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GeometryQuestion and Answers: Page 105

Question Number 25139 Answers: 1 Comments: 0

What is the real and the imaginary part of the complex number z = (− 1)^(1000003)

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{real}\:\mathrm{and}\:\mathrm{the}\:\mathrm{imaginary}\:\mathrm{part}\:\mathrm{of}\:\mathrm{the}\:\mathrm{complex}\:\mathrm{number}\:\:\:\mathrm{z}\:=\:\left(−\:\mathrm{1}\right)^{\mathrm{1000003}} \\ $$

Question Number 24831 Answers: 1 Comments: 0

∫_1 ^2 ∫_1 ^2 ln(x+y)dx dy

$$ \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} \int_{\mathrm{1}} ^{\mathrm{2}} {ln}\left({x}+{y}\right){dx}\:{dy} \\ $$

Question Number 24778 Answers: 0 Comments: 4

Show that the shortest distance between two opposite edges a,d of a tetrahedron is 6V/adsin 𝛉, where θ is the angle between the edges and V is the volume of the tetrahedron.

$${Show}\:{that}\:{the}\:{shortest}\:{distance} \\ $$$${between}\:{two}\:{opposite}\:{edges}\:\boldsymbol{{a}},\boldsymbol{{d}}\: \\ $$$${of}\:{a}\:{tetrahedron}\:{is}\:\mathrm{6}\boldsymbol{{V}}/\boldsymbol{{ad}}\mathrm{sin}\:\boldsymbol{\theta}, \\ $$$${where}\:\theta\:{is}\:{the}\:{angle}\:{between}\:{the} \\ $$$${edges}\:{and}\:{V}\:{is}\:{the}\:{volume}\:{of}\:{the} \\ $$$${tetrahedron}. \\ $$

Question Number 24684 Answers: 0 Comments: 0

Let ABCD be a square and M, N points on sides AB, BC respectably, such that ∠MDN = 45°. If R is the midpoint of MN show that RP = RQ where P, Q are the points of intersection of AC with the lines MD, ND.

$$\mathrm{Let}\:{ABCD}\:\mathrm{be}\:\mathrm{a}\:\mathrm{square}\:\mathrm{and}\:{M},\:{N}\:\mathrm{points} \\ $$$$\mathrm{on}\:\mathrm{sides}\:{AB},\:{BC}\:\mathrm{respectably},\:\mathrm{such}\:\mathrm{that} \\ $$$$\angle{MDN}\:=\:\mathrm{45}°.\:\mathrm{If}\:{R}\:\mathrm{is}\:\mathrm{the}\:\mathrm{midpoint}\:\mathrm{of} \\ $$$${MN}\:\mathrm{show}\:\mathrm{that}\:{RP}\:=\:{RQ}\:\mathrm{where}\:{P},\:{Q} \\ $$$$\mathrm{are}\:\mathrm{the}\:\mathrm{points}\:\mathrm{of}\:\mathrm{intersection}\:\mathrm{of}\:{AC}\:\mathrm{with} \\ $$$$\mathrm{the}\:\mathrm{lines}\:{MD},\:{ND}. \\ $$

Question Number 24604 Answers: 2 Comments: 0

x^2 −xsin x−cos x=0

$${x}^{\mathrm{2}} −{x}\mathrm{sin}\:{x}−\mathrm{cos}\:{x}=\mathrm{0} \\ $$

Question Number 24303 Answers: 1 Comments: 0

Assertion: Enthalpy of combustion is negative. Reason: Combustion reaction can be exothermic or endothermic.

$$\boldsymbol{\mathrm{Assertion}}:\:\mathrm{Enthalpy}\:\mathrm{of}\:\mathrm{combustion}\:\mathrm{is} \\ $$$$\mathrm{negative}. \\ $$$$\boldsymbol{\mathrm{Reason}}:\:\mathrm{Combustion}\:\mathrm{reaction}\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{exothermic}\:\mathrm{or}\:\mathrm{endothermic}. \\ $$

Question Number 24150 Answers: 0 Comments: 0

if y is a function of t then solve this y′′=ksiny diff.equ

$${if}\:{y}\:{is}\:{a}\:{function}\:{of}\:{t}\:{then}\:{solve}\:{this}\:{y}''={ksiny}\:{diff}.{equ} \\ $$

