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Question Number 92700    Answers: 0   Comments: 0

∫_(−(π/2)) ^(π/2) ((tan (x^2 )dx)/x) ?

$$\underset{−\frac{\pi}{\mathrm{2}}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{\mathrm{tan}\:\left({x}^{\mathrm{2}} \right){dx}}{{x}}\:?\: \\ $$

Question Number 92691    Answers: 1   Comments: 3

find x in eq tan^(−1) (x)= cos^(−1) (x)

$$\mathrm{find}\:{x}\:{in}\:{eq}\:\mathrm{tan}^{−\mathrm{1}} \left({x}\right)=\:\mathrm{cos}^{−\mathrm{1}} \left({x}\right) \\ $$

Question Number 92687    Answers: 0   Comments: 0

In a Gregorian calendar, a year finishing with 00 is a leap year if only it′s vintage is divisible by 400. Also, the 1^(st) January 1900 was a Monday. 1\ Show that a year with the vintage finishing with 00 cannot begin on a Sunday 2\ Show that for a person born between 1900−2071, his 28 anniversary will occur on the same day of birth.

$$\mathrm{In}\:\mathrm{a}\:\mathrm{Gregorian}\:\mathrm{calendar},\:\mathrm{a}\:\mathrm{year}\:\mathrm{finishing}\:\mathrm{with} \\ $$$$\mathrm{00}\:\mathrm{is}\:\mathrm{a}\:\mathrm{leap}\:\mathrm{year}\:\mathrm{if}\:\mathrm{only}\:\mathrm{it}'\mathrm{s}\:\mathrm{vintage}\:\mathrm{is}\:\mathrm{divisible} \\ $$$$\mathrm{by}\:\mathrm{400}.\:\mathrm{Also},\:\mathrm{the}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{January}\:\mathrm{1900}\:\mathrm{was}\:\mathrm{a} \\ $$$$\mathrm{Monday}. \\ $$$$\mathrm{1}\backslash\:\mathrm{Show}\:\mathrm{that}\:\mathrm{a}\:\mathrm{year}\:\mathrm{with}\:\mathrm{the}\:\mathrm{vintage}\:\mathrm{finishing}\:\mathrm{with} \\ $$$$\mathrm{00}\:\mathrm{cannot}\:\mathrm{begin}\:\mathrm{on}\:\mathrm{a}\:\mathrm{Sunday} \\ $$$$\mathrm{2}\backslash\:\mathrm{Show}\:\mathrm{that}\:\mathrm{for}\:\mathrm{a}\:\mathrm{person}\:\mathrm{born}\:\mathrm{between}\:\mathrm{1900}−\mathrm{2071}, \\ $$$$\mathrm{his}\:\mathrm{28}\:\mathrm{anniversary}\:\mathrm{will}\:\mathrm{occur}\:\mathrm{on}\:\mathrm{the}\:\mathrm{same} \\ $$$$\mathrm{day}\:\mathrm{of}\:\mathrm{birth}. \\ $$

Question Number 92684    Answers: 1   Comments: 4

Question Number 92682    Answers: 0   Comments: 5

While uninstalling/reinstalling app please use backup/import to keep your saved equation on device. backup: before unintalling import: after reinstalling

$$\mathrm{While}\:\mathrm{uninstalling}/\mathrm{reinstalling}\:\mathrm{app} \\ $$$$\mathrm{please}\:\mathrm{use}\:\mathrm{backup}/\mathrm{import}\:\mathrm{to}\:\mathrm{keep} \\ $$$$\mathrm{your}\:\mathrm{saved}\:\mathrm{equation}\:\mathrm{on}\:\mathrm{device}. \\ $$$$\mathrm{backup}:\:\mathrm{before}\:\mathrm{unintalling} \\ $$$$\mathrm{import}:\:\mathrm{after}\:\mathrm{reinstalling} \\ $$

Question Number 92712    Answers: 0   Comments: 2

find m for fix function f(x)=(((m−1)x+3)/(x−1))

$${find}\:\:\boldsymbol{{m}}\:{for}\:{fix}\:{function} \\ $$$${f}\left({x}\right)=\frac{\left(\boldsymbol{{m}}−\mathrm{1}\right){x}+\mathrm{3}}{{x}−\mathrm{1}} \\ $$

