Question and Answers Forum

All Questions   Topic List

AllQuestion and Answers: Page 210

Question Number 195066    Answers: 0   Comments: 0

Question Number 195065    Answers: 2   Comments: 0

Question Number 195064    Answers: 0   Comments: 0

Question Number 195046    Answers: 2   Comments: 1

Question Number 195043    Answers: 2   Comments: 1

Question Number 195042    Answers: 1   Comments: 0

Question Number 195038    Answers: 0   Comments: 1

prove that x+y=(1/(x−y))

$${prove}\:{that}\:{x}+{y}=\frac{\mathrm{1}}{{x}−{y}} \\ $$

Question Number 195035    Answers: 2   Comments: 0

(√i)=?

$$\sqrt{{i}}=? \\ $$

Question Number 195033    Answers: 1   Comments: 0

any point is the function is not continous f(x)=(4x+8)^((ln45)/8) a) −8 b) −2 c) no one d) 5

$${any}\:{point}\:{is}\:{the}\:{function}\:{is} \\ $$$${not}\:{continous} \\ $$$${f}\left({x}\right)=\left(\mathrm{4}{x}+\mathrm{8}\right)^{\frac{{ln}\mathrm{45}}{\mathrm{8}}} \\ $$$$\left.{a}\left.\right)\:−\mathrm{8}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{b}\right)\:−\mathrm{2} \\ $$$$\left.{c}\left.\right)\:{no}\:{one}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{d}\right)\:\mathrm{5} \\ $$

Question Number 195029    Answers: 1   Comments: 0

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

Question Number 195027    Answers: 1   Comments: 2

a_3 x^3 −x^2 +a_1 x−7=0 is a cubic polynomial in x whose Roots are α , β , γ positive real numbers satisfying ((225α^2 )/(α^2 +7))=((144β^2 )/(β^2 +7))=((100γ^2 )/(γ^2 +7)) find (a_1 )

$${a}_{\mathrm{3}} {x}^{\mathrm{3}} −{x}^{\mathrm{2}} +{a}_{\mathrm{1}} {x}−\mathrm{7}=\mathrm{0}\:{is}\:{a}\:{cubic}\:{polynomial}\:{in}\:{x} \\ $$$${whose}\:{Roots}\:{are}\:\alpha\:,\:\beta\:,\:\gamma\:{positive}\:{real}\:{numbers} \\ $$$${satisfying} \\ $$$$\frac{\mathrm{225}\alpha^{\mathrm{2}} }{\alpha^{\mathrm{2}} +\mathrm{7}}=\frac{\mathrm{144}\beta^{\mathrm{2}} }{\beta^{\mathrm{2}} +\mathrm{7}}=\frac{\mathrm{100}\gamma^{\mathrm{2}} }{\gamma^{\mathrm{2}} +\mathrm{7}} \\ $$$${find}\:\left({a}_{\mathrm{1}} \right) \\ $$

Question Number 195021    Answers: 1   Comments: 0

lim_(x→0) ((x−sin x)/x^n )=¿ (n∈N^∗ )

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}−{sin}\:{x}}{{x}^{{n}} }=¿\:\left({n}\in{N}^{\ast} \right) \\ $$

Question Number 195020    Answers: 1   Comments: 0

y=x^x^(x...) => (dy/dx)=¿

$${y}={x}^{{x}^{{x}...} } \\ $$$$=>\:\frac{{dy}}{{dx}}=¿ \\ $$

Question Number 195017    Answers: 3   Comments: 0

(x+1)^3 =1 x=?

$$\left({x}+\mathrm{1}\right)^{\mathrm{3}} =\mathrm{1}\:\:\:\:\:\:\:\:\:\:{x}=? \\ $$

Question Number 195015    Answers: 1   Comments: 0

Question Number 195013    Answers: 1   Comments: 0

lim_(x→3) (((2((√6)−(√(2x)) +(√(6−2x))))/( (√(36−4x^2 )))) )

$$\:\:\:\:\:\:\underset{\mathrm{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\left(\frac{\mathrm{2}\left(\sqrt{\mathrm{6}}−\sqrt{\mathrm{2x}}\:+\sqrt{\mathrm{6}−\mathrm{2x}}\right)}{\:\sqrt{\mathrm{36}−\mathrm{4x}^{\mathrm{2}} }}\:\right) \\ $$

Question Number 195012    Answers: 2   Comments: 0

Question Number 195003    Answers: 1   Comments: 0

y=lnx^x^x y^′ =?

$${y}={lnx}^{{x}^{{x}} } \:\:\:\:\:\:\:\:\:{y}^{'} =? \\ $$

Question Number 195002    Answers: 1   Comments: 0

x^(√x) =((√x))^x x=?

