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Question Number 220694 Answers: 1 Comments: 1

Question Number 220693 Answers: 3 Comments: 2

Question Number 220624 Answers: 0 Comments: 0

Question Number 220607 Answers: 1 Comments: 1

symplify (((√3)+(√(5+))(√9)+(√(15)))/( (√1)+(√5)+(√(12))))

$${symplify} \\ $$$$\frac{\sqrt{\mathrm{3}}+\sqrt{\mathrm{5}+}\sqrt{\mathrm{9}}+\sqrt{\mathrm{15}}}{\:\sqrt{\mathrm{1}}+\sqrt{\mathrm{5}}+\sqrt{\mathrm{12}}} \\ $$

Question Number 220606 Answers: 1 Comments: 1

Question Number 220577 Answers: 1 Comments: 0

Calculate the perimeter of a rectangle whose area is represented by the polynomial 25x^2 −35x+12(Given that the length and breath are not constant)

$${Calculate}\:{the}\:{perimeter}\:{of}\:{a}\:{rectangle} \\ $$$${whose}\:{area}\:{is}\:{represented}\:{by}\:{the}\:{polynomial} \\ $$$$\mathrm{25}{x}^{\mathrm{2}} −\mathrm{35}{x}+\mathrm{12}\left({Given}\:{that}\:{the}\:{length}\:{and}\:{breath}\:{are}\:{not}\:{constant}\right) \\ $$

Question Number 220579 Answers: 1 Comments: 0

A=7×19×31×43×.....upto 29 terms find the last four digits of A.

$$\:{A}=\mathrm{7}×\mathrm{19}×\mathrm{31}×\mathrm{43}×.....{upto}\:\mathrm{29}\:{terms} \\ $$$$\:{find}\:{the}\:{last}\:{four}\:{digits}\:{of}\:{A}. \\ $$

Question Number 220540 Answers: 2 Comments: 0

Find: 𝛀 = Σ_(n=1) ^∞ (((−1)^(n+1) )/(n^3 ∙(n + 1)^3 ∙(2n + 1)^2 )) = ?

$$\mathrm{Find}:\:\:\:\boldsymbol{\Omega}\:=\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}+\mathrm{1}} }{\mathrm{n}^{\mathrm{3}} \centerdot\left(\mathrm{n}\:+\:\mathrm{1}\right)^{\mathrm{3}} \centerdot\left(\mathrm{2n}\:+\:\mathrm{1}\right)^{\mathrm{2}} }\:=\:? \\ $$

Question Number 220519 Answers: 0 Comments: 1

Solve in R ((15)/(x^2 - 3x + 4)) + (7/(x^2 + 7x)) + ((10)/(x^2 + 4x - 21)) + 1 = 0

$$\mathrm{Solve}\:\mathrm{in}\:\:\:\mathbb{R} \\ $$$$\frac{\mathrm{15}}{\mathrm{x}^{\mathrm{2}} \:-\:\mathrm{3x}\:+\:\mathrm{4}}\:\:+\:\:\frac{\mathrm{7}}{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{7x}}\:\:+\:\:\frac{\mathrm{10}}{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{4x}\:-\:\mathrm{21}}\:\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$

Question Number 220486 Answers: 1 Comments: 2

z ∈ C and λ > 0 Then prove that: ∣z + 2λ∣ + ∣z + λ∣ ≥ ∣z + ((3λ − λ(√3)i)/2)∣

$$\mathrm{z}\:\in\:\mathbb{C}\:\:\:\mathrm{and}\:\:\:\lambda\:>\:\mathrm{0} \\ $$$$\mathrm{Then}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\mid\mathrm{z}\:+\:\mathrm{2}\lambda\mid\:+\:\mid\mathrm{z}\:+\:\lambda\mid\:\geqslant\:\mid\mathrm{z}\:+\:\frac{\mathrm{3}\lambda\:−\:\lambda\sqrt{\mathrm{3}}\mathrm{i}}{\mathrm{2}}\mid \\ $$