Question Number 92673    Answers: 1   Comments: 2

Show that if 3 prime numbers, all greater than 3, form an arithmetic progression then the common difference of the progression is divisible by 6.

$$\mathrm{Show}\:\mathrm{that}\:\mathrm{if}\:\mathrm{3}\:\mathrm{prime}\:\mathrm{numbers},\:\mathrm{all}\:\mathrm{greater} \\ $$$$\mathrm{than}\:\mathrm{3},\:\mathrm{form}\:\mathrm{an}\:\mathrm{arithmetic}\:\mathrm{progression}\:\mathrm{then}\:\mathrm{the}\:\mathrm{common} \\ $$$$\mathrm{difference}\:\mathrm{of}\:\mathrm{the}\:\mathrm{progression}\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{6}. \\ $$

Question Number 92668    Answers: 1   Comments: 2

∫_0 ^p (√(x/(c−x))) dx ?

$$\underset{\mathrm{0}} {\overset{\mathrm{p}} {\int}}\:\sqrt{\frac{\mathrm{x}}{\mathrm{c}−\mathrm{x}}}\:\mathrm{dx}\:?\: \\ $$

Question Number 92646    Answers: 0   Comments: 22

Question Number 92626    Answers: 0   Comments: 1

Question Number 92624    Answers: 0   Comments: 0

Question Number 92621    Answers: 0   Comments: 1

Question Number 92619    Answers: 0   Comments: 0

find x in closset interval [ −4,3 ]of function f(x)= 3x−(9/2)sin^(−1) (((x−3)/3))− (1/2)(((x−3)/3))(√(9−(x−3)^2 )) maximum

$$\mathrm{find}\:\mathrm{x}\:\mathrm{in}\:\mathrm{closset}\:\mathrm{interval}\: \\ $$$$\left[\:−\mathrm{4},\mathrm{3}\:\right]\mathrm{of}\:\mathrm{function}\: \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\:\mathrm{3x}−\frac{\mathrm{9}}{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{x}−\mathrm{3}}{\mathrm{3}}\right)− \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{x}−\mathrm{3}}{\mathrm{3}}\right)\sqrt{\mathrm{9}−\left(\mathrm{x}−\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$\mathrm{maximum} \\ $$

Question Number 92608    Answers: 0   Comments: 4

Question Number 92605    Answers: 1   Comments: 4

If ((1+x)/(1+(√(1+x)))) +((1−x)/(1−(√(1−x)))) =1 find x

$${If}\:\frac{\mathrm{1}+{x}}{\mathrm{1}+\sqrt{\mathrm{1}+{x}}}\:+\frac{\mathrm{1}−{x}}{\mathrm{1}−\sqrt{\mathrm{1}−{x}}}\:=\mathrm{1} \\ $$$${find}\:{x} \\ $$

Question Number 92635    Answers: 0   Comments: 9

Question Number 92597    Answers: 0   Comments: 1

If f(x)=(1/(x−1)) and g(x)=(√x) find domain and range of g(f(x)) .

$$\mathrm{If}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{x}−\mathrm{1}}\:\mathrm{and}\:\mathrm{g}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}}\: \\ $$$$\mathrm{find}\:\mathrm{domain}\:\mathrm{and}\:\mathrm{range}\:\mathrm{of}\: \\ $$$$\mathrm{g}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\:. \\ $$

Question Number 92594    Answers: 0   Comments: 5

Question Number 92593    Answers: 0   Comments: 0

using the squeeze theorem show that lim_(x→a) (√x) = (√a)

$$\mathrm{using}\:\mathrm{the}\:\mathrm{squeeze}\:\mathrm{theorem}\: \\ $$$$\mathrm{show}\:\mathrm{that} \\ $$$$\:\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:\sqrt{{x}}\:=\:\sqrt{{a}}\: \\ $$

Question Number 92585    Answers: 0   Comments: 1

sin 2z + 5(sin z+cos z)+1=0

$$\mathrm{sin}\:\mathrm{2z}\:+\:\mathrm{5}\left(\mathrm{sin}\:\mathrm{z}+\mathrm{cos}\:\mathrm{z}\right)+\mathrm{1}=\mathrm{0} \\ $$