$${x}^{\sqrt{{x}}} =\left(\sqrt{{x}}\right)^{{x}} \:\:\:\:\:{x}=? \\ $$

Question Number 194998    Answers: 0   Comments: 2

Resolution du probldme pose par sonukgindia (16.7.2023) voir Q194819 △ABC AM=AN=ADcos (𝛂/2) { ((AC=AM+MC=17 (1))),((AB=AN+NB =18 (2))) :} AB−AC=1=NB−MC (3) △CDE cos (C/2)=((CM)/(CD))=((CE)/(CD)) ⇒CM=CE △BDE cos (B/2)=((BE)/(BD))=((BN)/(BD))⇒BN=BE ⇒BE−CE=1 BC=BE+CE=2BE−1 △BEF BF=9 cos B=((BE)/9) BE=9cos B ⇒BC=18cos B−1 Posons BC=x x=18cos B−1 d apres triangle ABC AC^2 =AB^2 +BC^2 −2AB.BCcos B ⇒17^2 =18^2 +x^2 −36(((x+1)/(18))) x^2 +2x−35=0 alors x=5

$$\mathrm{Resolution}\:\mathrm{du}\:\mathrm{probldme}\:\mathrm{pose}\:\mathrm{par}\: \\ $$$$\mathrm{sonukgindia}\:\left(\mathrm{16}.\mathrm{7}.\mathrm{2023}\right) \\ $$$$\mathrm{voir}\:\:\mathrm{Q194819} \\ $$$$\bigtriangleup\boldsymbol{\mathrm{ABC}}\:\:\boldsymbol{\mathrm{AM}}=\boldsymbol{\mathrm{AN}}=\boldsymbol{\mathrm{AD}}\mathrm{cos}\:\frac{\boldsymbol{\alpha}}{\mathrm{2}} \\ $$$$\begin{cases}{\boldsymbol{\mathrm{AC}}=\boldsymbol{\mathrm{AM}}+\boldsymbol{\mathrm{MC}}=\mathrm{17}\:\:\:\:\:\:\:\:\:\:\left(\mathrm{1}\right)}\\{\boldsymbol{\mathrm{AB}}=\boldsymbol{\mathrm{AN}}+\boldsymbol{\mathrm{NB}}\:\:=\mathrm{18}\:\:\:\:\:\:\:\:\:\:\left(\mathrm{2}\right)}\end{cases} \\ $$$$\:\:\boldsymbol{\mathrm{AB}}−\boldsymbol{\mathrm{AC}}=\mathrm{1}=\boldsymbol{\mathrm{NB}}−\boldsymbol{\mathrm{MC}}\:\:\:\left(\mathrm{3}\right) \\ $$$$\bigtriangleup\boldsymbol{\mathrm{CDE}}\:\:\:\mathrm{cos}\:\frac{\boldsymbol{\mathrm{C}}}{\mathrm{2}}=\frac{\boldsymbol{\mathrm{CM}}}{\boldsymbol{\mathrm{CD}}}=\frac{\boldsymbol{\mathrm{CE}}}{\boldsymbol{\mathrm{CD}}}\:\Rightarrow\boldsymbol{\mathrm{CM}}=\boldsymbol{\mathrm{CE}} \\ $$$$\bigtriangleup\boldsymbol{\mathrm{BDE}}\:\:\:\mathrm{cos}\:\frac{\boldsymbol{\mathrm{B}}}{\mathrm{2}}=\frac{\boldsymbol{\mathrm{BE}}}{\boldsymbol{\mathrm{BD}}}=\frac{\boldsymbol{\mathrm{BN}}}{\boldsymbol{\mathrm{BD}}}\Rightarrow\boldsymbol{\mathrm{BN}}=\boldsymbol{\mathrm{BE}} \\ $$$$\Rightarrow\boldsymbol{\mathrm{BE}}−\boldsymbol{\mathrm{CE}}=\mathrm{1} \\ $$$$\:\:\:\:\boldsymbol{\mathrm{BC}}=\boldsymbol{\mathrm{BE}}+\boldsymbol{\mathrm{CE}}=\mathrm{2}\boldsymbol{\mathrm{BE}}−\mathrm{1} \\ $$$$\:\bigtriangleup\boldsymbol{\mathrm{BEF}}\:\:\:\boldsymbol{\mathrm{BF}}=\mathrm{9}\:\:\:\:\mathrm{cos}\:\boldsymbol{\mathrm{B}}=\frac{\boldsymbol{\mathrm{BE}}}{\mathrm{9}}\:\: \\ $$$$\:\:\boldsymbol{\mathrm{BE}}=\mathrm{9cos}\:\boldsymbol{\mathrm{B}}\:\:\:\:\Rightarrow\boldsymbol{\mathrm{BC}}=\mathrm{18cos}\:\boldsymbol{\mathrm{B}}−\mathrm{1} \\ $$$$\:\:\boldsymbol{\mathrm{Posons}}\:\:\boldsymbol{\mathrm{BC}}=\boldsymbol{\mathrm{x}}\:\:\:\:\boldsymbol{\mathrm{x}}=\mathrm{18cos}\:\boldsymbol{\mathrm{B}}−\mathrm{1} \\ $$$$ \\ $$$$\:\:\boldsymbol{\mathrm{d}}\:\boldsymbol{\mathrm{apres}}\:\boldsymbol{\mathrm{triangle}}\:\:\boldsymbol{\mathrm{ABC}} \\ $$$$\:\boldsymbol{\mathrm{AC}}^{\mathrm{2}} =\boldsymbol{\mathrm{AB}}^{\mathrm{2}} +\boldsymbol{\mathrm{BC}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{\mathrm{AB}}.\boldsymbol{\mathrm{BC}}\mathrm{cos}\:\boldsymbol{\mathrm{B}} \\ $$$$\:\:\:\: \\ $$$$\Rightarrow\mathrm{17}^{\mathrm{2}} =\mathrm{18}^{\mathrm{2}} +\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{36}\left(\frac{\boldsymbol{\mathrm{x}}+\mathrm{1}}{\mathrm{18}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{\mathrm{x}}−\mathrm{35}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{alors}}\:\:\:\:\:\:\boldsymbol{\mathrm{x}}=\mathrm{5} \\ $$