Question Number 220474 Answers: 0 Comments: 1

Question Number 220473 Answers: 0 Comments: 0

lim_(n→∞) (tan((π/4)+(1/n)))^n →^(t=(1/n)) =lim_(t→0) [tan((π/4)+t)]^(1/t) ⇒lim_(t→0) [tan((π/4)+t)−1]×(1/t) =lim_(t→0) (((1+tant)/(1−tant))−1)×(1/t) =lim_(t→0) (((2tant)/(1−tant)))×(1/t)=2 ⇒Ans=e^2 ✓

$${lim}_{{n}\rightarrow\infty} \left({tan}\left(\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{{n}}\right)\right)^{{n}} \:\:\:\:\:\overset{{t}=\frac{\mathrm{1}}{{n}}} {\rightarrow} \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}} \left[{tan}\left(\frac{\pi}{\mathrm{4}}+{t}\right)\right]^{\frac{\mathrm{1}}{{t}}} \\ $$$$\Rightarrow{lim}_{{t}\rightarrow\mathrm{0}} \left[{tan}\left(\frac{\pi}{\mathrm{4}}+{t}\right)−\mathrm{1}\right]×\frac{\mathrm{1}}{{t}} \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}} \left(\frac{\mathrm{1}+{tant}}{\mathrm{1}−{tant}}−\mathrm{1}\right)×\frac{\mathrm{1}}{{t}} \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}} \left(\frac{\mathrm{2}{tant}}{\mathrm{1}−{tant}}\right)×\frac{\mathrm{1}}{{t}}=\mathrm{2} \\ $$$$\Rightarrow{Ans}={e}^{\mathrm{2}} \:\:\checkmark \\ $$$$ \\ $$

Question Number 220468 Answers: 3 Comments: 0

a+b+c=1, a^2 +b^2 +c^2 =1 (a,b,c ∈R) a^(10) +b^(10) +c^(10) =1, a^4 +b^4 +c^4 =?

$$\:{a}+{b}+{c}=\mathrm{1},\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{1}\:\:\left({a},{b},{c}\:\in{R}\right) \\ $$$$\:{a}^{\mathrm{10}} +{b}^{\mathrm{10}} +{c}^{\mathrm{10}} =\mathrm{1},\:{a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} =? \\ $$

Question Number 220459 Answers: 3 Comments: 0

find Σ_(n=1) ^∞ (1/(n^2 +a^2 ))=? (a∈R)

$${find}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{a}^{\mathrm{2}} }=?\:\:\:\:\:\:\:\:\:\:\left({a}\in{R}\right) \\ $$

Question Number 220457 Answers: 0 Comments: 0

Find: 𝛀 = ∫_0 ^( ∞) ∫_0 ^( (𝛑/2)) (((x+1)sin^2 (x)ln(y^3 +1))/(xy(y^2 +1))) dxdy = ?

$$\mathrm{Find}:\:\:\:\boldsymbol{\Omega}\:=\:\int_{\mathrm{0}} ^{\:\infty} \int_{\mathrm{0}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\left(\mathrm{x}+\mathrm{1}\right)\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)\mathrm{ln}\left(\mathrm{y}^{\mathrm{3}} +\mathrm{1}\right)}{\mathrm{xy}\left(\mathrm{y}^{\mathrm{2}} +\mathrm{1}\right)}\:\mathrm{dxdy}\:=\:? \\ $$

Question Number 220456 Answers: 0 Comments: 0

a, b, c, d ≥ 1 ; a + b + c = d show that; ab + bc + ca + (1/a) + (1/b) + (1/c) ≥ 2d − 3 + (9/d)

$$\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:{a},\:{b},\:{c},\:{d}\:\geqslant\:\mathrm{1}\:\:\:\:\:\:;\:\:\:{a}\:+\:{b}\:+\:{c}\:=\:{d} \\ $$$$\:\:\:\:\:\:\:\:{show}\:{that};\: \\ $$$$\:\:\:{ab}\:+\:{bc}\:+\:{ca}\:+\:\frac{\mathrm{1}}{{a}}\:+\:\frac{\mathrm{1}}{{b}}\:+\:\frac{\mathrm{1}}{{c}}\:\geqslant\:\mathrm{2}{d}\:−\:\mathrm{3}\:+\:\frac{\mathrm{9}}{{d}}\:\:\:\: \\ $$

Question Number 220405 Answers: 2 Comments: 0

9^x^2 −3^(x+1) =0

$$\mathrm{9}^{{x}^{\mathrm{2}} } −\mathrm{3}^{{x}+\mathrm{1}} =\mathrm{0} \\ $$

Question Number 220395 Answers: 2 Comments: 0

Find: Ω =Σ_(x=1) ^∞ Σ_(y=1) ^∞ (1/(x^2 y^3 (x^2 + 1)(y + 2))) = ?

$$\mathrm{Find}:\:\:\:\Omega\:=\underset{\boldsymbol{\mathrm{x}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\underset{\boldsymbol{\mathrm{y}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} \:\mathrm{y}^{\mathrm{3}} \:\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\left(\mathrm{y}\:+\:\mathrm{2}\right)}\:=\:? \\ $$

