Question and Answers Forum

AllQuestion and Answers: Page 1202

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.

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

Question Number 92539 Answers: 0 Comments: 0

1) find ∫ (dx/((x+2)^3 (x^2 −1)^2 )) 2) calculate ∫_2 ^(+∞) (dx/((x+2)^3 (x^2 −1)^2 ))

$$\left.\mathrm{1}\right)\:{find}\:\int\:\:\:\:\:\frac{{dx}}{\left({x}+\mathrm{2}\right)^{\mathrm{3}} \left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{2}} ^{+\infty} \:\frac{{dx}}{\left({x}+\mathrm{2}\right)^{\mathrm{3}} \left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} } \\ $$

Question Number 92529 Answers: 0 Comments: 0

two smooth spheres of masses 2m and 3m have velocites (−12i + 8j) u ms^(−1) and (5i + 12j)u , respectively where u is a constant. The spheres collide with thier line of centre of parallel to j. Given that the coefficient of restitution between the spheres is (1/4), find the loss in kinetic energy due to impact.

$$\mathrm{two}\:\mathrm{smooth}\:\mathrm{spheres}\:\mathrm{of}\:\mathrm{masses}\:\mathrm{2}{m}\:\mathrm{and}\:\mathrm{3}{m}\:\mathrm{have}\:\mathrm{velocites} \\ $$$$\left(−\mathrm{12}\boldsymbol{\mathrm{i}}\:+\:\mathrm{8}\boldsymbol{\mathrm{j}}\right)\:{u}\:\mathrm{ms}^{−\mathrm{1}} \:\mathrm{and}\:\left(\mathrm{5}\boldsymbol{\mathrm{i}}\:+\:\mathrm{12}\boldsymbol{\mathrm{j}}\right){u}\:,\:\mathrm{respectively}\:\mathrm{where}\:{u}\: \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{constant}.\:\mathrm{The}\:\mathrm{spheres}\:\mathrm{collide}\:\mathrm{with}\:\mathrm{thier}\:\mathrm{line}\:\mathrm{of}\:\mathrm{centre}\:\mathrm{of} \\ $$$$\mathrm{parallel}\:\mathrm{to}\:\boldsymbol{\mathrm{j}}.\:\mathrm{Given}\:\mathrm{that}\:\mathrm{the}\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{restitution}\:\mathrm{between} \\ $$$$\mathrm{the}\:\mathrm{spheres}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{4}},\:\mathrm{find}\:\mathrm{the}\:\mathrm{loss}\:\mathrm{in}\:\mathrm{kinetic}\:\mathrm{energy}\:\mathrm{due}\:\mathrm{to}\:\mathrm{impact}. \\ $$

Question Number 92503 Answers: 0 Comments: 3

lim_(x→∞) (e^(10x) −8x)^(1/(2x)) =

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{e}^{\mathrm{10}{x}} −\mathrm{8}{x}\right)^{\frac{\mathrm{1}}{\mathrm{2}{x}}} \:=\: \\ $$

Question Number 92489 Answers: 0 Comments: 0

(3+((cq)/(12b)))s^2 +((6c)/b)s+(((8c^2 )/(3b^2 ))−b)=0 (1+((cq)/(4b)))s^2 +((3c)/b)(1−((cq)/(12b)))s+(((12c^2 )/b^2 )−b)=0 solve simultaneously for q and s in terms of b and c.

$$\left(\mathrm{3}+\frac{{cq}}{\mathrm{12}{b}}\right){s}^{\mathrm{2}} +\frac{\mathrm{6}{c}}{{b}}{s}+\left(\frac{\mathrm{8}{c}^{\mathrm{2}} }{\mathrm{3}{b}^{\mathrm{2}} }−{b}\right)=\mathrm{0} \\ $$$$\left(\mathrm{1}+\frac{{cq}}{\mathrm{4}{b}}\right){s}^{\mathrm{2}} +\frac{\mathrm{3}{c}}{{b}}\left(\mathrm{1}−\frac{{cq}}{\mathrm{12}{b}}\right){s}+\left(\frac{\mathrm{12}{c}^{\mathrm{2}} }{{b}^{\mathrm{2}} }−{b}\right)=\mathrm{0} \\ $$$${solve}\:{simultaneously}\:{for}\:\boldsymbol{{q}}\:{and}\:\boldsymbol{{s}} \\ $$$${in}\:{terms}\:{of}\:{b}\:{and}\:{c}. \\ $$

Question Number 92488 Answers: 0 Comments: 11

(3x)^(log_b 3) = (5x)^(log_b 5) x = ?

$$\:\left(\mathrm{3x}\right)^{\mathrm{log}_{\mathrm{b}} \:\mathrm{3}} \:=\:\left(\mathrm{5x}\right)^{\mathrm{log}_{\mathrm{b}} \:\mathrm{5}} \\ $$$$\: \\ $$$$\:\mathrm{x}\:=\:? \\ $$

Question Number 92486 Answers: 0 Comments: 0

The smallest interval [a, b] such that ∫_( 0) ^1 (1/(√(1+x^4 ))) dx ∈ [a, b] is given by

$$\mathrm{The}\:\mathrm{smallest}\:\mathrm{interval}\:\left[{a},\:{b}\right]\:\mathrm{such}\:\mathrm{that} \\ $$$$\:\underset{\:\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\mathrm{1}}{\sqrt{\mathrm{1}+{x}^{\mathrm{4}} }}\:{dx}\:\in\:\left[{a},\:{b}\right]\:\:\mathrm{is}\:\mathrm{given}\:\mathrm{by} \\ $$

Pg 1197 Pg 1198 Pg 1199 Pg 1200 Pg 1201 Pg 1202 Pg 1203 Pg 1204 Pg 1205 Pg 1206

Contact: info@tinkutara.com