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AllQuestion and Answers: Page 281

Question Number 192044 Answers: 0 Comments: 0

Question Number 192045 Answers: 0 Comments: 0

(1/((1+(√2))(1+(2)^(1/4) )(1+(2)^(1/8) )(1+(2)^(1/(16)) )))×(((1−(2)^(1/(16)) )/(1−(2_ )^(1/(16)) ))) ⇒_ ((1−(2)^(1/(16)) )/((1+(√2))(1+(2)^(1/4) )(1+(2)^(1/8) )(1−(2)^(1/8) )))⇒(1/((1+(√2))(1+(2)^(1/4) )(1−(2)^(1/4) )))

Question Number 192024 Answers: 1 Comments: 0

Question Number 192023 Answers: 2 Comments: 0

Question Number 192046 Answers: 2 Comments: 0

tow type creams are in a box that one type of these have 25gr mass and another ones have 37gr mass, if the total mass of those is 870gr then find the number of each type of creams.

$${tow}\:{type}\:{creams}\:{are}\:{in}\:{a}\:{box}\:{that}\:{one}\: \\ $$$${type}\:{of}\:{these}\:{have}\:\mathrm{25}{gr}\:{mass}\:{and}\:{another} \\ $$$${ones}\:{have}\:\mathrm{37}{gr}\:{mass},\:{if}\:{the}\:{total}\:{mass} \\ $$$${of}\:{those}\:{is}\:\mathrm{870}{gr}\:{then}\:{find}\:{the}\:{number} \\ $$$${of}\:{each}\:{type}\:{of}\:{creams}. \\ $$

Question Number 192039 Answers: 0 Comments: 0

let x,y,z be positive real number such that x^4 +y^4 +z^4 = 1 find the minimum value of (x^3 /(1−x^8 )) + (y^3 /(1−y^8 )) + (z^3 /(1−z^8 ))

$$\mathrm{let}\:{x},{y},{z}\:\mathrm{be}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{number}\:\mathrm{such}\:\mathrm{that} \\ $$$$\:{x}^{\mathrm{4}} +{y}^{\mathrm{4}} +{z}^{\mathrm{4}} \:=\:\mathrm{1}\:\mathrm{find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value} \\ $$$$\mathrm{of} \\ $$$$\:\:\:\:\frac{{x}^{\mathrm{3}} }{\mathrm{1}−{x}^{\mathrm{8}} }\:+\:\frac{{y}^{\mathrm{3}} }{\mathrm{1}−{y}^{\mathrm{8}} }\:+\:\frac{{z}^{\mathrm{3}} }{\mathrm{1}−{z}^{\mathrm{8}} } \\ $$

Question Number 192015 Answers: 1 Comments: 0

Question Number 192013 Answers: 0 Comments: 0

∫e^(at^b ) dt with a,b as a constant how to evaluate this

$$ \\ $$$$\:\int{e}^{{at}^{{b}} } {dt} \\ $$$$\:{with}\:{a},{b}\:{as}\:{a}\:{constant} \\ $$$$\:{how}\:{to}\:{evaluate}\:{this} \\ $$$$ \\ $$

Question Number 192009 Answers: 2 Comments: 0

if a>1 , show ((Σ_(k=1) ^(a^2 −1) (√(a+(√k))))/(Σ_(k=1) ^(a^2 −1) (√(a−(√k))))) = (√2) + 1

$$\:\:\:\:\:{if}\:\:{a}>\mathrm{1}\:,\:{show} \\ $$$$\:\:\:\:\:\frac{\underset{{k}=\mathrm{1}} {\overset{{a}^{\mathrm{2}} −\mathrm{1}} {\sum}}\:\:\sqrt{{a}+\sqrt{{k}}}}{\underset{{k}=\mathrm{1}} {\overset{{a}^{\mathrm{2}} −\mathrm{1}} {\sum}}\:\:\sqrt{{a}−\sqrt{{k}}}}\:\:\:=\:\:\:\sqrt{\mathrm{2}}\:\:+\:\:\mathrm{1} \\ $$

Question Number 192001 Answers: 2 Comments: 1

Question Number 191997 Answers: 1 Comments: 2

Show that C={−1,1,−ı,ı} where ı=(√(−1)) with addition operation is a group. Help!

$$\mathrm{Show}\:\mathrm{that}\:\mathbb{C}=\left\{−\mathrm{1},\mathrm{1},−\imath,\imath\right\}\:\mathrm{where} \\ $$$$\imath=\sqrt{−\mathrm{1}}\:\mathrm{with}\:\mathrm{addition}\:\mathrm{operation}\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{group}. \\ $$$$ \\ $$$$\mathrm{Help}! \\ $$

Question Number 191993 Answers: 0 Comments: 0

Question Number 191986 Answers: 1 Comments: 0

Ques. 1 Let (G,∗) be a group, then show that for each a∈G, ∃ a unique element e∈G ∣ a∗e=e∗a=a Ques. 2 If a∈G ⇒ x∈G and x is unique show that if x∗a=e, then a∗x=e. Hello!

