Question and Answers Forum

All Questions   Topic List

AllQuestion and Answers: Page 77

Question Number 215708    Answers: 1   Comments: 0

Question Number 215704    Answers: 1   Comments: 1

Question Number 215710    Answers: 1   Comments: 0

3 different integer numbers are chosen from 0 to 10. what is the probability that they form 1 Cluster 3 Clusters 2 Clusters A cluster is a set of numbers that has maximum range of 2. for example 0,1,2 forms only one cluster. 0,1,4 forms 2 {0,1} and {4}. 0,1,3 also forms 2 {0,1} and {1,2}

$$ \\ $$3 different integer numbers are chosen from 0 to 10. what is the probability that they form 1 Cluster 3 Clusters 2 Clusters A cluster is a set of numbers that has maximum range of 2. for example 0,1,2 forms only one cluster. 0,1,4 forms 2 {0,1} and {4}. 0,1,3 also forms 2 {0,1} and {1,2}

Question Number 215696    Answers: 2   Comments: 0

Question Number 215687    Answers: 2   Comments: 0

Question Number 215679    Answers: 2   Comments: 0

Question Number 215764    Answers: 2   Comments: 0

∫_0 ^1 x^(99) cos^4 (x)dx

$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\mathrm{99}} \mathrm{cos}^{\mathrm{4}} \left(\mathrm{x}\right)\mathrm{dx} \\ $$

Question Number 215667    Answers: 2   Comments: 0

lim_(x→1) sec((π/2)x)(arctanx−(π/4))=?

$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}{sec}\left(\frac{\pi}{\mathrm{2}}{x}\right)\left({arctanx}−\frac{\pi}{\mathrm{4}}\right)=? \\ $$

Question Number 215757    Answers: 0   Comments: 0

Question Number 215659    Answers: 2   Comments: 0

Question Number 215656    Answers: 1   Comments: 0

3x^3 −2x^2 −12x+8=0

$$\mathrm{3}{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{8}=\mathrm{0} \\ $$

Question Number 215650    Answers: 2   Comments: 2

Question Number 215641    Answers: 1   Comments: 0

Question Number 215640    Answers: 2   Comments: 0

If 2025^(sin^2 x) − 2025^(cos^2 x) = (√(2025)) then 2025^(cos2x) + (1/(2025^(cos2x) )) = ?

$${If}\:\:\mathrm{2025}^{{sin}^{\mathrm{2}} {x}} \:−\:\mathrm{2025}^{{cos}^{\mathrm{2}} {x}} \:=\:\sqrt{\mathrm{2025}} \\ $$$${then}\:\mathrm{2025}^{{cos}\mathrm{2}{x}} \:+\:\frac{\mathrm{1}}{\mathrm{2025}^{{cos}\mathrm{2}{x}} }\:=\:? \\ $$

Question Number 215639    Answers: 2   Comments: 0

(1+(1/z))^((1+z)) =2

$$\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{z}}\right)^{\left(\mathrm{1}+\mathrm{z}\right)} =\mathrm{2} \\ $$

Question Number 215625    Answers: 1   Comments: 5

x^4 +x^3 −8x^2 +2x+4=0 x=1 ∨ x=2 ∨ x=2±(√2) Is this right? I have not enough time to edit my solution

$${x}^{\mathrm{4}} +{x}^{\mathrm{3}} −\mathrm{8}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}=\mathrm{0} \\ $$$${x}=\mathrm{1}\:\vee\:{x}=\mathrm{2}\:\vee\:{x}=\mathrm{2}\pm\sqrt{\mathrm{2}} \\ $$$$\mathrm{Is}\:\mathrm{this}\:\mathrm{right}?\:\mathrm{I}\:\mathrm{have}\:\mathrm{not}\:\mathrm{enough}\:\mathrm{time}\:\mathrm{to}\:\mathrm{edit}\:\mathrm{my}\:\mathrm{solution} \\ $$