Question Number 194985    Answers: 0   Comments: 0

lim_(n→+∞) [((1/n))^n +((2/n))^n +...(((n−1)/n))^n ]=?

$$\underset{\mathrm{n}\rightarrow+\infty} {\mathrm{lim}}\left[\left(\frac{\mathrm{1}}{\mathrm{n}}\right)^{\mathrm{n}} +\left(\frac{\mathrm{2}}{\mathrm{n}}\right)^{\mathrm{n}} +...\left(\frac{\mathrm{n}−\mathrm{1}}{\mathrm{n}}\right)^{\mathrm{n}} \right]=? \\ $$

Question Number 194991    Answers: 0   Comments: 4

a_3 x^3 −x^2 +a_1 x−7=0 is a cubic polynomial in x whose Roots are α , β , γ positive real numbers satisfying ((225α^2 )/(α^2 +7))=((144β^2 )/(β^2 +7))=((100γ^2 )/(γ^2 +7)) Find (a_1 )

$$ \\ $$$${a}_{\mathrm{3}} {x}^{\mathrm{3}} −{x}^{\mathrm{2}} +{a}_{\mathrm{1}} {x}−\mathrm{7}=\mathrm{0}\:{is}\:{a}\:{cubic}\:{polynomial}\:{in}\:{x} \\ $$$${whose}\:{Roots}\:{are}\:\alpha\:,\:\beta\:,\:\gamma\:{positive}\:{real}\:{numbers} \\ $$$${satisfying} \\ $$$$\frac{\mathrm{225}\alpha^{\mathrm{2}} }{\alpha^{\mathrm{2}} +\mathrm{7}}=\frac{\mathrm{144}\beta^{\mathrm{2}} }{\beta^{\mathrm{2}} +\mathrm{7}}=\frac{\mathrm{100}\gamma^{\mathrm{2}} }{\gamma^{\mathrm{2}} +\mathrm{7}} \\ $$$${Find}\:\left({a}_{\mathrm{1}} \right) \\ $$

Question Number 194990    Answers: 0   Comments: 0

If (f(x))=x^2 −x+1 find the value of f(1971)+f(2021)+f(50)

$$\mathrm{I}{f}\:\left({f}\left({x}\right)\right)={x}^{\mathrm{2}} −{x}+\mathrm{1}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$${f}\left(\mathrm{1971}\right)+{f}\left(\mathrm{2021}\right)+{f}\left(\mathrm{50}\right) \\ $$

Question Number 194988    Answers: 1   Comments: 0

ABC=CBF AB

$${ABC}={CBF} \\ $$$${AB} \\ $$

Question Number 194971    Answers: 1   Comments: 2

Question Number 194968    Answers: 2   Comments: 0

  Pg 205      Pg 206      Pg 207      Pg 208      Pg 209      Pg 210      Pg 211      Pg 212      Pg 213      Pg 214   

Terms of Service

Privacy Policy

Contact: info@tinkutara.com