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Question Number 79998    Answers: 0   Comments: 3

given 3x + 4y+1 = 3(√x) + 2(√y) find the value of (√(x.y))

$$\mathrm{given}\:\mathrm{3x}\:+\:\mathrm{4y}+\mathrm{1}\:=\:\mathrm{3}\sqrt{\mathrm{x}}\:+\:\mathrm{2}\sqrt{\mathrm{y}}\: \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\sqrt{\mathrm{x}.\mathrm{y}}\: \\ $$

Question Number 79978    Answers: 3   Comments: 0

Given for x,y,z>0: 2^x =3^y =5^z Arrange 2x, 3y, 5z in increasing order.

$${Given}\:{for}\:{x},{y},{z}>\mathrm{0}: \\ $$$$\mathrm{2}^{{x}} =\mathrm{3}^{{y}} =\mathrm{5}^{{z}} \\ $$$${Arrange}\:\mathrm{2}{x},\:\mathrm{3}{y},\:\mathrm{5}{z}\:{in}\:{increasing}\:{order}. \\ $$

Question Number 79974    Answers: 1   Comments: 0

Question Number 80004    Answers: 1   Comments: 2

x and y any integer satisfy equation (x−2004)(x−2006)=2^y the greatest possible value of x+y

$${x}\:{and}\:{y}\:{any}\:{integer}\:{satisfy} \\ $$$${equation}\:\left({x}−\mathrm{2004}\right)\left({x}−\mathrm{2006}\right)=\mathrm{2}^{{y}} \\ $$$${the}\:{greatest}\:{possible}\:{value} \\ $$$${of}\:{x}+{y} \\ $$

Question Number 79932    Answers: 1   Comments: 1

Question Number 79883    Answers: 0   Comments: 9

to Sir Jagoll (and of course everybody else) (1) y=((x^2 −x−6)/(x^2 −3x−4))= =(((x−3)(x+2))/((x−4)(x+1))) ⇒ ⇒ { ((zeros at x=−2; x=3)),((vertical asymptotes at x=−1; x=4)) :} defined for x∈R\{−1; 4} range: transforming y=((x^2 −x−6)/(x^2 −3x−4)) to x^2 −((3y−1)/(y−1))x−((2(2y−3))/(y−1))=0 D=((25y^2 −46y+25)/(4(y−1)^2 ))>0∀x∈R ⇒ ⇒ { ((range=R)),((horizontal asymptote at y=1 (∗))) :} (∗) doesn′t mean y=1 is not within range!!! x=1 ⇒ y=1 y′=−((2(x^2 −2x+7))/((x−4)^2 (x+1)^2 )) no real zeros ⇒ ⇒ no local extremes y′′=((4(x^3 −3x^2 +21x−25))/((x−4)^3 (x+1)^3 ))=0 at x≈1.33131 ⇒ ⇒ turning point (2) y=((x^2 −x−6)/(x^2 −3x+4))= =(((x−3)(x+2))/(x^2 −3x+4)); x^2 −3x+4=0 no real zeros ⇒ ⇒ { ((zeros at x=−2; x=3)),((no vertical asymptote)) :} defined for x∈R range: transforming y=((x^2 −x−6)/(x^2 −3x+4)) to x^2 −((3y−1)/(y−1))x+((2(2y+3))/(y−1))=0 D=−((7y^2 +14y−25)/(4(y−1)^2 ))≥0 for −1−((4(√(14)))/7)≤y≤−1+((4(√(14)))/7) ⇒ ⇒ { ((range=[−1−((4(√(14)))/7); −1+((4(√(14)))/7)])),((horizontal asymptote at y=1 (∗))) :} (∗) doesn′t mean y=1 is not within range!!! x=5 ⇒ y=1 y′=−((2(x^2 −10x+11))/((x^2 −3x+4)^2 ))=0 at x=5±(√(14)) y^(′′) =((4(x^3 −15x^2 +33x−13))/((x^2 −3x+4)^3 )) y′′ { ((>0 at x=5−(√(14)) ⇒ local minimum)),((<0 at x=5+(√(14)) ⇒ local maximum)),((=0 at { ((x≈.506699)),((x≈2.06421)),((x≈12.4291)) :} ⇒ 3 turning points )) :} (3) y=((x^2 −x+6)/(x^2 −3x−4))= =((x^2 −x+6)/((x−4)(x+1))) ⇒ ⇒ { ((no real zeros)),((vertical asymptotes at x=−1; x=4)) :} defined for x∈R\{−1; 4} range: transforming y=((x^2 −x+6)/(x^2 −3x−4)) to x^2 −((3y−1)/(y−1))x−((2(2y+3))/(y−1))=0 D=((25y^2 +2y−23)/(4(y−1)^2 ))<0 for −1<y<((23)/(25)) ⇒ ⇒ { ((range=R\]−1; ((23)/(25))[)),((horizontal asymptote at y=1 (∗))) :} (∗) doesn′t mean y=1 is not within range!!! x=−5 ⇒ y=1 y′=−((2(x^2 +10x−11))/((x−4)^2 (x+1)^2 ))=0 at x=−11; x=1 y′′=((4(x^3 +15x^2 −33x+53))/((x−4)^3 (x+1)^3 )) y′′ { ((>0 at x=−11 ⇒ local minimum)),((<0 at x=1 ⇒ local maximum)),((=0 at x≈−17.1098 ⇒ turning point)) :}