Question Number 24055 Answers: 0 Comments: 2

Any Architect in the house? please i need your help

$${Any}\:{Architect}\:{in}\:{the}\:{house}? \\ $$$$ \\ $$$${please}\:{i}\:{need}\:{your}\:{help} \\ $$

Question Number 23871 Answers: 0 Comments: 0

Let ABC be a triangle and B′ be the reflection of B in the line CA and C′ be reflection of C in the line AB. Prove that ΔABC′ ≅ ΔACB′ ≅ ΔABC.

$$\mathrm{Let}\:{ABC}\:\mathrm{be}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{and}\:{B}'\:\mathrm{be}\:\mathrm{the} \\ $$$$\mathrm{reflection}\:\mathrm{of}\:{B}\:\mathrm{in}\:\mathrm{the}\:\mathrm{line}\:{CA}\:\mathrm{and}\:{C}'\:\mathrm{be} \\ $$$$\mathrm{reflection}\:\mathrm{of}\:{C}\:\mathrm{in}\:\mathrm{the}\:\mathrm{line}\:{AB}.\:\mathrm{Prove} \\ $$$$\mathrm{that}\:\Delta{ABC}'\:\cong\:\Delta{ACB}'\:\cong\:\Delta{ABC}. \\ $$

Question Number 23856 Answers: 0 Comments: 4

The value of (C_0 + C_1 )(C_1 + C_2 ).... (C_(n−1) + C_n ) is (1) (((n + 1)^n )/(n!)) ∙ C_1 C_2 .....C_n (2) (((n − 1)^n )/(n!)) ∙ C_1 C_2 .....C_n (3) (((n)^n )/((n + 1)!)) ∙ C_1 C_2 .....C_n (4) (((n)^n )/(n!)) ∙ C_1 C_2 .....C_n

$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\left({C}_{\mathrm{0}} \:+\:{C}_{\mathrm{1}} \right)\left({C}_{\mathrm{1}} \:+\:{C}_{\mathrm{2}} \right).... \\ $$$$\left({C}_{{n}−\mathrm{1}} \:+\:{C}_{{n}} \right)\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\frac{\left({n}\:+\:\mathrm{1}\right)^{{n}} }{{n}!}\:\centerdot\:{C}_{\mathrm{1}} {C}_{\mathrm{2}} .....{C}_{{n}} \\ $$$$\left(\mathrm{2}\right)\:\frac{\left({n}\:−\:\mathrm{1}\right)^{{n}} }{{n}!}\:\centerdot\:{C}_{\mathrm{1}} {C}_{\mathrm{2}} .....{C}_{{n}} \\ $$$$\left(\mathrm{3}\right)\:\frac{\left({n}\right)^{{n}} }{\left({n}\:+\:\mathrm{1}\right)!}\:\centerdot\:{C}_{\mathrm{1}} {C}_{\mathrm{2}} .....{C}_{{n}} \\ $$$$\left(\mathrm{4}\right)\:\frac{\left({n}\right)^{{n}} }{{n}!}\:\centerdot\:{C}_{\mathrm{1}} {C}_{\mathrm{2}} .....{C}_{{n}} \\ $$

Question Number 23769 Answers: 1 Comments: 9

guys , how was kvpy ( SA)?? : tinkutara , physicslover,etc....... i screwd in bio completely. how much you guys are expecting and do you have any idea of cutoff ?

$$\mathrm{guys}\:,\:\mathrm{how}\:\mathrm{was}\:\mathrm{kvpy}\:\left(\:\mathrm{SA}\right)?? \\ $$$$:\:\mathrm{tinkutara}\:,\:\mathrm{physicslover},\mathrm{etc}....... \\ $$$$\mathrm{i}\:\mathrm{screwd}\:\mathrm{in}\:\mathrm{bio}\:\mathrm{completely}. \\ $$$$\mathrm{how}\:\mathrm{much}\:\mathrm{you}\:\mathrm{guys}\:\mathrm{are}\:\mathrm{expecting} \\ $$$$\mathrm{and}\:\mathrm{do}\:\mathrm{you}\:\mathrm{have}\:\mathrm{any}\:\mathrm{idea}\:\mathrm{of}\: \\ $$$$\mathrm{cutoff}\:? \\ $$

