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Question Number 112162 Answers: 1 Comments: 3

Question Number 112159 Answers: 1 Comments: 0

(1)Find the equation of hyperbola with centre point at (1,−2) and coordinates of foci is (6,−2) and (−4,−2) (2) If hyperbola ((x^2 −2nx+n^2 )/(25))−((y^2 −2my+m^2 )/(16))=1 have a asympyotes passes through at (0,1), then 5m−4n =

$$\left(\mathrm{1}\right)\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{hyperbola}\:\mathrm{with}\: \\ $$$$\mathrm{centre}\:\mathrm{point}\:\mathrm{at}\:\left(\mathrm{1},−\mathrm{2}\right)\:\mathrm{and}\:\mathrm{coordinates} \\ $$$$\mathrm{of}\:\mathrm{foci}\:\mathrm{is}\:\left(\mathrm{6},−\mathrm{2}\right)\:\mathrm{and}\:\left(−\mathrm{4},−\mathrm{2}\right) \\ $$$$\left(\mathrm{2}\right)\:\mathrm{If}\:\mathrm{hyperbola}\:\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{2nx}+\mathrm{n}^{\mathrm{2}} }{\mathrm{25}}−\frac{\mathrm{y}^{\mathrm{2}} −\mathrm{2my}+\mathrm{m}^{\mathrm{2}} }{\mathrm{16}}=\mathrm{1} \\ $$$$\mathrm{have}\:\mathrm{a}\:\mathrm{asympyotes}\:\mathrm{passes}\:\mathrm{through}\:\mathrm{at}\: \\ $$$$\left(\mathrm{0},\mathrm{1}\right),\:\mathrm{then}\:\mathrm{5m}−\mathrm{4n}\:=\: \\ $$

Question Number 112145 Answers: 1 Comments: 0

4^(tan^2 x) + 2^(1/(cos^2 x)) − 80 = 0

$$\mathrm{4}^{\mathrm{tan}\:^{\mathrm{2}} {x}} \:+\:\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{cos}\:^{\mathrm{2}} {x}}} \:−\:\mathrm{80}\:=\:\mathrm{0} \\ $$

Question Number 112136 Answers: 2 Comments: 0

find (5/(6+7sin 2β)) , if tan β = 0.2

$${find}\:\:\:\frac{\mathrm{5}}{\mathrm{6}+\mathrm{7sin}\:\mathrm{2}\beta}\:,\:\:\:{if}\:\:\:\mathrm{tan}\:\beta\:\:=\:\:\mathrm{0}.\mathrm{2} \\ $$

Question Number 112133 Answers: 1 Comments: 0

Question Number 112128 Answers: 5 Comments: 2

Question Number 112318 Answers: 2 Comments: 2

A triangle ABC has the following properties BC=1, AB=AC and that the angle bisector from vertex B is also a median. Find all possible triangle(s) with its/their side−lengths and angles.

$$\mathrm{A}\:\mathrm{triangle}\:\mathrm{ABC}\:\mathrm{has}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{properties}\:\mathrm{BC}=\mathrm{1},\:\boldsymbol{\mathrm{AB}}=\boldsymbol{\mathrm{AC}}\:\mathrm{and}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{angle}\:\mathrm{bisector}\:\mathrm{from}\:\mathrm{vertex}\:\mathrm{B}\:\mathrm{is} \\ $$$$\mathrm{also}\:\mathrm{a}\:\mathrm{median}.\:\mathrm{Find}\:\mathrm{all}\:\mathrm{possible} \\ $$$$\mathrm{triangle}\left(\mathrm{s}\right)\:\mathrm{with}\:\mathrm{its}/\mathrm{their} \\ $$$$\mathrm{side}−\mathrm{lengths}\:\mathrm{and}\:\mathrm{angles}. \\ $$

Question Number 112135 Answers: 3 Comments: 0

solve ∫_(−1) ^1 ∣3^x −2^x ∣dx

$${solve} \\ $$$$\int_{−\mathrm{1}} ^{\mathrm{1}} \mid\mathrm{3}^{{x}} −\mathrm{2}^{{x}} \mid{dx} \\ $$

Question Number 112119 Answers: 1 Comments: 0

....calculus.... prove that::: if Ω =∫_(0 ) ^( 1) ln(ln(1−(√x) ))dx then Re(Ω) := −γ + ln(2).... m.n. july 1970#

$$\:\:\:\:\:\:\:\:\:\:\:\:\:....{calculus}.... \\ $$$${prove}\:{that}::: \\ $$$${if}\:\:\:\Omega\:=\int_{\mathrm{0}\:\:} ^{\:\mathrm{1}} {ln}\left({ln}\left(\mathrm{1}−\sqrt{{x}}\:\right)\right){dx} \\ $$$${then} \\ $$$$\mathscr{R}{e}\left(\Omega\right)\::=\:−\gamma\:+\:{ln}\left(\mathrm{2}\right).... \\ $$$$ \\ $$$${m}.{n}.\:{july}\:\mathrm{1970}# \\ $$