$$\mathrm{to}\:\mathrm{Sir}\:\mathrm{Jagoll}\:\left({and}\:{of}\:{course}\:{everybody}\:{else}\right) \\ $$$$ \\ $$$$\left(\mathrm{1}\right) \\ $$$${y}=\frac{{x}^{\mathrm{2}} −{x}−\mathrm{6}}{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}}= \\ $$$$=\frac{\left({x}−\mathrm{3}\right)\left({x}+\mathrm{2}\right)}{\left({x}−\mathrm{4}\right)\left({x}+\mathrm{1}\right)}\:\Rightarrow \\ $$$$\Rightarrow\:\begin{cases}{\mathrm{zeros}\:\mathrm{at}\:{x}=−\mathrm{2};\:{x}=\mathrm{3}}\\{\mathrm{vertical}\:\mathrm{asymptotes}\:\mathrm{at}\:{x}=−\mathrm{1};\:{x}=\mathrm{4}}\end{cases} \\ $$$$\mathrm{defined}\:\mathrm{for}\:{x}\in\mathbb{R}\backslash\left\{−\mathrm{1};\:\mathrm{4}\right\} \\ $$$$\mathrm{range}:\:\mathrm{transforming}\:{y}=\frac{{x}^{\mathrm{2}} −{x}−\mathrm{6}}{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}}\:\mathrm{to} \\ $$$${x}^{\mathrm{2}} −\frac{\mathrm{3}{y}−\mathrm{1}}{{y}−\mathrm{1}}{x}−\frac{\mathrm{2}\left(\mathrm{2}{y}−\mathrm{3}\right)}{{y}−\mathrm{1}}=\mathrm{0} \\ $$$${D}=\frac{\mathrm{25}{y}^{\mathrm{2}} −\mathrm{46}{y}+\mathrm{25}}{\mathrm{4}\left({y}−\mathrm{1}\right)^{\mathrm{2}} }>\mathrm{0}\forall{x}\in\mathbb{R}\:\Rightarrow \\ $$$$\Rightarrow\:\begin{cases}{\mathrm{range}=\mathbb{R}}\\{\mathrm{horizontal}\:\mathrm{asymptote}\:\mathrm{at}\:{y}=\mathrm{1}\:\left(\ast\right)}\end{cases} \\ $$$$\left(\ast\right)\:\mathrm{doesn}'\mathrm{t}\:\mathrm{mean}\:{y}=\mathrm{1}\:\mathrm{is}\:\mathrm{not}\:\mathrm{within}\:\mathrm{range}!!! \\ $$$$\:\:\:\:\:\:\:\:{x}=\mathrm{1}\:\Rightarrow\:{y}=\mathrm{1} \\ $$$${y}'=−\frac{\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{7}\right)}{\left({x}−\mathrm{4}\right)^{\mathrm{2}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:\mathrm{no}\:\mathrm{real}\:\mathrm{zeros}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{no}\:\mathrm{local}\:\mathrm{extremes} \\ $$$${y}''=\frac{\mathrm{4}\left({x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{21}{x}−\mathrm{25}\right)}{\left({x}−\mathrm{4}\right)^{\mathrm{3}} \left({x}+\mathrm{1}\right)^{\mathrm{3}} }=\mathrm{0}\:\mathrm{at}\:{x}\approx\mathrm{1}.