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Question Number 215718 Answers: 1 Comments: 0

lim_(x→∞) (ln(9/2))^((1/2)x) =?

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left({ln}\frac{\mathrm{9}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}{x}} =? \\ $$$$ \\ $$

Question Number 215715 Answers: 1 Comments: 1

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

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

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