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Question Number 194359    Answers: 1   Comments: 1

v2.282 has been published. This update fixes issues with missed notifications and adds an arbitarary precision scientific calculator.

$${v}\mathrm{2}.\mathrm{282}\:\mathrm{has}\:\mathrm{been}\:\mathrm{published}.\:\mathrm{This} \\ $$$$\mathrm{update}\:\mathrm{fixes}\:\mathrm{issues}\:\mathrm{with}\:\mathrm{missed} \\ $$$$\mathrm{notifications}\:\mathrm{and}\:\mathrm{adds}\:\mathrm{an} \\ $$$$\mathrm{arbitarary}\:\mathrm{precision}\: \\ $$$$\mathrm{scientific}\:\mathrm{calculator}. \\ $$

Question Number 194352    Answers: 0   Comments: 0

$$\:\:\:\:\Subset \\ $$

Question Number 194350    Answers: 2   Comments: 0

Question Number 194344    Answers: 2   Comments: 0

Question Number 194343    Answers: 3   Comments: 0

⌊9.9^− ⌋ = ?

$$\lfloor\mathrm{9}.\overset{−} {\mathrm{9}}\rfloor\:=\:? \\ $$

Question Number 194338    Answers: 2   Comments: 0

Question Number 194335    Answers: 1   Comments: 0

Question Number 194334    Answers: 0   Comments: 2

^()

$$\:\:\:\:\underbrace{ ^{} } \\ $$

Question Number 194331    Answers: 1   Comments: 0

Question Number 194327    Answers: 1   Comments: 0

If ta4θ = 1, find the values of θ.

$$\mathrm{If}\:\mathrm{ta4}\theta\:=\:\mathrm{1},\:\mathrm{find}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:\theta. \\ $$

Question Number 194326    Answers: 1   Comments: 0

(√(((√(x^2 +66^2 +x))/x) )) −(√(x(√(x^2 +66^2 ))−x^2 )) = 5

$$\:\:\sqrt{\frac{\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{66}^{\mathrm{2}} +\mathrm{x}}}{\mathrm{x}}\:}\:−\sqrt{\mathrm{x}\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{66}^{\mathrm{2}} }−\mathrm{x}^{\mathrm{2}} }\:=\:\mathrm{5}\: \\ $$

Question Number 194325    Answers: 1   Comments: 0

X

$$\:\:\cancel{\mathcal{X}} \\ $$

Question Number 194322    Answers: 0   Comments: 0

Question Number 194321    Answers: 1   Comments: 0

find min −(((x−y)(((xy)/4)−4)^2 )/(xy)) s. t. x>0>y

$$\mathrm{find}\:\mathrm{min}\:\:−\frac{\left({x}−{y}\right)\left(\frac{{xy}}{\mathrm{4}}−\mathrm{4}\right)^{\mathrm{2}} }{{xy}} \\ $$$$\mathrm{s}.\:\mathrm{t}.\:\:{x}>\mathrm{0}>{y} \\ $$

Question Number 194317    Answers: 0   Comments: 0

Question Number 194316    Answers: 0   Comments: 1

if x∈R & x^x^6 =((√2))^(√2) ⇒ x=?

$${if}\:\:{x}\in{R}\:\:\&\:\:{x}^{{x}^{\mathrm{6}} } =\left(\sqrt{\mathrm{2}}\right)^{\sqrt{\mathrm{2}}} \:\Rightarrow\:\:{x}=? \\ $$

Question Number 194315    Answers: 1   Comments: 0

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

Question Number 194312    Answers: 0   Comments: 0

Question Number 194304    Answers: 1   Comments: 0

Question Number 194301    Answers: 1   Comments: 0

Resolution de l exercice du 28.6.23 (envoye par universe ) Q194116

$$\boldsymbol{\mathrm{Resolution}}\:\boldsymbol{\mathrm{de}}\:\boldsymbol{\mathrm{l}}\:\boldsymbol{\mathrm{exercice}}\:\boldsymbol{\mathrm{du}}\:\mathrm{28}.\mathrm{6}.\mathrm{23} \\ $$$$\:\:\left({env}\mathrm{o}{ye}\:{par}\:{universe}\:\right) \\ $$$$\boldsymbol{{Q}}\mathrm{194116} \\ $$$$ \\ $$

Question Number 194297    Answers: 1   Comments: 0

Let a , b , c be real positive numbers & abc=1 prove that ((ab)/(a^5 +b^5 +ab))+((bc)/(b^5 +c^5 +bc))+((ac)/(a^5 +c^5 +ac))≤1

$${Let}\:{a}\:,\:{b}\:,\:{c}\:{be}\:\:{real}\:{positive}\:{numbers}\:\&\: \\ $$$${abc}=\mathrm{1}\: \\ $$$${prove}\:{that} \\ $$$$\frac{{ab}}{{a}^{\mathrm{5}} +{b}^{\mathrm{5}} +{ab}}+\frac{{bc}}{{b}^{\mathrm{5}} +{c}^{\mathrm{5}} +{bc}}+\frac{{ac}}{{a}^{\mathrm{5}} +{c}^{\mathrm{5}} +{ac}}\leqslant\mathrm{1} \\ $$

Question Number 194295    Answers: 1   Comments: 0

Question Number 194292    Answers: 0   Comments: 0

Question Number 194286    Answers: 2   Comments: 0

find lim_(x→0) ⌊ ((tan(x))/x)⌋

$$ \\ $$$$\:\:\boldsymbol{{find}}\:\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{lim}}}\:\lfloor\:\frac{\boldsymbol{{tan}}\left(\boldsymbol{{x}}\right)}{\boldsymbol{{x}}}\rfloor \\ $$

Question Number 194282    Answers: 1   Comments: 0

f(f(x)) = ax + b 1. show that f(ax+b) = af(x) + b deduce f ′(ax + b) 2. Show that f ′(x) is a constant hence deduce f

$${f}\left({f}\left({x}\right)\right)\:=\:{ax}\:+\:{b} \\ $$$$\mathrm{1}.\:{show}\:{that}\:{f}\left({ax}+{b}\right)\:=\:{af}\left({x}\right)\:+\:{b} \\ $$$${deduce}\:{f}\:'\left({ax}\:+\:{b}\right) \\ $$$$\mathrm{2}.\:{Show}\:{that}\:{f}\:'\left({x}\right)\:{is}\:{a}\:{constant}\: \\ $$$${hence}\:{deduce}\:{f} \\ $$

Question Number 194279    Answers: 1   Comments: 0

$$\:\underbrace{ } \\ $$

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