\mathrm{33131}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{turning}\:\mathrm{point} \\ $$$$ \\ $$$$\left(\mathrm{2}\right) \\ $$$${y}=\frac{{x}^{\mathrm{2}} −{x}−\mathrm{6}}{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}}= \\ $$$$=\frac{\left({x}−\mathrm{3}\right)\left({x}+\mathrm{2}\right)}{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}};\:{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}=\mathrm{0}\:\mathrm{no}\:\mathrm{real}\:\mathrm{zeros}\:\Rightarrow \\ $$$$\Rightarrow\:\begin{cases}{\mathrm{zeros}\:\mathrm{at}\:{x}=−\mathrm{2};\:{x}=\mathrm{3}}\\{\mathrm{no}\:\mathrm{vertical}\:\mathrm{asymptote}}\end{cases} \\ $$$$\mathrm{defined}\:\mathrm{for}\:{x}\in\mathbb{R} \\ $$$$\mathrm{range}:\:\mathrm{transforming}\:{y}=\frac{{x}^{\mathrm{2}} −{x}−\mathrm{6}}{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}}\:\mathrm{to} \\ $$$${x}^{\mathrm{2}} −\frac{\mathrm{3}{y}−\mathrm{1}}{{y}−\mathrm{1}}{x}+\frac{\mathrm{2}\left(\mathrm{2}{y}+\mathrm{3}\right)}{{y}−\mathrm{1}}=\mathrm{0} \\ $$$${D}=−\frac{\mathrm{7}{y}^{\mathrm{2}} +\mathrm{14}{y}−\mathrm{25}}{\mathrm{4}\left({y}−\mathrm{1}\right)^{\mathrm{2}} }\geqslant\mathrm{0}\:\mathrm{for}\:−\mathrm{1}−\frac{\mathrm{4}\sqrt{\mathrm{14}}}{\mathrm{7}}\leqslant{y}\leqslant−\mathrm{1}+\frac{\mathrm{4}\sqrt{\mathrm{14}}}{\mathrm{7}}\:\Rightarrow \\ $$$$\Rightarrow\:\begin{cases}{\mathrm{range}=\left[−\mathrm{1}−\frac{\mathrm{4}\sqrt{\mathrm{14}}}{\mathrm{7}};\:−\mathrm{1}+\frac{\mathrm{4}\sqrt{\mathrm{14}}}{\mathrm{7}}\right]}\\{\mathrm{horizontal}\:\mathrm{asymptote}\:\mathrm{at}\:{y}=\mathrm{1}\:\left(\ast\right)}\end{cases} \\ $$$$\left(\ast\right)\:\mathrm{doesn}'\mathrm{t}\:\mathrm{mean}\:{y}=\mathrm{1}\:\mathrm{is}\:\mathrm{not}\:\mathrm{within}\:\mathrm{range}!!! \\ $$$$\:\:\:\:\:\:\:\:{x}=\mathrm{5}\:\Rightarrow\:{y}=\mathrm{1} \\ $$$${y}'=−\frac{\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{11}\right)}{\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}\right)^{\mathrm{2}} }=\mathrm{0}\:\mathrm{at}\:{x}=\mathrm{5}\pm\sqrt{\mathrm{14}} \\ $$$${y}^{''} =\frac{\mathrm{4}\left({x}^{\mathrm{3}} −\mathrm{15}{x}^{\mathrm{2}} +\mathrm{33}{x}−\mathrm{13}\right)}{\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}\right)^{\mathrm{3}} } \\ $$$${y}''\begin{cases}{>\mathrm{0}\:\mathrm{at}\:{x}=\mathrm{5}−\sqrt{\mathrm{14}}\:\Rightarrow\:\mathrm{local}\:\mathrm{minimum}}\\{<\mathrm{0}\:\mathrm{at}\:{x}=\mathrm{5}+\sqrt{\mathrm{14}}\:\Rightarrow\:\mathrm{local}\:\mathrm{maximum}}\\{=\mathrm{0}\:\mathrm{at}\:\begin{cases}{{x}\approx.\mathrm{506699}}\\{{x}\approx\mathrm{2}.\mathrm{06421}}\\{{x}\approx\mathrm{12}.