Question Number 215608    Answers: 1   Comments: 3

Question Number 215604    Answers: 1   Comments: 0

Question Number 215598    Answers: 1   Comments: 1

Question Number 215593    Answers: 2   Comments: 2

Question Number 215589    Answers: 0   Comments: 0

jarak pada peta = skala × jarak sebenarnya

$${jarak}\:{pada}\:{peta}\:=\:{skala}\:×\:{jarak}\:{sebenarnya} \\ $$

Question Number 215588    Answers: 0   Comments: 1

jarak sebenarnya= ((jarak pada peta)/(skala))

$${jarak}\:{sebenarnya}=\:\frac{{jarak}\:{pada}\:{peta}}{{skala}} \\ $$

Question Number 215587    Answers: 0   Comments: 1

skala=((jarak di peta)/(jarak sebenarnya))

$${skala}=\frac{{jarak}\:{di}\:{peta}}{{jarak}\:{sebenarnya}} \\ $$

Question Number 215564    Answers: 2   Comments: 0

given; x(√y)+y(√x)=630 y(√y)+x(√x)=604 find x and y

$$\boldsymbol{\mathrm{given}};\:\:\boldsymbol{\mathrm{x}}\sqrt{\boldsymbol{\mathrm{y}}}+\boldsymbol{\mathrm{y}}\sqrt{\boldsymbol{\mathrm{x}}}=\mathrm{630} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{y}}\sqrt{\boldsymbol{\mathrm{y}}}+\boldsymbol{\mathrm{x}}\sqrt{\boldsymbol{\mathrm{x}}}=\mathrm{604} \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{x}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{y}} \\ $$

Question Number 215559    Answers: 2   Comments: 0

(√y) + (√x) = 5 (√x) ∙ (√y) = 8 (((√x) y − x (√y))/(y − x)) = ?

$$\sqrt{\mathrm{y}}\:+\:\sqrt{\mathrm{x}}\:=\:\mathrm{5} \\ $$$$\sqrt{\mathrm{x}}\:\centerdot\:\sqrt{\mathrm{y}}\:=\:\mathrm{8} \\ $$$$\frac{\sqrt{\mathrm{x}}\:\mathrm{y}\:−\:\mathrm{x}\:\sqrt{\mathrm{y}}}{\mathrm{y}\:−\:\mathrm{x}}\:=\:? \\ $$

Question Number 215550    Answers: 1   Comments: 0

Let u^((1)) ,u^((2)) s.t. { ((u_(tt) ^((1)) =((∂^2 /∂x_1 ^2 )+(∂^2 /∂x_i ^2 ))u^((1)) )),((u^((1)) (x_1 ,x_2 ,0)=𝛙(x_1 ,x_2 ))),((u^((1)) (x_1 ,x_2 ,0)=0)) :}, { ((u_(tt) ^((2)) =((∂^2 /∂x_1 ^2 )+(∂^2 /∂x_2 ^2 )+c^2 )u^((2)) )),((u^((2)) (x_1 x_2 ,0)=0)),((u_t ^((2)) (x_1 ,x_2 ,0)=𝛙(x_1 ,x_2 ))) :} prove:u^((2)) (x_1 ,x_2 ,t)=(1/(2𝛑))∫∫_(𝛏_1 ^2 +𝛏_2 ^2 ≤t^2 ) ((e^(𝛏_2 c) u^((1)) (x_1 ,x_2 ,𝛏_1 )d𝛏_1 d𝛏_2 )/( (√(t^2 −𝛏_1 ^2 −𝛏_2 ^2 ))))