Question Number 23758 Answers: 1 Comments: 0

solve ∫tan^(−1) x ln (1+x^2 )dx

$${solve} \\ $$$$\int\mathrm{tan}^{−\mathrm{1}} {x}\:\mathrm{ln}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right){dx} \\ $$

Question Number 23752 Answers: 1 Comments: 0

∫_1 ^2 x^3 +1=?

$$\int_{\mathrm{1}} ^{\mathrm{2}} {x}^{\mathrm{3}} +\mathrm{1}=? \\ $$

Question Number 23679 Answers: 2 Comments: 0

Question Number 23677 Answers: 0 Comments: 0

solve lim_(x→inf+) ∫^(2(√x)) _(2sin(1/x)) ((2t^4 +1)/((t−3)(t^3 +3))) dt

$${solve} \\ $$$$\:\:\:\:\:\:\:\:\:\: \\ $$$$\underset{{x}\rightarrow{inf}+} {\mathrm{li}{m}}\:\:\underset{\mathrm{2}{sin}\frac{\mathrm{1}}{{x}}} {\int}^{\mathrm{2}\sqrt{{x}}} \frac{\mathrm{2}{t}^{\mathrm{4}} +\mathrm{1}}{\left({t}−\mathrm{3}\right)\left({t}^{\mathrm{3}} +\mathrm{3}\right)}\:{dt} \\ $$

Question Number 23663 Answers: 1 Comments: 3

solve ∫^1_ _(−1) x^2 d(lnx)

$${solve} \\ $$$$ \\ $$$$\underset{−\mathrm{1}} {\int}^{\mathrm{1}_{} } {x}^{\mathrm{2}} {d}\left({lnx}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Question Number 23592 Answers: 1 Comments: 0

Let ABC be a triangle with AB = AC and ∠BAC = 30°. Let A′ be the reflection of A in the line BC; B′ be the reflection of B in the line CA; C′ be the reflection of C in the line AB. Show that A′, B′, C′ form the vertices of an equilateral triangle.

$$\mathrm{Let}\:{ABC}\:\mathrm{be}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{with}\:{AB}\:=\:{AC} \\ $$$$\mathrm{and}\:\angle{BAC}\:=\:\mathrm{30}°.\:\mathrm{Let}\:{A}'\:\mathrm{be}\:\mathrm{the}\:\mathrm{reflection} \\ $$$$\mathrm{of}\:{A}\:\mathrm{in}\:\mathrm{the}\:\mathrm{line}\:{BC};\:{B}'\:\mathrm{be}\:\mathrm{the}\:\mathrm{reflection} \\ $$$$\mathrm{of}\:{B}\:\mathrm{in}\:\mathrm{the}\:\mathrm{line}\:{CA};\:{C}'\:\mathrm{be}\:\mathrm{the}\:\mathrm{reflection} \\ $$$$\mathrm{of}\:{C}\:\mathrm{in}\:\mathrm{the}\:\mathrm{line}\:{AB}.\:\mathrm{Show}\:\mathrm{that}\:{A}',\:{B}',\:{C}' \\ $$$$\mathrm{form}\:\mathrm{the}\:\mathrm{vertices}\:\mathrm{of}\:\mathrm{an}\:\mathrm{equilateral} \\ $$$$\mathrm{triangle}. \\ $$

Question Number 23539 Answers: 1 Comments: 0

∫tan^6 x dx

$$\int\mathrm{tan}\:^{\mathrm{6}} \mathrm{x}\:\mathrm{dx} \\ $$

Question Number 23477 Answers: 0 Comments: 0

Find the value of x, ∫_(−∞) ^x dx = ∫∣± sinh cot ln (15−(√(33+x)))∣ dx

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{x}, \\ $$$$\:\:\:\:\:\:\:\int_{−\infty} ^{{x}} {d}\mathrm{x}\:=\:\int\mid\pm\:\mathrm{sinh}\:\mathrm{cot}\:\mathrm{ln}\:\left(\mathrm{15}−\sqrt{\mathrm{33}+{x}}\right)\mid\:\mathrm{dx} \\ $$

Question Number 23418 Answers: 1 Comments: 0

∫sec^2 (√x) /(√x) dx

$$\int\mathrm{sec}\:^{\mathrm{2}} \sqrt{\mathrm{x}}\:/\sqrt{\mathrm{x}}\:\mathrm{dx} \\ $$

Question Number 23408 Answers: 1 Comments: 0