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Question Number 217186    Answers: 3   Comments: 0

((x−4051)/(2024))+((x−4050)/(2025))+((x−4049)/(2026))=3

$$\frac{{x}−\mathrm{4051}}{\mathrm{2024}}+\frac{{x}−\mathrm{4050}}{\mathrm{2025}}+\frac{{x}−\mathrm{4049}}{\mathrm{2026}}=\mathrm{3} \\ $$

Question Number 217178    Answers: 2   Comments: 0

Find: 100-99+98-97+96-95+...+2-1 = ?

$$\mathrm{Find}: \\ $$$$\mathrm{100}-\mathrm{99}+\mathrm{98}-\mathrm{97}+\mathrm{96}-\mathrm{95}+...+\mathrm{2}-\mathrm{1}\:=\:? \\ $$

Question Number 217164    Answers: 1   Comments: 0

Question Number 217163    Answers: 2   Comments: 0

If a+b=b+c=4 find: a^2 −b^2 −8c = ?

$$\mathrm{If}\:\:\:\mathrm{a}+\mathrm{b}=\mathrm{b}+\mathrm{c}=\mathrm{4} \\ $$$$\mathrm{find}:\:\:\:\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} −\mathrm{8c}\:=\:? \\ $$

Question Number 217159    Answers: 3   Comments: 0

Solve for x: ((x+3)/(x−2))+((2x−5)/(x+4))=((4x+1)/(x^2 +2x−8))

$${Solve}\:{for}\:{x}: \\ $$$$\frac{{x}+\mathrm{3}}{{x}−\mathrm{2}}+\frac{\mathrm{2}{x}−\mathrm{5}}{{x}+\mathrm{4}}=\frac{\mathrm{4}{x}+\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{8}} \\ $$

Question Number 217158    Answers: 1   Comments: 0

circle= (x−3)^2 +(y−4)^2 =1 parabola= ax(x−10)=y what is the values of a where the parabola is tangent to the circle

$$\mathrm{circle}= \\ $$$$\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\left({y}−\mathrm{4}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{parabola}= \\ $$$${ax}\left({x}−\mathrm{10}\right)={y} \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:{a}\:\mathrm{where} \\ $$$$\mathrm{the}\:\mathrm{parabola}\:\mathrm{is}\:\mathrm{tangent}\:\mathrm{to}\:\mathrm{the}\:\mathrm{circle} \\ $$$$ \\ $$

Question Number 217149    Answers: 1   Comments: 0

Question Number 217148    Answers: 2   Comments: 0

I have seen a relationship in the curve path of a thrown object at β while the total passed distance D_v and highest point had passedD_u then β = arctan(((4D_u )/D_v )) but cant find the proof. I would like to say would anyone like to proove it?then please.

$${I}\:{have}\:{seen}\:{a}\:{relationship}\:{in}\:{the}\:{curve} \\ $$$${path}\:{of}\:{a}\:{thrown}\:{object}\:{at}\:\beta\: \\ $$$${while}\:{the}\:{total}\:{passed}\:{distance}\:{D}_{{v}} \:{and} \\ $$$${highest}\:{point}\:{had}\:{passedD}_{{u}} \\ $$$${then}\:\beta\:=\:{arctan}\left(\frac{\mathrm{4}{D}_{{u}} }{{D}_{{v}} }\right) \\ $$$${but}\:{cant}\:{find}\:{the}\:{proof}. \\ $$$${I}\:{would}\:{like}\:{to}\:{say}\:{would}\:{anyone}\:{like} \\ $$$${to}\:{proove}\:{it}?{then}\:{please}. \\ $$

Question Number 217146    Answers: 1   Comments: 1

determiner le cote du care ABCD inscrit dans l elipse {(−3,+3):(−8,+8)}

$$\mathrm{determiner}\:\mathrm{le}\:\mathrm{cote}\:\mathrm{du}\:\mathrm{care}\:\boldsymbol{\mathrm{ABCD}} \\ $$$$\mathrm{inscrit}\:\mathrm{dans}\:\mathrm{l}\:\mathrm{elipse}\:\left\{\left(−\mathrm{3},+\mathrm{3}\right):\left(−\mathrm{8},+\mathrm{8}\right)\right\} \\ $$

Question Number 217140    Answers: 1   Comments: 0

if a, b, c are three digits, abc and bca are two numbers. where abc +cba = 444, b =2. find the value of a+b+c.

if a, b, c are three digits, abc and bca are two numbers. where abc +cba = 444, b =2. find the value of a+b+c.