Question Number 220391 Answers: 3 Comments: 0

Question Number 220266 Answers: 1 Comments: 0

2^a + 2^b + 2^c = 148

$$\mathrm{2}^{\mathrm{a}} \:\:+\:\:\mathrm{2}^{\mathrm{b}} \:\:+\:\:\mathrm{2}^{\mathrm{c}} \:\:=\:\:\mathrm{148} \\ $$

Question Number 220247 Answers: 2 Comments: 0

Question Number 220221 Answers: 0 Comments: 2

Q.The density of an object of mass M is δ and the density of the air is ρ. the mass of of the object is measured with the help of a metal weight of mass m . the density of the metal weight is d. if ρ≪δ them show that the real mass M will be m(1−(ρ/d) )(1+(ρ/δ)) I have managed to M=((m(1−(ρ/d)))/((1−(ρ/δ)))) but I can not figure it to the end please help

$${Q}.{The}\:{density}\:{of}\:{an}\:{object}\:{of}\:{mass}\:{M}\:{is}\:\delta\:{and}\:{the}\:{density}\:{of}\:{the}\:{air}\:{is}\:\rho. \\ $$$${the}\:{mass}\:{of}\:{of}\:{the}\:{object}\:{is}\:{measured}\:{with}\:\:{the}\:{help}\:{of}\:{a}\:{metal}\:{weight}\:{of}\:{mass}\:{m}\:. \\ $$$${the}\:{density}\:{of}\:{the}\:{metal}\:{weight}\:{is}\:{d}. \\ $$$${if}\:\rho\ll\delta\:{them}\:{show}\:{that}\:{the}\:{real}\:{mass}\:{M}\:{will}\:{be} \\ $$$${m}\left(\mathrm{1}−\frac{\rho}{{d}}\:\right)\left(\mathrm{1}+\frac{\rho}{\delta}\right) \\ $$$${I}\:{have}\:{managed}\:{to}\:{M}=\frac{{m}\left(\mathrm{1}−\frac{\rho}{{d}}\right)}{\left(\mathrm{1}−\frac{\rho}{\delta}\right)} \\ $$$${but}\:{I}\:{can}\:{not}\:{figure}\:{it}\:{to}\:{the}\:{end} \\ $$$${please}\:{help} \\ $$

Question Number 220193 Answers: 3 Comments: 0

x^8 =21x+13 ; x∈R x=?

$$\:\:\:\:\:\:\:\:\boldsymbol{{x}}^{\mathrm{8}} =\mathrm{21}\boldsymbol{{x}}+\mathrm{13}\:\:\:\:\:\:\:\:\:\:;\:\:\:\:\boldsymbol{{x}}\in{R} \\ $$$$\:\:\:\:\boldsymbol{{x}}=? \\ $$

Question Number 220164 Answers: 2 Comments: 0

if α^2 −5α+2=0 & β^2 −5β+2=0 then ((4α+β^5 )/(5β^2 ))=?

$${if}\:\:\alpha^{\mathrm{2}} −\mathrm{5}\alpha+\mathrm{2}=\mathrm{0}\:\:\&\:\:\beta^{\mathrm{2}} −\mathrm{5}\beta+\mathrm{2}=\mathrm{0} \\ $$$${then}\:\:\frac{\mathrm{4}\alpha+\beta^{\mathrm{5}} }{\mathrm{5}\beta^{\mathrm{2}} }=? \\ $$

Question Number 220101 Answers: 0 Comments: 0

Question Number 220097 Answers: 1 Comments: 0

x ∈ Q ; x ≠ 1 (7/(x − 1)) + (6/x) − (4/(x + 1)) + ((3x + 5)/(x^2 − 1)) = (1/x)

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\:\in\:\mathbb{Q}\:\:\:;\:\:\:\:{x}\:\neq\:\mathrm{1} \\ $$$$\:\frac{\mathrm{7}}{{x}\:−\:\mathrm{1}}\:+\:\frac{\mathrm{6}}{{x}}\:−\:\frac{\mathrm{4}}{{x}\:+\:\mathrm{1}}\:+\:\frac{\mathrm{3}{x}\:+\:\mathrm{5}}{{x}^{\mathrm{2}} \:−\:\mathrm{1}}\:=\:\frac{\mathrm{1}}{{x}} \\ $$$$\:\:\:\:\:\:\: \\ $$

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