=0, y(L)=0 has only the trival solution y=0 for the cases λ=0 and λ<0. let L be a non−zero real number. ?

$$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{boundary}−\mathrm{value} \\ $$$$\mathrm{problem}\:\mathrm{y}''+\lambda\mathrm{y}=\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\mathrm{y}\left(\mathrm{0}\right)=\mathrm{0}, \\ $$$$\mathrm{y}\left(\mathrm{L}\right)=\mathrm{0}\:\mathrm{has}\:\mathrm{only}\:\mathrm{the}\:\mathrm{trival}\:\mathrm{solution} \\ $$$$\mathrm{y}=\mathrm{0}\:\mathrm{for}\:\mathrm{the}\:\mathrm{cases}\:\lambda=\mathrm{0}\:\mathrm{and}\:\lambda<\mathrm{0}. \\ $$$$\mathrm{let}\:\mathrm{L}\:\mathrm{be}\:\mathrm{a}\:\mathrm{non}−\mathrm{zero}\:\mathrm{real}\:\mathrm{number}. \\ $$$$ \\ $$$$ \\ $$$$? \\ $$

Question Number 184306    Answers: 1   Comments: 0

Consider the boundary value problem y^(′′) −2y′+2y=0, y(a)=c ,y(b)=d. 1) If this problem has a unique solution, how are a and b related? 2) If this problem has no solution, how are a,b,c and d related? Help!

$$\mathrm{Consider}\:\mathrm{the}\:\mathrm{boundary}\:\mathrm{value}\: \\ $$$$\mathrm{problem}\:\mathrm{y}^{''} −\mathrm{2y}'+\mathrm{2y}=\mathrm{0},\:\:\:\:\:\:\:\mathrm{y}\left(\mathrm{a}\right)=\mathrm{c} \\ $$$$,\mathrm{y}\left(\mathrm{b}\right)=\mathrm{d}. \\ $$$$\left.\mathrm{1}\right)\:\mathrm{If}\:\mathrm{this}\:\mathrm{problem}\:\mathrm{has}\:\mathrm{a}\:\mathrm{unique} \\ $$$$\mathrm{solution},\:\mathrm{how}\:\mathrm{are}\:\mathrm{a}\:\mathrm{and}\:\mathrm{b}\:\mathrm{related}? \\ $$$$\left.\mathrm{2}\right)\:\mathrm{If}\:\mathrm{this}\:\mathrm{problem}\:\mathrm{has}\:\mathrm{no}\:\mathrm{solution}, \\ $$$$\mathrm{how}\:\mathrm{are}\:\mathrm{a},\mathrm{b},\mathrm{c}\:\mathrm{and}\:\mathrm{d}\:\mathrm{related}? \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Help}! \\ $$

Question Number 184264    Answers: 1   Comments: 0

Question Number 184218    Answers: 1   Comments: 0

Question Number 184188    Answers: 1   Comments: 0

Differentiate, y = x^(x−1) hi

$$\mathrm{Differentiate},\:\mathrm{y}\:=\:\mathrm{x}^{\mathrm{x}−\mathrm{1}} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{hi} \\ $$

Question Number 184187    Answers: 1   Comments: 0

Differentiate, y=e^x + x^x M.m

$$\mathrm{Differentiate},\:\mathrm{y}=\mathrm{e}^{\mathrm{x}} \:+\:\mathrm{x}^{\mathrm{x}} \\ $$$$ \\ $$$$\mathrm{M}.\mathrm{m} \\ $$

Question Number 184185    Answers: 1   Comments: 0

Differentiate, y=(log_e x)^x M.m

$$\mathrm{Differentiate},\:\mathrm{y}=\left(\mathrm{log}_{\mathrm{e}} \mathrm{x}\right)^{\mathrm{x}} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{M}.\mathrm{m} \\ $$

Question Number 184184    Answers: 1   Comments: 0

y=(sinx)^x Differentiate

$$\mathrm{y}=\left(\mathrm{sinx}\right)^{\mathrm{x}} \\ $$$$ \\ $$$$\mathrm{Differentiate} \\ $$

Question Number 184126    Answers: 1   Comments: 0

Question Number 184125    Answers: 0   Comments: 0

Question Number 184048    Answers: 5   Comments: 0

{ ((u_0 = 2)),((u_(n+1) = ((2u_n −1)/u_n ))) :} Find u_n .

$$\:\:\begin{cases}{{u}_{\mathrm{0}} \:=\:\mathrm{2}}\\{{u}_{{n}+\mathrm{1}} \:=\:\frac{\mathrm{2}{u}_{{n}} \:−\mathrm{1}}{{u}_{{n}} }}\end{cases} \\ $$$$\:\:\:{Find}\:{u}_{{n}} . \\ $$

