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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

Question Number 111988 Answers: 2 Comments: 1

Question Number 111975 Answers: 1 Comments: 0

Question Number 111974 Answers: 3 Comments: 0

Question Number 111973 Answers: 0 Comments: 2

(1/(5!!)) + ((1.3)/(7!!)) + ((1.3.5)/(9!!)) + ... + ((1.3.5.7....95)/(99!!)) =?

$$\frac{\mathrm{1}}{\mathrm{5}!!}\:+\:\frac{\mathrm{1}.\mathrm{3}}{\mathrm{7}!!}\:+\:\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}}{\mathrm{9}!!}\:+\:...\:+\:\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}....\mathrm{95}}{\mathrm{99}!!}\:=? \\ $$

Question Number 111965 Answers: 1 Comments: 0

....advanced calculus... evaluate: i:: ∫_0 ^( 1) xH_x dx =??? ii::Σ_(n=1 ) ^∞ (H_n /(n^2 2^(n ) )) =??? iii:: ∫_0 ^( 1) ((ln(((x+1))^(1/3) ))/((x+1)(x+2)))=??? m.n. july 1970...#

$$\:\:\:\:\:\:....{advanced}\:\:{calculus}... \\ $$$${evaluate}: \\ $$$$ \\ $$$$\:\:\:\:{i}::\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {xH}_{{x}} {dx}\:=???\:\: \\ $$$$\:\:\:\:\:{ii}::\underset{{n}=\mathrm{1}\:} {\overset{\infty} {\sum}}\frac{{H}_{{n}} }{{n}^{\mathrm{2}} \mathrm{2}^{{n}\:} }\:=??? \\ $$$$\:\:\:\:\:{iii}::\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}\right)}{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)}=??? \\ $$$$\:\:\:\:\:\:\:\:\:{m}.{n}.\:{july}\:\mathrm{1970}...# \\ $$

Question Number 111960 Answers: 1 Comments: 0

(1)lim_(x→(π/4)) sin ((π/4)−x).tan (x+(π/4)) ? (2)lim_(x→(π/2)) ((π(π−2x)tan (x−(π/2)))/(2(x−π)cos^2 x)) ? (3)lim_(x→0) ((3x(cos 3x−cos 7x))/( (√(sin 2x+1)) −(√(tan 2x+1)))) ? (4)lim_(x→0) ((sin 2x)/(3−(√(3x+9)))) ?

$$\left(\mathrm{1}\right)\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}−{x}\right).\mathrm{tan}\:\left({x}+\frac{\pi}{\mathrm{4}}\right)\:? \\ $$$$\left(\mathrm{2}\right)\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\pi\left(\pi−\mathrm{2}{x}\right)\mathrm{tan}\:\left({x}−\frac{\pi}{\mathrm{2}}\right)}{\mathrm{2}\left({x}−\pi\right)\mathrm{cos}\:^{\mathrm{2}} {x}}\:? \\ $$$$\left(\mathrm{3}\right)\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3}{x}\left(\mathrm{cos}\:\mathrm{3}{x}−\mathrm{cos}\:\mathrm{7}{x}\right)}{\:\sqrt{\mathrm{sin}\:\mathrm{2}{x}+\mathrm{1}}\:−\sqrt{\mathrm{tan}\:\mathrm{2}{x}+\mathrm{1}}}\:? \\ $$$$\left(\mathrm{4}\right)\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{3}−\sqrt{\mathrm{3}{x}+\mathrm{9}}}\:? \\ $$$$ \\ $$

Question Number 111957 Answers: 0 Comments: 31

“MATHEMATICS” CONTAINS ALL THE LETTERS OF “ETHICS”. IS THERE ANY LESSON FOR US IN ABOVE SAYING? FOR “math-lovers”? FOR “math-giants”? FOR “overflow-mathematicians”? ........ ...... _( BTW this saying belongs to me)

Question Number 111954 Answers: 1 Comments: 0

Proof for the following : n! ≥ 2×3^(n−2)

$${P}\mathrm{roof}\:\mathrm{for}\:\mathrm{the}\:\mathrm{following}\:: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:{n}!\:\geqslant\:\mathrm{2}×\mathrm{3}^{{n}−\mathrm{2}} \\ $$$$ \\ $$

Question Number 111953 Answers: 0 Comments: 0

Question Number 112004 Answers: 0 Comments: 0

Suppose x,y,z ∈N, (yz+x) is prime (yz+x)∣(zx+y), (yz+x)∣(xy+z). Find all possible values of (((xy+z)(zx+y))/((yz+x)^2 )).