$$\boldsymbol{\mathrm{Let}}\:\boldsymbol{{u}}^{\left(\mathrm{1}\right)} ,\boldsymbol{{u}}^{\left(\mathrm{2}\right)} \boldsymbol{\mathrm{s}}.\boldsymbol{\mathrm{t}}.\begin{cases}{\boldsymbol{{u}}_{\boldsymbol{{tt}}} ^{\left(\mathrm{1}\right)} =\left(\frac{\partial^{\mathrm{2}} }{\partial\boldsymbol{{x}}_{\mathrm{1}} ^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} }{\partial\boldsymbol{{x}}_{{i}} ^{\mathrm{2}} }\right)\boldsymbol{{u}}^{\left(\mathrm{1}\right)} }\\{\boldsymbol{{u}}^{\left(\mathrm{1}\right)} \left(\boldsymbol{{x}}_{\mathrm{1}} ,\boldsymbol{{x}}_{\mathrm{2}} ,\mathrm{0}\right)=\boldsymbol{\psi}\left(\boldsymbol{{x}}_{\mathrm{1}} ,\boldsymbol{{x}}_{\mathrm{2}} \right)}\\{\boldsymbol{{u}}^{\left(\mathrm{1}\right)} \left(\boldsymbol{{x}}_{\mathrm{1}} ,\boldsymbol{{x}}_{\mathrm{2}} ,\mathrm{0}\right)=\mathrm{0}}\end{cases},\begin{cases}{\boldsymbol{{u}}_{\boldsymbol{{tt}}} ^{\left(\mathrm{2}\right)} =\left(\frac{\partial^{\mathrm{2}} }{\partial\boldsymbol{{x}}_{\mathrm{1}} ^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} }{\partial\boldsymbol{{x}}_{\mathrm{2}} ^{\mathrm{2}} }+\boldsymbol{{c}}^{\mathrm{2}} \right)\boldsymbol{{u}}^{\left(\mathrm{2}\right)} }\\{\boldsymbol{{u}}^{\left(\mathrm{2}\right)} \left(\boldsymbol{{x}}_{\mathrm{1}} \boldsymbol{{x}}_{\mathrm{2}} ,\mathrm{0}\right)=\mathrm{0}}\\{\boldsymbol{{u}}_{\boldsymbol{{t}}} ^{\left(\mathrm{2}\right)} \left(\boldsymbol{{x}}_{\mathrm{1}} ,\boldsymbol{{x}}_{\mathrm{2}} ,\mathrm{0}\right)=\boldsymbol{\psi}\left(\boldsymbol{{x}}_{\mathrm{1}} ,\boldsymbol{{x}}_{\mathrm{2}} \right)}\end{cases} \\ $$$$\mathrm{prove}:\boldsymbol{{u}}^{\left(\mathrm{2}\right)} \left(\boldsymbol{{x}}_{\mathrm{1}} ,\boldsymbol{{x}}_{\mathrm{2}} ,\boldsymbol{{t}}\right)=\frac{\mathrm{1}}{\mathrm{2}\boldsymbol{\pi}}\int\int_{\boldsymbol{\xi}_{\mathrm{1}} ^{\mathrm{2}} +\boldsymbol{\xi}_{\mathrm{2}} ^{\mathrm{2}} \leq\boldsymbol{{t}}^{\mathrm{2}} } \frac{\boldsymbol{{e}}^{\boldsymbol{\xi}_{\mathrm{2}} \boldsymbol{{c}}} \boldsymbol{{u}}^{\left(\mathrm{1}\right)} \left(\boldsymbol{{x}}_{\mathrm{1}} ,\boldsymbol{{x}}_{\mathrm{2}} ,\boldsymbol{\xi}_{\mathrm{1}} \right)\boldsymbol{{d}\xi}_{\mathrm{1}} \boldsymbol{{d}\xi}_{\mathrm{2}} }{\:\sqrt{\boldsymbol{{t}}^{\mathrm{2}} −\boldsymbol{\xi}_{\mathrm{1}} ^{\mathrm{2}} −\boldsymbol{\xi}_{\mathrm{2}} ^{\mathrm{2}} }} \\ $$

  Pg 72      Pg 73      Pg 74      Pg 75      Pg 76      Pg 77      Pg 78      Pg 79      Pg 80      Pg 81   

Terms of Service

Privacy Policy

Contact: info@tinkutara.com