Question Number 112114 Answers: 0 Comments: 0

solution of Φ=∫_0 ^1 xH_x dx =^(γ+ψ(x+1)) ∫_0 ^( 1) x(γ+ψ(x+1))dx =^(ψ(x+1)=(1/x)+ψ(x)) ∫_0 ^( 1) x(γ+(1/x)+ψ(x))dx =(γ/2)+1+ ∫_0 ^( 1) x.(d/dx)(ln(Γ(x))) =(γ/2)+1+[xln(Γ(x))]_0 ^1 −∫_0 ^( 1) ln(Γ(x))dx we know (why?) ∫_0 ^( 1) ln(Γ(x))dx=ln((√(2π)) ) and lim_(x→0^+ ) (xln(Γ(x)))=0 (why??) finally Φ =∫_0 ^( 1) xH_x dx=(γ/2)+1−ln((√(2π))) ✓ m.n.july 1970....

$$\:\:{solution}\:{of} \\ $$$$\Phi=\int_{\mathrm{0}} ^{\mathrm{1}} {xH}_{{x}} {dx}\:\overset{\gamma+\psi\left({x}+\mathrm{1}\right)} {=}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}\left(\gamma+\psi\left({x}+\mathrm{1}\right)\right){dx}\: \\ $$$$\:\overset{\psi\left({x}+\mathrm{1}\right)=\frac{\mathrm{1}}{{x}}+\psi\left({x}\right)} {=}\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}\left(\gamma+\frac{\mathrm{1}}{{x}}+\psi\left({x}\right)\right){dx} \\ $$$$=\frac{\gamma}{\mathrm{2}}+\mathrm{1}+\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}.\frac{{d}}{{dx}}\left({ln}\left(\Gamma\left({x}\right)\right)\right) \\ $$$$=\frac{\gamma}{\mathrm{2}}+\mathrm{1}+\left[{xln}\left(\Gamma\left({x}\right)\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left(\Gamma\left({x}\right)\right){dx} \\ $$$$\:\:{we}\:\:{know}\:\left({why}?\right)\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left(\Gamma\left({x}\right)\right){dx}={ln}\left(\sqrt{\mathrm{2}\pi}\:\right) \\ $$$$\:{and}\:\:{lim}_{{x}\rightarrow\mathrm{0}^{+} } \left({xln}\left(\Gamma\left({x}\right)\right)\right)=\mathrm{0}\:\left({why}??\right) \\ $$$${finally}\:\Phi\:=\int_{\mathrm{0}} ^{\:\mathrm{1}} {xH}_{{x}} {dx}=\frac{\gamma}{\mathrm{2}}+\mathrm{1}−{ln}\left(\sqrt{\mathrm{2}\pi}\right)\:\checkmark \\ $$$$\:\:{m}.{n}.{july}\:\mathrm{1970}.... \\ $$$$\: \\ $$

Question Number 112096 Answers: 3 Comments: 0

Question Number 112097 Answers: 0 Comments: 2

∣∣x−1∣+1∣ < x

$$\:\:\:\:\:\:\mid\mid{x}−\mathrm{1}\mid+\mathrm{1}\mid\:<\:{x} \\ $$

Question Number 112087 Answers: 2 Comments: 0

a(√a) −3 = 10(√a) → (√a) + (√a^(−1) ) =?

$$\:\:\:\mathrm{a}\sqrt{\mathrm{a}}\:−\mathrm{3}\:=\:\mathrm{10}\sqrt{\mathrm{a}}\:\rightarrow\:\sqrt{\mathrm{a}}\:+\:\sqrt{\mathrm{a}^{−\mathrm{1}} }\:=? \\ $$

Question Number 112083 Answers: 2 Comments: 0

lim_(n→∞) 4^n (1−cos ((α/2^n ))) ? (√(bemath))

$$\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{4}^{{n}} \:\left(\mathrm{1}−\mathrm{cos}\:\left(\frac{\alpha}{\mathrm{2}^{{n}} }\right)\right)\:?\: \\ $$$$\:\:\sqrt{{bemath}} \\ $$

Question Number 112076 Answers: 1 Comments: 3

Question Number 112075 Answers: 0 Comments: 2

Are there infinitely many solutions to sin((β/2))=cos(90−(β/2)) ?