Question Number 92577    Answers: 1   Comments: 3

lim_(x→∞) ((ln x)/x)

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\mathrm{x}}{\mathrm{x}} \\ $$

Question Number 92575    Answers: 0   Comments: 6

hello i see a lot of questions about special functions in the forum so this book is a good book to learn special functions for anyone want to learn

$${hello}\: \\ $$$${i}\:{see}\:{a}\:{lot}\:{of}\:{questions}\:{about}\:{special} \\ $$$${functions}\:{in}\:{the}\:{forum}\:{so}\:{this}\:{book} \\ $$$${is}\:{a}\:{good}\:{book}\:{to}\:{learn}\:{special}\:{functions} \\ $$$${for}\:{anyone}\:{want}\:{to}\:{learn} \\ $$

Question Number 92566    Answers: 0   Comments: 2

Posting Question with Images Preferably you should type the question. However if you are using pictures then please do the following steps which posting a photo of printed question. A. Use camscanner app to take pictures (search for camscanner in playstore). B. Crop picture so that you only have specifc question that you want to ask in the image.

$$\mathrm{Posting}\:\mathrm{Question}\:\mathrm{with}\:\mathrm{Images} \\ $$$$\mathrm{Preferably}\:\mathrm{you}\:\mathrm{should}\:\mathrm{type}\:\mathrm{the}\:\mathrm{question}. \\ $$$$\mathrm{However}\:\mathrm{if}\:\mathrm{you}\:\mathrm{are}\:\mathrm{using}\:\mathrm{pictures}\:\mathrm{then} \\ $$$$\mathrm{please}\:\mathrm{do}\:\mathrm{the}\:\mathrm{following}\:\mathrm{steps} \\ $$$$\mathrm{which}\:\mathrm{posting}\:\mathrm{a}\:\mathrm{photo}\:\mathrm{of}\:\mathrm{printed} \\ $$$$\mathrm{question}. \\ $$$$\mathrm{A}.\:\mathrm{Use}\:\mathrm{camscanner}\:\mathrm{app}\:\mathrm{to}\:\mathrm{take}\: \\ $$$$\mathrm{pictures}\:\left(\mathrm{search}\:\mathrm{for}\:\mathrm{camscanner}\:\mathrm{in}\right. \\ $$$$\left.\mathrm{playstore}\right).\: \\ $$$$\mathrm{B}.\:\mathrm{Crop}\:\mathrm{picture}\:\mathrm{so}\:\mathrm{that}\:\mathrm{you}\:\mathrm{only} \\ $$$$\mathrm{have}\:\mathrm{specifc}\:\mathrm{question}\:\mathrm{that}\:\mathrm{you}\:\mathrm{want} \\ $$$$\mathrm{to}\:\mathrm{ask}\:\mathrm{in}\:\mathrm{the}\:\mathrm{image}. \\ $$$$ \\ $$

Question Number 92557    Answers: 0   Comments: 3

If a_1 = 5, a_2 = 13 and a_(n + 2) = 5a_(n + 1) − 6a_n . Find a_n

$$\mathrm{If}\:\:\:\mathrm{a}_{\mathrm{1}} \:\:=\:\:\mathrm{5},\:\:\:\:\:\mathrm{a}_{\mathrm{2}} \:\:=\:\:\mathrm{13}\:\:\:\:\:\mathrm{and}\:\:\:\:\mathrm{a}_{\mathrm{n}\:\:+\:\:\mathrm{2}} \:\:\:=\:\:\mathrm{5a}_{\mathrm{n}\:\:+\:\:\mathrm{1}} \:−\:\:\mathrm{6a}_{\mathrm{n}} . \\ $$$$\mathrm{Find}\:\:\:\:\:\mathrm{a}_{\mathrm{n}} \\ $$

Question Number 92555    Answers: 0   Comments: 5

Σ_(n=1) ^∞ 1/n^2 =π^2 /6

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{1}/{n}^{\mathrm{2}} =\pi^{\mathrm{2}} /\mathrm{6} \\ $$