\mathrm{4291}}\end{cases}\:\Rightarrow\:\mathrm{3}\:\mathrm{turning}\:\mathrm{points}\:}\end{cases} \\ $$$$ \\ $$$$\left(\mathrm{3}\right) \\ $$$${y}=\frac{{x}^{\mathrm{2}} −{x}+\mathrm{6}}{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}}= \\ $$$$=\frac{{x}^{\mathrm{2}} −{x}+\mathrm{6}}{\left({x}−\mathrm{4}\right)\left({x}+\mathrm{1}\right)}\:\Rightarrow \\ $$$$\Rightarrow\:\begin{cases}{\mathrm{no}\:\mathrm{real}\:\mathrm{zeros}}\\{\mathrm{vertical}\:\mathrm{asymptotes}\:\mathrm{at}\:{x}=−\mathrm{1};\:{x}=\mathrm{4}}\end{cases} \\ $$$$\mathrm{defined}\:\mathrm{for}\:{x}\in\mathbb{R}\backslash\left\{−\mathrm{1};\:\mathrm{4}\right\} \\ $$$$\mathrm{range}:\:\mathrm{transforming}\:{y}=\frac{{x}^{\mathrm{2}} −{x}+\mathrm{6}}{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}}\:\mathrm{to} \\ $$$${x}^{\mathrm{2}} −\frac{\mathrm{3}{y}−\mathrm{1}}{{y}−\mathrm{1}}{x}−\frac{\mathrm{2}\left(\mathrm{2}{y}+\mathrm{3}\right)}{{y}−\mathrm{1}}=\mathrm{0} \\ $$$${D}=\frac{\mathrm{25}{y}^{\mathrm{2}} +\mathrm{2}{y}−\mathrm{23}}{\mathrm{4}\left({y}−\mathrm{1}\right)^{\mathrm{2}} }<\mathrm{0}\:\mathrm{for}\:−\mathrm{1}<{y}<\frac{\mathrm{23}}{\mathrm{25}}\:\Rightarrow \\ $$$$\Rightarrow\:\begin{cases}{\left.\mathrm{range}=\mathbb{R}\backslash\right]−\mathrm{1};\:\frac{\mathrm{23}}{\mathrm{25}}\left[\right.}\\{\mathrm{horizontal}\:\mathrm{asymptote}\:\mathrm{at}\:{y}=\mathrm{1}\:\left(\ast\right)}\end{cases} \\ $$$$\left(\ast\right)\:\mathrm{doesn}'\mathrm{t}\:\mathrm{mean}\:{y}=\mathrm{1}\:\mathrm{is}\:\mathrm{not}\:\mathrm{within}\:\mathrm{range}!!! \\ $$$$\:\:\:\:\:\:\:\:{x}=−\mathrm{5}\:\Rightarrow\:{y}=\mathrm{1} \\ $$$${y}'=−\frac{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{10}{x}−\mathrm{11}\right)}{\left({x}−\mathrm{4}\right)^{\mathrm{2}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{0}\:\mathrm{at}\:{x}=−\mathrm{11};\:{x}=\mathrm{1} \\ $$$${y}''=\frac{\mathrm{4}\left({x}^{\mathrm{3}} +\mathrm{15}{x}^{\mathrm{2}} −\mathrm{33}{x}+\mathrm{53}\right)}{\left({x}−\mathrm{4}\right)^{\mathrm{3}} \left({x}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${y}''\begin{cases}{>\mathrm{0}\:\mathrm{at}\:{x}=−\mathrm{11}\:\Rightarrow\:\mathrm{local}\:\mathrm{minimum}}\\{<\mathrm{0}\:\mathrm{at}\:{x}=\mathrm{1}\:\Rightarrow\:\mathrm{local}\:\mathrm{maximum}}\\{=\mathrm{0}\:\mathrm{at}\:{x}\approx−\mathrm{17}.\mathrm{1098}\:\Rightarrow\:\mathrm{turning}\:\mathrm{point}}\end{cases} \\ $$