Question Number 217101    Answers: 0   Comments: 0

is this right when (a+bi)^(c+di) =∣a+bi∣^(c+di) e^(i(c+di)arg(a+bi)) ? I had let arg(a+bi)= { ((tan^(−1) ((b/a))),(a≥0 and b≥0)),((π−tan^(−1) (−(b/a))),(a<0 and b≥0)),((−(π−tan^(−1) ((b/a)))),(a<0 and b<0)),((−tan^(−1) ((b/a))),(a≥0 and b<0)) :} before I solved it (a+bi)^(c+di) =∣a+bi∣^(c+di) e^(i(c+di)arg(a+bi)) =∣a+bi∣^c ∣a+bi∣^di e^(ic∙arg(a+bi)) e^(−d∙arg(a+bi)) =∣a+bi∣^c (c^di )^(ln∣a+bi∣) e^(ic∙arg(a+bi)) e^(−d∙arg(a+bi))

$$\mathrm{is}\:\mathrm{this}\:\mathrm{right}\:\mathrm{when}\:\left({a}+{bi}\right)^{{c}+{di}} =\mid{a}+{bi}\mid^{{c}+{di}} {e}^{{i}\left({c}+{di}\right)\mathrm{arg}\left({a}+{bi}\right)} ? \\ $$$$\mathrm{I}\:\mathrm{had}\:\mathrm{let}\:\mathrm{arg}\left({a}+{bi}\right)=\begin{cases}{\mathrm{tan}^{−\mathrm{1}} \left(\frac{{b}}{{a}}\right)}&{{a}\geqslant\mathrm{0}\:\mathrm{and}\:{b}\geqslant\mathrm{0}}\\{\pi−\mathrm{tan}^{−\mathrm{1}} \left(−\frac{{b}}{{a}}\right)}&{{a}<\mathrm{0}\:\mathrm{and}\:{b}\geqslant\mathrm{0}}\\{−\left(\pi−\mathrm{tan}^{−\mathrm{1}} \left(\frac{{b}}{{a}}\right)\right)}&{{a}<\mathrm{0}\:\mathrm{and}\:{b}<\mathrm{0}}\\{−\mathrm{tan}^{−\mathrm{1}} \left(\frac{{b}}{{a}}\right)}&{{a}\geqslant\mathrm{0}\:\mathrm{and}\:{b}<\mathrm{0}}\end{cases}\:\mathrm{before}\:\mathrm{I}\:\mathrm{solved}\:\mathrm{it} \\ $$$$\left({a}+{bi}\right)^{{c}+{di}} =\mid{a}+{bi}\mid^{{c}+{di}} {e}^{{i}\left({c}+{di}\right)\mathrm{arg}\left({a}+{bi}\right)} \\ $$$$=\mid{a}+{bi}\mid^{{c}} \mid{a}+{bi}\mid^{{di}} {e}^{{ic}\centerdot\mathrm{arg}\left({a}+{bi}\right)} {e}^{−{d}\centerdot\mathrm{arg}\left({a}+{bi}\right)} \\ $$$$=\mid{a}+{bi}\mid^{{c}} \left({c}^{{di}} \right)^{\mathrm{ln}\mid{a}+{bi}\mid} {e}^{{ic}\centerdot\mathrm{arg}\left({a}+{bi}\right)} {e}^{−{d}\centerdot\mathrm{arg}\left({a}+{bi}\right)} \\ $$

Question Number 217122    Answers: 1   Comments: 0

∫ ((√(cos 2x))/(cos x)) dx =?

$$\:\:\:\:\:\int\:\frac{\sqrt{\mathrm{cos}\:\mathrm{2x}}}{\mathrm{cos}\:\mathrm{x}}\:\mathrm{dx}\:=? \\ $$

Question Number 217121    Answers: 0   Comments: 0

Question Number 217137    Answers: 0   Comments: 3

Please do not post totally meaningless questions/answers. If you dont know the answer leave it unanswered. If some has ideas they will post either partially or full answers. Do not post meaningless answers that cancel operator or function from numerator/denominator.