Question Number 184030    Answers: 1   Comments: 0

How many words can be made from 5 letters if (a) all letters are different (b) 2 letters are identical (c) all letters are different but 2 partucular letters cannot be adjacent. M.m

$$\mathrm{How}\:\mathrm{many}\:\mathrm{words}\:\mathrm{can}\:\mathrm{be}\:\mathrm{made}\: \\ $$$$\mathrm{from}\:\mathrm{5}\:\mathrm{letters}\:\mathrm{if} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{all}\:\mathrm{letters}\:\mathrm{are}\:\mathrm{different} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{2}\:\mathrm{letters}\:\mathrm{are}\:\mathrm{identical} \\ $$$$\left(\mathrm{c}\right)\:\mathrm{all}\:\mathrm{letters}\:\mathrm{are}\:\mathrm{different}\:\mathrm{but}\:\mathrm{2} \\ $$$$\mathrm{partucular}\:\mathrm{letters}\:\mathrm{cannot}\:\mathrm{be} \\ $$$$\mathrm{adjacent}. \\ $$$$ \\ $$$$ \\ $$$$\mathrm{M}.\mathrm{m} \\ $$

Question Number 184028    Answers: 0   Comments: 0

H^(A^P P) Y Y_(E_A R) ! ⌊e⌋⌊i-i⌋⌊e⌋⌊𝛑⌋^(−)

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{H}}^{\boldsymbol{\mathrm{A}}^{\boldsymbol{\mathrm{P}}} \boldsymbol{\mathrm{P}}} \boldsymbol{\mathrm{Y}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{Y}}_{\boldsymbol{\mathrm{E}}_{\boldsymbol{\mathrm{A}}} \boldsymbol{\mathrm{R}}} \:! \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\overline {\lfloor\boldsymbol{\mathrm{e}}\rfloor\lfloor\boldsymbol{\mathrm{i}}-\boldsymbol{\mathrm{i}}\rfloor\lfloor\boldsymbol{\mathrm{e}}\rfloor\lfloor\boldsymbol{\pi}\rfloor}\:\: \\ $$

Question Number 184010    Answers: 0   Comments: 3

determinant ((( determinant (((2023))) )))_( is^ _(a number_(which is divisible_(by_(•_• ) ) ) ) ) (i)its sum of digits & (ii)its sum of squares of digits

$$\:\:\:\:\:\:\:\:\:\:\underset{\:\underset{\underset{\underset{\underset{\underset{\bullet} {\bullet}} {\boldsymbol{\mathrm{by}}}} {\boldsymbol{\mathrm{which}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{divisible}}}} {\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{number}}}} {\boldsymbol{\mathrm{is}}^{\:} }} {\begin{array}{|c|}{\:\begin{array}{|c|}{\mathrm{2023}}\\\hline\end{array}\:}\\\hline\end{array}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\boldsymbol{\mathrm{i}}\right)\boldsymbol{\mathrm{its}}\:\boldsymbol{\mathrm{sum}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{digits}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\& \\ $$$$\:\:\:\:\:\:\:\left(\boldsymbol{\mathrm{ii}}\right)\boldsymbol{\mathrm{its}}\:\boldsymbol{\mathrm{sum}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{squares}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{digits}} \\ $$

Question Number 183913    Answers: 0   Comments: 0

Question Number 183737    Answers: 2   Comments: 0

Question Number 183687    Answers: 3   Comments: 0

∫(1/(lnx))dx Help out

$$\int\frac{\mathrm{1}}{\mathrm{lnx}}\mathrm{dx} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Help}\:\mathrm{out} \\ $$

Question Number 183667    Answers: 0   Comments: 8

If w is one of the complex cube roots of unity, show that (a+wb+w^2 c)(a+w^2 b+wc) is equal to (α^2 +b^2 +c^2 −ab−bc−cα). Kindly help me out, Thank you.

$$\mathrm{If}\:\mathrm{w}\:\mathrm{is}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{complex}\:\mathrm{cube}\: \\ $$$$\mathrm{roots}\:\mathrm{of}\:\mathrm{unity},\:\mathrm{show}\:\mathrm{that} \\ $$$$\left(\mathrm{a}+\mathrm{wb}+\mathrm{w}^{\mathrm{2}} \mathrm{c}\right)\left(\mathrm{a}+\mathrm{w}^{\mathrm{2}} \mathrm{b}+\mathrm{wc}\right)\:\mathrm{is}\:\mathrm{equal} \\ $$$$\mathrm{to}\:\left(\alpha^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} −\mathrm{ab}−\mathrm{bc}−\mathrm{c}\alpha\right). \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Kindly}\:\mathrm{help}\:\mathrm{me}\:\mathrm{out},\:\mathrm{Thank}\:\mathrm{you}. \\ $$

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