Question Number 79876    Answers: 1   Comments: 1

f is derivable in R. 1) Demonstrate that if f is pair , f ′ is odd(unpair). 1) Demonstrate that if f is unpair , f ′ is pair. Please help me sirs

$${f}\:{is}\:{derivable}\:{in}\:\mathbb{R}. \\ $$$$\left.\mathrm{1}\right)\:\mathrm{Demonstrate}\:\mathrm{that}\:\mathrm{if}\:{f}\:\mathrm{is}\:\mathrm{pair}\:,\:{f}\:'\:\mathrm{is}\:\mathrm{odd}\left(\mathrm{unpair}\right). \\ $$$$\left.\mathrm{1}\right)\:\mathrm{Demonstrate}\:\mathrm{that}\:\mathrm{if}\:{f}\:{is}\:\mathrm{unpair}\:,\:{f}\:'\:\mathrm{is}\:\mathrm{pair}. \\ $$$$ \\ $$$$\mathrm{Please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{sirs} \\ $$

Question Number 79947    Answers: 0   Comments: 4

Find ((ln 2)/(2!))+((ln 3)/(3!))+((ln 4)/(4!))+...+((ln n)/(n!))+...=?

$${Find} \\ $$$$\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}!}+\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{3}!}+\frac{\mathrm{ln}\:\mathrm{4}}{\mathrm{4}!}+...+\frac{\mathrm{ln}\:{n}}{{n}!}+...=? \\ $$

Question Number 79946    Answers: 2   Comments: 0

Find ∫_0 ^( n) [(x)^(1/3) ]dx=? in terms of n. (n∈N)

$${Find}\: \\ $$$$\int_{\mathrm{0}} ^{\:{n}} \left[\sqrt[{\mathrm{3}}]{{x}}\right]{dx}=?\: \\ $$$${in}\:{terms}\:{of}\:{n}.\:\left({n}\in\mathbb{N}\right) \\ $$

Question Number 79864    Answers: 0   Comments: 0

Question Number 79861    Answers: 0   Comments: 1

Question Number 79826    Answers: 0   Comments: 7

Question Number 79807    Answers: 2   Comments: 2

Question Number 79792    Answers: 1   Comments: 1

JUST FOR FUN (1/2), (2/3), 1, (8/5), (8/3), ? what do you think is the next number ? why?

$${JUST}\:{FOR}\:{FUN} \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\mathrm{2}},\:\frac{\mathrm{2}}{\mathrm{3}},\:\mathrm{1},\:\frac{\mathrm{8}}{\mathrm{5}},\:\frac{\mathrm{8}}{\mathrm{3}},\:? \\ $$$${what}\:{do}\:{you}\:{think}\:{is}\:{the}\:{next}\:{number}\:? \\ $$$${why}? \\ $$

Question Number 79757    Answers: 0   Comments: 1

And equation of a circle is x^2 +y^2 −2x+4y=0. (T) is his his tangent line at M(x_0 ;y_0 ) passing by D(2;1). a) Show that y verify y_0 ^2 +y_0 =0 b) deduct others tangent′s equations to Circle passing by D.

$$\mathrm{And}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{is} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{4}{y}=\mathrm{0}. \\ $$$$\left({T}\right)\:\mathrm{is}\:\mathrm{his}\:\mathrm{his}\:\mathrm{tangent}\:\mathrm{line}\:\mathrm{at}\:\mathrm{M}\left({x}_{\mathrm{0}} ;{y}_{\mathrm{0}} \right) \\ $$$${passing}\:{by}\:{D}\left(\mathrm{2};\mathrm{1}\right). \\ $$$$ \\ $$$$\left.\mathrm{a}\right)\:\mathrm{Show}\:\mathrm{that}\:\mathrm{y}\:\mathrm{verify}\:\mathrm{y}_{\mathrm{0}} ^{\mathrm{2}} +\mathrm{y}_{\mathrm{0}} =\mathrm{0} \\ $$$$\left.\mathrm{b}\right)\:\mathrm{deduct}\:\mathrm{others}\:\mathrm{tangent}'\mathrm{s}\: \\ $$$$\mathrm{equations}\:\mathrm{to}\:\mathrm{Circle}\:\mathrm{passing}\:\mathrm{by}\:\mathrm{D}. \\ $$

Question Number 79635    Answers: 0   Comments: 9

Sum: (1/1) + (1/(1 + 2)) + (1/(1 + 2 + 3)) + ... + (1/(1 + 2 + 3 + ... + 8016))

$$\mathrm{Sum}:\:\:\frac{\mathrm{1}}{\mathrm{1}}\:+\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{2}\:+\:\mathrm{3}}\:+\:...\:\:+\:\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{2}\:+\:\mathrm{3}\:+\:...\:+\:\mathrm{8016}} \\ $$

Question Number 79571    Answers: 1   Comments: 5

Solve for x: ((x + ((x + ((x + ...))^(1/3) ))^(1/3) ))^(1/3) = ((x ((x ((x ....))^(1/3) ))^(1/3) ))^(1/3)

