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Question Number 223228 Answers: 1 Comments: 2

Question Number 223224 Answers: 0 Comments: 1

Σ_(k=1) ^∞ (1/k)=∞ Σ_(p prime) (1/p)=∞ (1+(1/2)+(1/3)+(1/4)....)−((1/2)+(1/3)+(1/5)+...)=??

$$\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{k}}=\infty \\ $$$$\underset{\mathrm{p}\:\mathrm{prime}} {\sum}\:\frac{\mathrm{1}}{{p}}=\infty \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}....\right)−\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}+...\right)=?? \\ $$

Question Number 223230 Answers: 2 Comments: 0

Maksimum (((x^2 − 4x + 1)/(x^2 + 1))) = ?

$$\mathrm{Maksimum}\:\left(\frac{\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{4x}\:+\:\mathrm{1}}{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{1}}\right)\:=\:? \\ $$

Question Number 223216 Answers: 2 Comments: 0

Question Number 223213 Answers: 2 Comments: 1

Question Number 223210 Answers: 1 Comments: 0

I have a theory. this may not be true and I cannot prove it . I think If you want to draw a closed shape in x^(th) dimention the minimum number of vertex you will need is x+1 like if you want to draw a closed shape in 2^(nd) dimention you will need atleast 3 vertex(2+1){triangle} you can not draw a closed shape in two dimention using 2 vertex Similarly if you want to draw a closed shape in 3rd fimention you will need atleast 4 vertex(3+1){tetrahedron} you can not make a closed shape in 3dimention using 3 or 2 vertex Does that mean you will need atleast 5 vertex to draw a 4d object?? I dont know. It is just an assumption based on some similarities I noticed I cannot give you any proof of my theory

$${I}\:{have}\:{a}\:{theory}.\:{this}\:{may}\:{not}\:{be}\:{true}\:{and} \\ $$$${I}\:{cannot}\:{prove}\:{it}\:.\:{I}\:{think} \\ $$$${If}\:{you}\:{want}\:{to}\:{draw} \\ $$$$\:{a}\:{closed}\:{shape}\:{in}\:{x}^{{th}} \:{dimention} \\ $$$${the}\:{minimum}\:{number}\:{of} \\ $$$$\:{vertex}\:{you}\:{will}\:{need}\:{is}\:{x}+\mathrm{1} \\ $$$$ \\ $$$$\:{like}\:{if}\:{you}\:{want}\:{to}\:{draw}\:{a}\:{closed}\:{shape} \\ $$$${in}\:\mathrm{2}^{{nd}} \:{dimention}\:{you}\:{will}\:{need}\:{atleast} \\ $$$$\mathrm{3}\:{vertex}\left(\mathrm{2}+\mathrm{1}\right)\left\{{triangle}\right\} \\ $$$${you}\:{can}\:{not}\:{draw}\:{a}\:{closed}\:{shape}\:{in}\:{two} \\ $$$${dimention}\:{using}\:\mathrm{2}\:{vertex} \\ $$$${Similarly}\:{if}\:{you}\:\:{want}\:{to}\:{draw} \\ $$$${a}\:{closed}\:{shape}\:{in}\:\mathrm{3}{rd}\:{fimention} \\ $$$${you}\:{will}\:{need}\:{atleast}\:\mathrm{4}\:{vertex}\left(\mathrm{3}+\mathrm{1}\right)\left\{{tetrahedron}\right\} \\ $$$${you}\:{can}\:{not}\:{make}\:{a}\:{closed}\:{shape}\: \\ $$$${in}\:\mathrm{3}{dimention}\:{using}\:\mathrm{3}\:{or}\:\mathrm{2}\:{vertex} \\ $$$${Does}\:{that}\:{mean}\:{you}\:{will}\:{need}\: \\ $$$${atleast}\:\mathrm{5}\:{vertex}\:{to}\:{draw}\:{a}\:\mathrm{4}{d}\:{object}?? \\ $$$${I}\:{dont}\:{know}. \\ $$$${It}\:{is}\:{just}\:{an}\:{assumption}\:{based}\:{on} \\ $$$${some}\:{similarities}\:{I}\:{noticed} \\ $$$${I}\:{cannot}\:{give}\:{you}\:{any}\:{proof}\:{of}\:{my}\:{theory}\: \\ $$

Question Number 223204 Answers: 1 Comments: 0

∫_(π/6) ^(π/3) (dx/(1+(√(tanx)))) =...?

$$\:\:\:\underset{\frac{\pi}{\mathrm{6}}} {\overset{\frac{\pi}{\mathrm{3}}} {\int}}\frac{\boldsymbol{{dx}}}{\mathrm{1}+\sqrt{\boldsymbol{{tanx}}}}\:=...? \\ $$

Question Number 223203 Answers: 2 Comments: 0

Find x and y ix+y=ix^3 −y^3 and xy(y−ix)=c

$${Find}\:{x}\:{and}\:{y} \\ $$$$\:\:\:\:{ix}+{y}={ix}^{\mathrm{3}} −{y}^{\mathrm{3}} \\ $$$$\:\:\:{and}\: \\ $$$$\:\:\:\:{xy}\left({y}−{ix}\right)={c} \\ $$

Question Number 223193 Answers: 0 Comments: 0

Prove that: ∫_( 0) ^( (π/2)) tan^(− 1) (r sin θ) dθ = 2𝛘_2 ((((√(1 + r^2 )) − 1)/r))

$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\:\:\:\:\:\:\:\int_{\:\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \:\mathrm{tan}^{−\:\mathrm{1}} \left(\mathrm{r}\:\mathrm{sin}\:\theta\right)\:\mathrm{d}\theta\:\:\:=\:\:\:\mathrm{2}\boldsymbol{\chi}_{\mathrm{2}} \left(\frac{\sqrt{\mathrm{1}\:\:+\:\:\mathrm{r}^{\mathrm{2}} }\:\:−\:\:\mathrm{1}}{\mathrm{r}}\right) \\ $$

Question Number 223192 Answers: 0 Comments: 0

Evaluate ; ∫_0 ^1 Π_(n=1) ^∞ (1−q^(24n) ) dq