$$\mathrm{Are}\:\mathrm{there}\:\mathrm{infinitely}\:\mathrm{many}\:\mathrm{solutions}\:\mathrm{to} \\ $$$$\mathrm{sin}\left(\frac{\beta}{\mathrm{2}}\right)=\mathrm{cos}\left(\mathrm{90}−\frac{\beta}{\mathrm{2}}\right)\:? \\ $$

Question Number 112067 Answers: 1 Comments: 4

(√(bemath −−−−■■)) find the value of Π_(m=1) ^6 tan (((mπ)/7)).

$$\:\:\sqrt{{bemath}\:−−−−\blacksquare\blacksquare} \\ $$$${find}\:{the}\:{value}\:{of}\:\underset{{m}=\mathrm{1}} {\overset{\mathrm{6}} {\prod}}\mathrm{tan}\:\left(\frac{{m}\pi}{\mathrm{7}}\right).\: \\ $$

Question Number 112065 Answers: 1 Comments: 0

Question Number 112059 Answers: 1 Comments: 0

Using the cosine rule(c^2 =a^2 +b^2 −2abcosC), prove the triangle inequality: if a,b and c are sides of a triangle ABC, then a+b≥c and explain when equality holds. Further prove that sin α + sin β ≥ sin(α+β) for 0° ≤α,β≤180°

$$\mathrm{Using}\:\mathrm{the}\:\mathrm{cosine} \\ $$$$\mathrm{rule}\left(\mathrm{c}^{\mathrm{2}} =\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} −\mathrm{2abcosC}\right),\:\mathrm{prove}\:\mathrm{the} \\ $$$$\mathrm{triangle}\:\mathrm{inequality}:\:\mathrm{if}\:\mathrm{a},\mathrm{b}\:\mathrm{and}\:\mathrm{c}\:\mathrm{are} \\ $$$$\mathrm{sides}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{ABC},\:\mathrm{then}\:\mathrm{a}+\mathrm{b}\geqslant\mathrm{c} \\ $$$$\mathrm{and}\:\mathrm{explain}\:\mathrm{when}\:\mathrm{equality}\:\mathrm{holds}. \\ $$$$\mathrm{Further}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{sin}\:\alpha\:+\:\mathrm{sin}\:\beta\:\geqslant \\ $$$$\mathrm{sin}\left(\alpha+\beta\right)\:\mathrm{for}\:\mathrm{0}°\:\leqslant\alpha,\beta\leqslant\mathrm{180}° \\ $$

Question Number 112058 Answers: 1 Comments: 0

Question Number 112053 Answers: 1 Comments: 1

solve for x, y, z ∈C: 2x^2 −3x=(√(13x^2 −52x+40)) 6y^2 −14x=(√(x^2 −220x+300)) z^2 −2z=(√(−12x^2 +72x−132)) [exact solutions possible in all cases]

$$\mathrm{solve}\:\mathrm{for}\:{x},\:{y},\:{z}\:\in\mathbb{C}: \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{3}{x}=\sqrt{\mathrm{13}{x}^{\mathrm{2}} −\mathrm{52}{x}+\mathrm{40}} \\ $$$$\mathrm{6}{y}^{\mathrm{2}} −\mathrm{14}{x}=\sqrt{{x}^{\mathrm{2}} −\mathrm{220}{x}+\mathrm{300}} \\ $$$${z}^{\mathrm{2}} −\mathrm{2}{z}=\sqrt{−\mathrm{12}{x}^{\mathrm{2}} +\mathrm{72}{x}−\mathrm{132}} \\ $$$$\left[\mathrm{exact}\:\mathrm{solutions}\:\mathrm{possible}\:\mathrm{in}\:\mathrm{all}\:\mathrm{cases}\right] \\ $$

Question Number 112050 Answers: 3 Comments: 0

Question Number 112011 Answers: 2 Comments: 0

Find ∣{n∈N∣n≤100, 4!∣2^n −n^3 }∣. (Note that ∣S∣ denotes the cardinality or number of elements of a set,S).

$$\mathrm{Find}\:\mid\left\{\mathrm{n}\in\mathbb{N}\mid\mathrm{n}\leqslant\mathrm{100},\:\mathrm{4}!\mid\mathrm{2}^{\mathrm{n}} −\mathrm{n}^{\mathrm{3}} \right\}\mid. \\ $$$$\left(\mathrm{Note}\:\mathrm{that}\:\mid\mathrm{S}\mid\:\mathrm{denotes}\:\mathrm{the}\:\mathrm{cardinality}\right. \\ $$$$\left.\mathrm{or}\:\mathrm{number}\:\mathrm{of}\:\mathrm{elements}\:\mathrm{of}\:\mathrm{a}\:\mathrm{set},\mathrm{S}\right). \\ $$

Question Number 112009 Answers: 0 Comments: 1

Question Number 112008 Answers: 0 Comments: 0

Question Number 112007 Answers: 0 Comments: 0

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