$$\mathrm{Ques}.\:\mathrm{1} \\ $$$$\mathrm{Let}\:\left(\mathrm{G},\ast\right)\:\mathrm{be}\:\mathrm{a}\:\mathrm{group},\:\mathrm{then}\:\mathrm{show} \\ $$$$\mathrm{that}\:\mathrm{for}\:\mathrm{each}\:\mathrm{a}\in\mathrm{G},\:\exists\:\mathrm{a}\:\mathrm{unique}\: \\ $$$$\mathrm{element}\:\mathrm{e}\in\mathrm{G}\:\mid\:\mathrm{a}\ast\mathrm{e}=\mathrm{e}\ast\mathrm{a}=\mathrm{a} \\ $$$$ \\ $$$$\mathrm{Ques}.\:\mathrm{2} \\ $$$$\mathrm{If}\:\mathrm{a}\in\mathrm{G}\:\Rightarrow\:\mathrm{x}\in\mathrm{G}\:\mathrm{and}\:\mathrm{x}\:\mathrm{is}\:\mathrm{unique} \\ $$$$\mathrm{show}\:\mathrm{that}\:\mathrm{if}\:\mathrm{x}\ast\mathrm{a}=\mathrm{e},\:\mathrm{then}\:\mathrm{a}\ast\mathrm{x}=\mathrm{e}. \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Hello}! \\ $$

Question Number 191966 Answers: 1 Comments: 0

Question Number 191962 Answers: 1 Comments: 0

Question Number 191961 Answers: 0 Comments: 0

Question Number 191960 Answers: 2 Comments: 0

Question Number 191958 Answers: 1 Comments: 2

Question Number 191950 Answers: 1 Comments: 0

(√a) = 1 + (1/a) find: a^2 −a−(√a) = ?

$$\sqrt{\mathrm{a}}\:=\:\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{a}}\:\:\:\mathrm{find}:\:\:\:\mathrm{a}^{\mathrm{2}} −\mathrm{a}−\sqrt{\mathrm{a}}\:=\:? \\ $$

Question Number 191946 Answers: 1 Comments: 0

((fof^(−1) (5)+fof^(−1) (15))/(fof^(−1) (5)))=?

$$\frac{\mathrm{fof}^{−\mathrm{1}} \left(\mathrm{5}\right)+\mathrm{fof}^{−\mathrm{1}} \left(\mathrm{15}\right)}{\mathrm{fof}^{−\mathrm{1}} \left(\mathrm{5}\right)}=? \\ $$

Question Number 191947 Answers: 2 Comments: 0

prove that (√(a+b(√(a−b(√(a+b(√(a−b(√(...)))))))))) = (((√(4a−3b^2 ))+b)/2)

$${prove}\:{that} \\ $$$$\sqrt{{a}+{b}\sqrt{{a}−{b}\sqrt{{a}+{b}\sqrt{{a}−{b}\sqrt{...}}}}}\:\:=\:\:\frac{\sqrt{\mathrm{4}{a}−\mathrm{3}{b}^{\mathrm{2}} }+{b}}{\mathrm{2}} \\ $$

Question Number 191937 Answers: 1 Comments: 0

Check whether (Q, ∙) is a group or not Hello bosses!

$$\mathrm{Check}\:\mathrm{whether}\:\left(\mathrm{Q},\:\centerdot\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{group}\:\mathrm{or} \\ $$$$\mathrm{not} \\ $$$$ \\ $$$$\mathrm{Hello}\:\mathrm{bosses}! \\ $$

Question Number 191935 Answers: 2 Comments: 0

Question Number 191934 Answers: 1 Comments: 0

Question Number 195092 Answers: 1 Comments: 0

lim_(x→1^+ ) ((4x+5)/(x−x^2 ))=?

$$\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}\:\frac{\mathrm{4}{x}+\mathrm{5}}{{x}−{x}^{\mathrm{2}} }=? \\ $$

Question Number 191930 Answers: 1 Comments: 2

What is the value of inside Area of (ABCDEF)? Such that: ∡AOB=120 ∡ANB=60;°R=ON (OA=OB=32cm) ArcAE=ArcBF(r=12cm) BASE is circulare (Aider le tailleur a savoir la surface du tissu necessaire pour couvrir l ′espace indique dans la figure?)

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{inside}\:\mathrm{Area}\:\mathrm{of} \\ $$$$\left(\mathrm{ABCDEF}\right)? \\ $$$$\mathrm{Such}\:\mathrm{that}:\:\measuredangle\mathrm{AOB}=\mathrm{120}\:\:\:\measuredangle\mathrm{ANB}=\mathrm{60};°\mathrm{R}=\mathrm{ON} \\ $$$$\:\left(\mathrm{OA}=\mathrm{OB}=\mathrm{32cm}\right)\:\mathrm{ArcAE}=\mathrm{ArcBF}\left(\mathrm{r}=\mathrm{12cm}\right) \\ $$$$\mathrm{BASE}\:\mathrm{is}\:\mathrm{circulare} \\ $$$$\left({Aider}\:{le}\:{tailleur}\:{a}\:{savoir}\:{la}\:{surface}\right. \\ $$$${du}\:{tissu}\:{necessaire}\:{pour}\:{couvrir}\: \\ $$$$\left.\:{l}\:'\mathrm{e}{space}\:{indique}\:{dans}\:{la}\:{figure}?\right) \\ $$

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