$$ \\ $$$$\mathrm{Please}\:\mathrm{do}\:\mathrm{not}\:\mathrm{post}\:\mathrm{totally}\:\mathrm{meaningless} \\ $$$$\mathrm{questions}/\mathrm{answers}. \\ $$$$\mathrm{If}\:\mathrm{you}\:\mathrm{dont}\:\mathrm{know}\:\mathrm{the}\:\mathrm{answer} \\ $$$$\mathrm{leave}\:\mathrm{it}\:\mathrm{unanswered}.\:\mathrm{If}\:\mathrm{some}\:\mathrm{has} \\ $$$$\mathrm{ideas}\:\mathrm{they}\:\mathrm{will}\:\mathrm{post}\:\mathrm{either}\:\mathrm{partially} \\ $$$$\mathrm{or}\:\mathrm{full}\:\mathrm{answers}. \\ $$$$\mathrm{Do}\:\mathrm{not}\:\mathrm{post}\:\mathrm{meaningless}\:\mathrm{answers}\:\mathrm{that} \\ $$$$\mathrm{cancel}\:\mathrm{operator}\:\mathrm{or}\:\mathrm{function}\:\mathrm{from} \\ $$$$\mathrm{numerator}/\mathrm{denominator}. \\ $$

Question Number 217132    Answers: 0   Comments: 0

Find all integers n> 1 such that n divides 2^(n−1) + 3^(n−1) .

$$ \\ $$$$\mathrm{Find}\:\mathrm{all}\:\mathrm{integers}\:\:\mathrm{n}>\:\mathrm{1}\:\mathrm{such}\:\mathrm{that} \\ $$$$\:\:\mathrm{n}\:\:\mathrm{divides}\:\:\mathrm{2}^{\mathrm{n}−\mathrm{1}} \:+\:\mathrm{3}^{\mathrm{n}−\mathrm{1}} . \\ $$

Question Number 217130    Answers: 0   Comments: 0

Prove that for every integer n≥2 the number n^4 + 4^n is composite.

$$ \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{for}\:\mathrm{every}\:\mathrm{integer}\:\:\mathrm{n}\geqslant\mathrm{2}\:\:\mathrm{the}\:\mathrm{number}\:\:\mathrm{n}^{\mathrm{4}} +\:\mathrm{4}^{{n}} \:\:\mathrm{is} \\ $$$$\mathrm{c}{o}\mathrm{mposite}. \\ $$

Question Number 217129    Answers: 1   Comments: 2

prove that if an integer n is not divisible by 2 or 3 then n^2 ≡1(mod 24)

$${prove}\:{that}\:{if}\:{an}\:{integer}\:{n}\:{is}\:{not}\:{divisible}\:{by}\:\mathrm{2}\:{or}\:\mathrm{3} \\ $$$$\:{then}\:{n}^{\mathrm{2}} \equiv\mathrm{1}\left({mod}\:\mathrm{24}\right) \\ $$

Question Number 217088    Answers: 1   Comments: 0

show that ∫_( n) ^( n + 1) ln(t) dt ≤ ln(n + (1/2)) Given u_n = (((4n)^n n!e^(−n) )/((2n)!)), ∀n ≥ 1 prove, using the preceding question that u_n is decreasing and convergent

$${show}\:{that}\:\int_{\:{n}} ^{\:{n}\:+\:\mathrm{1}} {ln}\left({t}\right)\:{dt}\:\leqslant\:{ln}\left({n}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${Given}\:{u}_{{n}} \:=\:\frac{\left(\mathrm{4}{n}\right)^{{n}} {n}!{e}^{−{n}} }{\left(\mathrm{2}{n}\right)!},\:\forall{n}\:\geqslant\:\mathrm{1} \\ $$$${prove},\:{using}\:{the}\:{preceding}\:{question}\:{that} \\ $$$${u}_{{n}} \:{is}\:{decreasing}\:{and}\:{convergent} \\ $$

Question Number 217085    Answers: 0   Comments: 2

Question Number 217084    Answers: 0   Comments: 0

Geometrie dans le plan. AB^(→) et CD^(→) sont deux vecteurs du plan. AB^(→) n′est pas nul. Demontre que si AB^(→) et CD^(→) sont colineaires alors il existe un nombre reel k tel que CD^(→) = k AB^(→) .