$$\mathrm{Suppose}\:\mathrm{x},\mathrm{y},\mathrm{z}\:\in\mathbb{N},\:\left(\mathrm{yz}+\mathrm{x}\right)\:\mathrm{is}\:\mathrm{prime} \\ $$$$\left(\mathrm{yz}+\mathrm{x}\right)\mid\left(\mathrm{zx}+\mathrm{y}\right),\:\left(\mathrm{yz}+\mathrm{x}\right)\mid\left(\mathrm{xy}+\mathrm{z}\right). \\ $$$$\mathrm{Find}\:\mathrm{all}\:\mathrm{possible}\:\mathrm{values}\:\mathrm{of}\: \\ $$$$\frac{\left(\mathrm{xy}+\mathrm{z}\right)\left(\mathrm{zx}+\mathrm{y}\right)}{\left(\mathrm{yz}+\mathrm{x}\right)^{\mathrm{2}} }. \\ $$

Question Number 112545 Answers: 2 Comments: 2

please solve : I=∫_0 ^( 1) xlog^2 (((1−x)/(1+x)))dx =??? ...m.n.july 1970.... good luck .

$$\:\:\:\:{please}\:{solve}\:: \\ $$$$ \\ $$$$\:\:\:\:\:\mathrm{I}=\int_{\mathrm{0}} ^{\:\mathrm{1}} {xlog}^{\mathrm{2}} \left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right){dx}\:=??? \\ $$$$ \\ $$$$\:\:\:\:\:\:...{m}.{n}.{july}\:\mathrm{1970}.... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{good}\:\:\:{luck}\:. \\ $$$$ \\ $$

Question Number 111939 Answers: 2 Comments: 0

Question Number 111937 Answers: 1 Comments: 0

Question Number 112203 Answers: 0 Comments: 2

proporsed by m.njuly 1970 ∫_0 ^1 ((ln(((x+1))^(1/3) ))/((x+1)(x+2)))dx solution let I=(1/3)∫_0 ^1 ((ln(x+1))/((x+1)(x+2)))dx I=(1/3)∫_0 ^1 ((ln(1+x))/(x+1))dx−∫_0 ^1 ((ln(x+1))/(x+2))dx=A−B let A=(1/3)∫_0 ^1 ((ln(x+1))/(x+1))dx ( using IBP) A=(1/3)[ln^2 (x+1)]_0 ^1 −(1/3)∫_0 ^1 ((ln(x+1))/(x+1))dx A(1+(1/3))=(1/3)ln^2 (2) A=∫_0 ^1 ((ln(x+1))/(x+1))dx=(1/3)ln^2 (2)......(1) then B=(1/3)∫_0 ^1 ((ln(x+1))/(x+2))dx B=(1/3)[ln(x+2)ln(x+1)]_0 ^1 −(1/3)∫_0 ^1 ((ln(x+2))/(x+1))dx let x+1=t B=(1/3)ln3ln 2−(1/3)∫_1 ^2 ((ln(1+t))/t)dt let t=−x B=(1/3)ln3ln2−(1/3)∫_(−1) ^(−2) ((ln(1−x))/x)dx but ∫((ln(1−x))/x)dx=−Li_2 (x) B=(1/3)ln3ln2+(1/3)[Li_2 (x)]_(−1) ^(−2) B=(1/3)ln3ln2+(1/3)Li_2 (−2)−(1/3)Li_2 (−1) B=(1/3)ln3ln2+(1/3)Li_2 (−2)+(π^2 /(36)) but I=A−B ∫_0 ^1 ((ln(((1+x))^(1/3) ))/((x+1)(x+2)))dx=(1/4)ln^2 (2)−(1/3)ln3ln2−(1/3)Li_2 (−2)−(π^2 /(36)) mathdave(06/09/2020)

Question Number 111934 Answers: 1 Comments: 0

Question Number 112533 Answers: 2 Comments: 2

Find the minimum number of n integers to be selected from S={1,2,3,...11} so that the difference of two of the n integers is 7.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{number}\:\mathrm{of}\:\mathrm{n} \\ $$$$\mathrm{integers}\:\mathrm{to}\:\mathrm{be}\:\mathrm{selected}\:\mathrm{from} \\ $$$$\mathrm{S}=\left\{\mathrm{1},\mathrm{2},\mathrm{3},...\mathrm{11}\right\}\:\mathrm{so}\:\mathrm{that}\:\mathrm{the}\:\mathrm{difference} \\ $$$$\mathrm{of}\:\mathrm{two}\:\mathrm{of}\:\mathrm{the}\:\mathrm{n}\:\mathrm{integers}\:\mathrm{is}\:\mathrm{7}. \\ $$

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