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Question Number 223348 Answers: 1 Comments: 1

Determine gcd(13a+19b,ab) given that gcd(a,19)=gcd(b,13)=1

$${Determine}\:{gcd}\left(\mathrm{13}{a}+\mathrm{19}{b},{ab}\right)\:{given}\:{that}\:{gcd}\left({a},\mathrm{19}\right)={gcd}\left({b},\mathrm{13}\right)=\mathrm{1} \\ $$

Question Number 223346 Answers: 1 Comments: 0

proof gcd(2^m −1,2^n −1)=2^(gcd(m,n)) −1

$${proof}\:{gcd}\left(\mathrm{2}^{{m}} −\mathrm{1},\mathrm{2}^{{n}} −\mathrm{1}\right)=\mathrm{2}^{{gcd}\left({m},{n}\right)} −\mathrm{1} \\ $$

Question Number 223340 Answers: 3 Comments: 1

Question Number 223317 Answers: 3 Comments: 0

Question Number 223315 Answers: 0 Comments: 4

Question Number 223304 Answers: 2 Comments: 0

lim_(x→0) ((∫_0 ^x^2 sin((√t))dt )/x^3 ) =...?

$$\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\underset{\mathrm{0}} {\overset{{x}^{\mathrm{2}} } {\int}}\boldsymbol{{sin}}\left(\sqrt{\boldsymbol{{t}}}\right)\boldsymbol{{dt}}\:}{\boldsymbol{{x}}^{\mathrm{3}} }\:=...? \\ $$

Question Number 223301 Answers: 2 Comments: 0

Question Number 223286 Answers: 2 Comments: 3

OC=4.5 & EF=2 & DE^⌢ =EC^⌢ & AC=2AF⇒AB=? ⇓⇓⇓

$${OC}=\mathrm{4}.\mathrm{5}\:\:\:\&\:\:{EF}=\mathrm{2}\:\:\&\:\:{D}\overset{\frown} {{E}}={E}\overset{\frown} {{C}}\:\:\&\:\:{AC}=\mathrm{2}{AF}\Rightarrow{AB}=? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Downarrow\Downarrow\Downarrow \\ $$

Question Number 223270 Answers: 2 Comments: 0

At a given instant, two cars are at distances 600m and 800m from the point of intersection of the straight roads crossing at right angles and approaching O at uniform speeds of 20 m/s and 30 m/s respectively. Find the shortest distance between the cars and the time taken to reach this position.

Question Number 223267 Answers: 3 Comments: 0

r,s,t ; are the roots of: x^3 +5x+1=0 find: (r^3 −1)(s^3 −1)(t^3 −1)

$$\boldsymbol{{r}},\boldsymbol{{s}},\boldsymbol{{t}}\:;\:\boldsymbol{{are}}\:\boldsymbol{{the}}\:\boldsymbol{{roots}}\:\boldsymbol{{of}}: \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{{x}}^{\mathrm{3}} +\mathrm{5}\boldsymbol{{x}}+\mathrm{1}=\mathrm{0} \\ $$$$\boldsymbol{{find}}: \\ $$$$\:\:\:\:\:\:\left(\boldsymbol{{r}}^{\mathrm{3}} −\mathrm{1}\right)\left(\boldsymbol{{s}}^{\mathrm{3}} −\mathrm{1}\right)\left(\boldsymbol{{t}}^{\mathrm{3}} −\mathrm{1}\right) \\ $$

Question Number 223261 Answers: 0 Comments: 1

Question Number 223255 Answers: 2 Comments: 0

Which of the following complex numbers is equivalent to ((3−5i)/(8+2i))?(Note: i=(√(−1))) A)(3/8)−((5i)/2) B)(3/8)+((5i)/2) C)(7/(34))−((23i)/(34)) D)(7/(34))+((23i)/(34))

$${Which}\:{of}\:{the}\:{following}\:{complex} \\ $$$${numbers}\:{is}\:{equivalent}\:{to}\:\frac{\mathrm{3}−\mathrm{5}{i}}{\mathrm{8}+\mathrm{2}{i}}?\left({Note}:\:{i}=\sqrt{\left.−\mathrm{1}\right)}\right. \\ $$$$\left.{A}\right)\frac{\mathrm{3}}{\mathrm{8}}−\frac{\mathrm{5}{i}}{\mathrm{2}} \\ $$$$ \\ $$$$\left.{B}\right)\frac{\mathrm{3}}{\mathrm{8}}+\frac{\mathrm{5}{i}}{\mathrm{2}} \\ $$$$\: \\ $$$$\left.{C}\right)\frac{\mathrm{7}}{\mathrm{34}}−\frac{\mathrm{23}{i}}{\mathrm{34}} \\ $$$$ \\ $$$$\left.{D}\right)\frac{\mathrm{7}}{\mathrm{34}}+\frac{\mathrm{23}{i}}{\mathrm{34}} \\ $$

Question Number 223254 Answers: 0 Comments: 0

Σ_(n = 1) ^∞ (− 1)^(n + 1) (H_n /n^2 )

$$\underset{\mathrm{n}\:\:=\:\:\mathrm{1}} {\overset{\infty} {\sum}}\left(−\:\:\mathrm{1}\right)^{\mathrm{n}\:\:+\:\:\mathrm{1}} \:\frac{\mathrm{H}_{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} } \\ $$

Question Number 223242 Answers: 4 Comments: 0

Question Number 223241 Answers: 2 Comments: 0

∫_( 0) ^( 1) ((ln(1 − x) ln(1 + x))/x) dx

$$\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{1}\:−\:\mathrm{x}\right)\:\mathrm{ln}\left(\mathrm{1}\:+\:\mathrm{x}\right)}{\mathrm{x}}\:\mathrm{dx} \\ $$

Question Number 223240 Answers: 1 Comments: 0

x^x =i number of solutions??

$${x}^{{x}} ={i} \\ $$$${number}\:{of}\:{solutions}?? \\ $$

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: 3 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

$$ \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{Evaluate}}\:;\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}−{q}^{\mathrm{24}{n}} \right)\:\mathrm{d}{q} \\ $$$$ \\ $$

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