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Question Number 219262 Answers: 0 Comments: 0

E^ lectric field strenth at any point in the space is defined as the force per unit charge at that point. It is a vector quantity whose magnitude is given by Coulomb^(s ) law and diection is in straight line loining the at that point. mathemstically

$$\overset{} {\mathrm{E}lectric}\:\mathrm{field}\:\mathrm{strenth}\:\mathrm{at}\:\mathrm{any}\:\mathrm{point}\:\mathrm{in}\:\mathrm{the}\:\mathrm{space} \\ $$$$\mathrm{is}\:\mathrm{defined}\:\mathrm{as}\:\mathrm{the}\:\mathrm{force}\:\mathrm{per}\:\mathrm{unit}\:\mathrm{charge}\:\mathrm{at}\:\mathrm{that}\:\mathrm{point}. \\ $$$$\:\mathrm{It}\:\mathrm{is}\:\mathrm{a}\:\mathrm{vector}\:\mathrm{quantity}\:\mathrm{whose}\:\mathrm{magnitude}\:\mathrm{is} \\ $$$$\mathrm{given}\:\mathrm{by}\:\mathrm{Coulomb}^{\mathrm{s}\:\:} \:\mathrm{law}\:\mathrm{and}\:\mathrm{diection}\:\mathrm{is}\:\mathrm{in}\: \\ $$$$\mathrm{straight}\:\mathrm{line}\:\mathrm{loining}\:\mathrm{the}\:\mathrm{at}\:\mathrm{that}\:\mathrm{point}. \\ $$$$\mathrm{mathemstically} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 219255 Answers: 6 Comments: 0

Question Number 219254 Answers: 4 Comments: 0

Question Number 219243 Answers: 3 Comments: 2

Question Number 219236 Answers: 1 Comments: 0

f(t) = (1/(2πi)) ∫_( c−i∞) ^( c+i∞) (e^(st) /(s^k )) ds , k ∈C

$$ \\ $$$$\:\:\:\:{f}\left({t}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}\pi{i}}\:\int_{\:{c}−{i}\infty} ^{\:{c}+{i}\infty} \:\frac{{e}^{{st}} }{{s}^{{k}} \:}\:\:{ds}\:\:\:,\:\:{k}\:\in\mathbb{C} \\ $$$$\: \\ $$

Question Number 219234 Answers: 0 Comments: 0

f(t)=∫_(0 ) ^( t) ((ζ(1/2 + iτ))/( (√(t − τ + 1)))) dτ

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:{f}\left({t}\right)=\int_{\mathrm{0}\:} ^{\:{t}} \:\frac{\zeta\left(\mathrm{1}/\mathrm{2}\:\:+\:\:{i}\tau\right)}{\:\sqrt{{t}\:−\:\tau\:\:+\:\mathrm{1}}}\:{d}\tau \\ $$$$ \\ $$

Question Number 219233 Answers: 3 Comments: 0

Question Number 219232 Answers: 4 Comments: 0

Question Number 219211 Answers: 2 Comments: 0

Question Number 219223 Answers: 2 Comments: 0

Question Number 219193 Answers: 2 Comments: 0

∫(√((x+1)/(x+2))) .(1/(x+3)) dx=?

$$\int\sqrt{\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}}}\:.\frac{\mathrm{1}}{{x}+\mathrm{3}}\:{dx}=? \\ $$

Question Number 219190 Answers: 2 Comments: 1

Question Number 219185 Answers: 5 Comments: 0

Question Number 219182 Answers: 1 Comments: 1

Question Number 219181 Answers: 3 Comments: 0

Question Number 219173 Answers: 4 Comments: 0

Question Number 219120 Answers: 2 Comments: 0

Question Number 219119 Answers: 4 Comments: 0

Question Number 219118 Answers: 5 Comments: 0

Question Number 219117 Answers: 6 Comments: 0

Question Number 219116 Answers: 5 Comments: 0

Question Number 219113 Answers: 3 Comments: 1

((a + 3b)/(a + b−1)) + ((a + 3b−1)/(a + b−3)) = 4 ⇒ a + b = ?

$$\frac{\mathrm{a}\:+\:\mathrm{3b}}{\mathrm{a}\:+\:\mathrm{b}−\mathrm{1}}\:+\:\frac{\mathrm{a}\:+\:\mathrm{3b}−\mathrm{1}}{\mathrm{a}\:+\:\mathrm{b}−\mathrm{3}}\:=\:\mathrm{4}\:\:\Rightarrow\:\:\mathrm{a}\:+\:\mathrm{b}\:=\:? \\ $$

Question Number 219112 Answers: 0 Comments: 0

Prove it: In triangle ABC, AB=c, BC=b, AC=a ab^2 c + abc^2 −a^2 bc ≥ tan (A/2) ((2S^3 )/((a+b)^2 −c^2 ))

$$\mathrm{Prove}\:\mathrm{it}: \\ $$$$\mathrm{In}\:\mathrm{triangle}\:\mathrm{ABC},\:\mathrm{AB}=\mathrm{c},\:\mathrm{BC}=\mathrm{b},\:\mathrm{AC}=\mathrm{a} \\ $$$$\mathrm{ab}^{\mathrm{2}} \mathrm{c}\:+\:\mathrm{abc}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} \mathrm{bc}\:\geqslant\:\mathrm{tan}\:\frac{\mathrm{A}}{\mathrm{2}}\:\frac{\mathrm{2S}^{\mathrm{3}} }{\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{2}} −\mathrm{c}^{\mathrm{2}} } \\ $$

Question Number 219110 Answers: 1 Comments: 0

find the nth term of x_(n+1) = x_n (2−x_n ) in x_1

$${find}\:{the}\:{nth}\:{term}\:{of}\:{x}_{{n}+\mathrm{1}} \:=\:{x}_{{n}} \left(\mathrm{2}−{x}_{{n}} \right) \\ $$$${in}\:{x}_{\mathrm{1}} \\ $$

Question Number 219098 Answers: 2 Comments: 0

ζ(α)=Σ_(n=1) ^(+∞) (1/n^α )

$$\zeta\left(\alpha\right)=\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\alpha} }\:\: \\ $$

Question Number 219093 Answers: 3 Comments: 1

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