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x}: \\ $$$$\:\sqrt[{\mathrm{3}}]{\mathrm{x}\:\:+\:\:\sqrt[{\mathrm{3}}]{\mathrm{x}\:\:+\:\:\sqrt[{\mathrm{3}}]{\mathrm{x}\:\:+\:\:...}}}\:\:\:\:\:\:\:=\:\:\:\:\:\sqrt[{\mathrm{3}}]{\mathrm{x}\:\:\sqrt[{\mathrm{3}}]{\mathrm{x}\:\:\sqrt[{\mathrm{3}}]{\mathrm{x}\:\:....}}} \\ $$

Question Number 79613    Answers: 1   Comments: 0

3xy(2x−y)−3bx+3c=0 3xy(x−2y)−3by−3c=0 find non-zero, real values of x,y if b,c∈R.

$$\mathrm{3}{xy}\left(\mathrm{2}{x}−{y}\right)−\mathrm{3}{bx}+\mathrm{3}{c}=\mathrm{0} \\ $$$$\mathrm{3}{xy}\left({x}−\mathrm{2}{y}\right)−\mathrm{3}{by}−\mathrm{3}{c}=\mathrm{0} \\ $$$${find}\:{non}-{zero},\:{real}\:{values} \\ $$$${of}\:{x},{y}\:\:{if}\:{b},{c}\in\mathbb{R}. \\ $$

Question Number 79560    Answers: 0   Comments: 1

(√(1+x)) ≤ ((5−x))^(1/(4 ))

$$\sqrt{\mathrm{1}+\mathrm{x}}\:\leqslant\:\sqrt[{\mathrm{4}\:}]{\mathrm{5}−\mathrm{x}} \\ $$

Question Number 79538    Answers: 1   Comments: 13

Question Number 79536    Answers: 0   Comments: 4

Question Number 79513    Answers: 0   Comments: 1

Q.solve if t^2 =n^2 cos^2 (x)+m^2 sin^2 (x) then show that: t+(d^2 t/dx^2 )=(((nm)^2 )/t^3 )

$${Q}.{solve} \\ $$$${if}\:{t}^{\mathrm{2}} ={n}^{\mathrm{2}} {cos}^{\mathrm{2}} \left({x}\right)+{m}^{\mathrm{2}} {sin}^{\mathrm{2}} \left({x}\right) \\ $$$$ \\ $$$${then}\:{show}\:{that}: \\ $$$${t}+\frac{{d}^{\mathrm{2}} {t}}{{dx}^{\mathrm{2}} }=\frac{\left({nm}\right)^{\mathrm{2}} }{{t}^{\mathrm{3}} } \\ $$$$ \\ $$

Question Number 79512    Answers: 0   Comments: 0

Q.solve if t^2 =n^2 cos^2 (x)+m^2 sin^2 (x) then show that: t+(d^2 t/dx^2 )=(((nm)^2 )/t^3 )

$${Q}.{solve} \\ $$$${if}\:{t}^{\mathrm{2}} ={n}^{\mathrm{2}} {cos}^{\mathrm{2}} \left({x}\right)+{m}^{\mathrm{2}} {sin}^{\mathrm{2}} \left({x}\right) \\ $$$$ \\ $$$${then}\:{show}\:{that}: \\ $$$${t}+\frac{{d}^{\mathrm{2}} {t}}{{dx}^{\mathrm{2}} }=\frac{\left({nm}\right)^{\mathrm{2}} }{{t}^{\mathrm{3}} } \\ $$$$ \\ $$

Question Number 79852    Answers: 2   Comments: 1

Question Number 79424    Answers: 0   Comments: 0

Derive the width of the diffraction pattern for the case of (i)single slits (ii)double slits

$${Derive}\:{the}\:{width}\:{of}\:{the} \\ $$$${diffraction}\:{pattern}\:{for} \\ $$$${the}\:{case}\:{of} \\ $$$$\left({i}\right){single}\:{slits} \\ $$$$\left({ii}\right){double}\:{slits} \\ $$

Question Number 79417    Answers: 2   Comments: 3

For x,y∈R find the minimum and maximum of 2x^2 −3x+4y if x^2 +2y^2 −xy−5x−7y−30=0.

$${For}\:{x},{y}\in\mathbb{R}\:{find}\:{the}\:{minimum}\:{and} \\ $$$${maximum}\:{of}\:\mathrm{2}{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}{y} \\ $$$${if}\:{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} −{xy}−\mathrm{5}{x}−\mathrm{7}{y}−\mathrm{30}=\mathrm{0}. \\ $$

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