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Question Number 171152 Answers: 1 Comments: 0

The average ages of three person is 27 years. Their ages are in the propor tion of 1:3:5. What is the age in years of the youngest one among them? Mastermind

$${The}\:{average}\:{ages}\:{of}\:{three}\:{person}\:{is} \\ $$$$\mathrm{27}\:{years}.\:{Their}\:{ages}\:{are}\:{in}\:{the}\:{propor} \\ $$$${tion}\:{of}\:\mathrm{1}:\mathrm{3}:\mathrm{5}.\:{What}\:{is}\:{the}\:{age}\:{in}\:{years} \\ $$$${of}\:{the}\:{youngest}\:{one}\:{among}\:{them}? \\ $$$$ \\ $$$${Mastermind} \\ $$

Question Number 171189 Answers: 2 Comments: 0

f(x)=((16^x )/(4+16^x )) faind volue of f((1/(20)))+f((2/(20)))+........+f(((19)/(20)))

$${f}\left({x}\right)=\frac{\mathrm{16}^{{x}} }{\mathrm{4}+\mathrm{16}^{{x}} }\:\:\:\:\:{faind}\:{volue}\:{of} \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{20}}\right)+{f}\left(\frac{\mathrm{2}}{\mathrm{20}}\right)+........+{f}\left(\frac{\mathrm{19}}{\mathrm{20}}\right) \\ $$

Question Number 171147 Answers: 0 Comments: 2

Question Number 171146 Answers: 0 Comments: 2

Question Number 171145 Answers: 0 Comments: 0

let p be a non constant polynomial with degree n such that: p(z)=a_n z^n +...+a_0 show that if z∈C with∣z∣≥max{1,(2/(∣a_n ∣))Σ_(i=1) ^(n−1) ∣a_i ∣} then (1/2)∣a_n ∣∣z∣^n ≤∣p(z)∣≤(3/2)∣a_n ∣∣z∣^n (i already did the right side of the inequality so try to show ∣p(z)∣≥(1/2)∣a_n ∣∣z∣^n )

$$\mathrm{let}\:\mathrm{p}\:\mathrm{be}\:\mathrm{a}\:\mathrm{non}\:\mathrm{constant}\:\mathrm{polynomial}\:\mathrm{with}\: \\ $$$$\mathrm{degree}\:\mathrm{n}\:\mathrm{such}\:\mathrm{that}:\:\mathrm{p}\left(\mathrm{z}\right)=\mathrm{a}_{\mathrm{n}} \mathrm{z}^{\mathrm{n}} +...+\mathrm{a}_{\mathrm{0}} \\ $$$$\mathrm{show}\:\mathrm{that}\:\mathrm{if}\:\mathrm{z}\in\mathbb{C}\:\mathrm{with}\mid\mathrm{z}\mid\geqslant\mathrm{max}\left\{\mathrm{1},\frac{\mathrm{2}}{\mid\mathrm{a}_{\mathrm{n}} \mid}\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\mid\mathrm{a}_{\mathrm{i}} \mid\right\} \\ $$$$\mathrm{then}\:\frac{\mathrm{1}}{\mathrm{2}}\mid\mathrm{a}_{\mathrm{n}} \mid\mid\mathrm{z}\mid^{\mathrm{n}} \leqslant\mid\mathrm{p}\left(\mathrm{z}\right)\mid\leqslant\frac{\mathrm{3}}{\mathrm{2}}\mid\mathrm{a}_{\mathrm{n}} \mid\mid\mathrm{z}\mid^{\mathrm{n}} \\ $$$$\left(\mathrm{i}\:\mathrm{already}\:\mathrm{did}\:\mathrm{the}\:\mathrm{right}\:\mathrm{side}\:\mathrm{of}\:\mathrm{the}\:\mathrm{inequality}\right. \\ $$$$\left.\mathrm{so}\:\mathrm{try}\:\mathrm{to}\:\mathrm{show}\:\mid\mathrm{p}\left(\mathrm{z}\right)\mid\geqslant\frac{\mathrm{1}}{\mathrm{2}}\mid\mathrm{a}_{\mathrm{n}} \mid\mid\mathrm{z}\mid^{\mathrm{n}} \right) \\ $$

Question Number 171136 Answers: 0 Comments: 4

Using Taylor′s theorem, prove that x−(x^3 /6)<sin x<x−(x^3 /6)+(x^5 /(120)) for x>0

$$\mathrm{Using}\:\mathrm{Taylor}'\mathrm{s}\:\mathrm{theorem},\:\mathrm{prove}\:\mathrm{that} \\ $$$${x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}<\mathrm{sin}\:{x}<{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}+\frac{{x}^{\mathrm{5}} }{\mathrm{120}}\:\:\:\mathrm{for}\:{x}>\mathrm{0} \\ $$

Question Number 171134 Answers: 0 Comments: 0