Question Number 92544    Answers: 0   Comments: 12

A new version of application with a few usability enhancement will be available on playstore in next few days. Key Changes a. Continous scrolling on forum. b. Clicking on images pop up image which can be easily scrolled. c. Ability to add a profile picture. d. Question menu can be clicked directly. No need to select/move to top to access menu. e. Hyperlinks in plaintext post can be click directly. Double tap not needed. f. Images can be cropped while uploading (in addition to rotate option currently present). g. Clicking on notification message directly tales to selected question. h. Editor now has a menu bar at top (in addition to keyboar shorrcuts) Please try out the new version and provide feedback. Thanks.

$$\mathrm{A}\:\mathrm{new}\:\mathrm{version}\:\mathrm{of}\:\mathrm{application}\:\mathrm{with} \\ $$$$\mathrm{a}\:\mathrm{few}\:\mathrm{usability}\:\mathrm{enhancement} \\ $$$$\mathrm{will}\:\:\mathrm{be}\:\mathrm{available}\:\mathrm{on}\:\mathrm{playstore} \\ $$$$\mathrm{in}\:\mathrm{next}\:\mathrm{few}\:\mathrm{days}. \\ $$$$\mathrm{Key}\:\mathrm{Changes} \\ $$$$\mathrm{a}.\:\mathrm{Continous}\:\mathrm{scrolling}\:\mathrm{on}\:\mathrm{forum}. \\ $$$$\mathrm{b}.\:\mathrm{Clicking}\:\mathrm{on}\:\mathrm{images}\:\mathrm{pop}\:\mathrm{up}\:\mathrm{image} \\ $$$$\mathrm{which}\:\mathrm{can}\:\mathrm{be}\:\mathrm{easily}\:\mathrm{scrolled}. \\ $$$$\mathrm{c}.\:\mathrm{Ability}\:\mathrm{to}\:\mathrm{add}\:\mathrm{a}\:\mathrm{profile}\:\mathrm{picture}. \\ $$$$\mathrm{d}.\:\mathrm{Question}\:\mathrm{menu}\:\mathrm{can}\:\mathrm{be}\:\mathrm{clicked} \\ $$$$\mathrm{directly}.\:\mathrm{No}\:\mathrm{need}\:\mathrm{to}\:\mathrm{select}/\mathrm{move}\:\mathrm{to}\:\mathrm{top} \\ $$$$\mathrm{to}\:\mathrm{access}\:\mathrm{menu}. \\ $$$$\mathrm{e}.\:\mathrm{Hyperlinks}\:\mathrm{in}\:\mathrm{plaintext}\:\mathrm{post}\:\mathrm{can} \\ $$$$\mathrm{be}\:\mathrm{click}\:\mathrm{directly}.\:\mathrm{Double}\:\mathrm{tap} \\ $$$$\mathrm{not}\:\mathrm{needed}. \\ $$$$\mathrm{f}.\:\mathrm{Images}\:\mathrm{can}\:\mathrm{be}\:\mathrm{cropped}\:\mathrm{while}\:\mathrm{uploading} \\ $$$$\left(\mathrm{in}\:\mathrm{addition}\:\mathrm{to}\:\mathrm{rotate}\:\mathrm{option}\:\mathrm{currently}\right. \\ $$$$\left.\mathrm{present}\right). \\ $$$$\mathrm{g}.\:\mathrm{Clicking}\:\mathrm{on}\:\mathrm{notification}\:\mathrm{message} \\ $$$$\mathrm{directly}\:\mathrm{tales}\:\mathrm{to}\:\mathrm{selected}\:\mathrm{question}. \\ $$$$\mathrm{h}.\:\mathrm{Editor}\:\mathrm{now}\:\mathrm{has}\:\mathrm{a}\:\mathrm{menu}\:\mathrm{bar}\:\mathrm{at}\:\mathrm{top} \\ $$$$\left(\mathrm{in}\:\mathrm{addition}\:\mathrm{to}\:\mathrm{keyboar}\:\mathrm{shorrcuts}\right) \\ $$$$\mathrm{Please}\:\mathrm{try}\:\mathrm{out}\:\mathrm{the}\:\mathrm{new}\:\mathrm{version}\:\mathrm{and} \\ $$$$\mathrm{provide}\:\mathrm{feedback}. \\ $$$$\mathrm{Thanks}. \\ $$

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