$$\boldsymbol{\mathrm{Geometr}}\mathrm{i}\boldsymbol{\mathrm{e}}\:\boldsymbol{\mathrm{dans}}\:\boldsymbol{\mathrm{le}}\:\boldsymbol{\mathrm{plan}}. \\ $$$$\overset{\rightarrow} {\mathrm{AB}}\:\mathrm{et}\:\overset{\rightarrow} {\mathrm{CD}}\:\mathrm{sont}\:\mathrm{deux}\:\mathrm{vecteurs}\:\mathrm{du}\:\mathrm{plan}. \\ $$$$\overset{\rightarrow} {\mathrm{AB}}\:\mathrm{n}'\mathrm{est}\:\mathrm{pas}\:\mathrm{nul}. \\ $$$$\mathrm{Demontre}\:\mathrm{que}\:\mathrm{si}\:\overset{\rightarrow} {\mathrm{AB}}\:\mathrm{et}\:\overset{\rightarrow} {\mathrm{CD}}\:\mathrm{sont}\:\mathrm{colineaires} \\ $$$$\mathrm{alors}\:\mathrm{il}\:\mathrm{existe}\:\mathrm{un}\:\mathrm{nombre}\:\mathrm{reel}\:\mathrm{k}\:\mathrm{tel}\:\mathrm{que} \\ $$$$\overset{\rightarrow} {\mathrm{CD}}\:=\:\mathrm{k}\:\overset{\rightarrow} {\mathrm{AB}}. \\ $$

Question Number 217080    Answers: 3   Comments: 1

Question Number 217068    Answers: 0   Comments: 0

Question Number 217066    Answers: 1   Comments: 0

Find all integer x,y such that x^2 −y^2 =100

$${Find}\:{all}\:{integer}\:{x},{y}\:{such}\:{that} \\ $$$${x}^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{100} \\ $$

Question Number 217065    Answers: 0   Comments: 9

To Tinku Tara sir, How can I import something in Latex form. •There is an option Get Latex but there is no “Paste Latex” option •When I tried to import something using option “Paste plain text” all the whitespaces/new line were lost in imported material. Please solve these issues as soon as possible

$${To}\:{Tinku}\:{Tara}\:{sir}, \\ $$$${How}\:{can}\:{I}\:{import}\:{something}\:{in}\:{Latex}\:{form}. \\ $$$$\bullet{There}\:{is}\:{an}\:{option}\:\mathrm{Get}\:\mathrm{Latex}\:{but}\:{there}\:{is}\:{no} \\ $$$$``\mathrm{Paste}\:\mathrm{Latex}''\:{option} \\ $$$$\bullet{When}\:{I}\:{tried}\:{to}\:{import}\:{something}\:{using}\:{option} \\ $$$$``\mathrm{Paste}\:\mathrm{plain}\:\mathrm{text}''\:{all}\:{the}\:{whitespaces}/{new}\:{line}\: \\ $$$${were}\:{lost}\:{in}\:{imported}\:{material}. \\ $$$$\boldsymbol{{Please}}\:{solve}\:{these}\:{issues}\:{as}\:{soon}\:{as}\:{possible} \\ $$

Question Number 217071    Answers: 1   Comments: 0

Find all two-digit numbers that are equal to four times the sum of their digits. Solve this using at least two different methods and verify your answers.

$$\mathrm{Find}\:\mathrm{all}\:\mathrm{two}-\mathrm{digit}\:\mathrm{numbers}\:\mathrm{that}\:\mathrm{are}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{four}\:\mathrm{times}\:\mathrm{the}\:\mathrm{sum}\: \\ $$$$\mathrm{of}\:\mathrm{their}\:\mathrm{digits}.\:\mathrm{Solve}\:\mathrm{this}\:\mathrm{using}\:\mathrm{at}\:\mathrm{least}\:\mathrm{two}\:\mathrm{different}\:\mathrm{methods}\: \\ $$$$\mathrm{and}\:\mathrm{verify}\:\mathrm{your}\:\mathrm{answers}. \\ $$

Question Number 217064    Answers: 1   Comments: 0

Two numbers differ by 6. The sum of their reciprocals is (2/(15)) . Determine the numbers.

$$\mathrm{Two}\:\mathrm{numbers}\:\mathrm{differ}\:\mathrm{by}\:\mathrm{6}.\:\mathrm{The}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{their}\:\mathrm{reciprocals}\:\mathrm{is}\:\frac{\mathrm{2}}{\mathrm{15}}\:. \\ $$$$\mathcal{D}{etermine}\:{the}\:{numbers}. \\ $$

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