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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 23870 Answers: 0 Comments: 0

Find a vector function F whose graph is the the line of intersection of the plane 4x−2y+z=3 and x+y−3y=1

$${Find}\:{a}\:{vector}\:{function}\:{F}\:\:{whose}\:{graph}\:{is}\:{the} \\ $$$${the}\:{line}\:{of}\:{intersection}\:{of}\:{the}\:{plane}\: \\ $$$$\mathrm{4}{x}−\mathrm{2}{y}+{z}=\mathrm{3}\:{and}\:{x}+{y}−\mathrm{3}{y}=\mathrm{1} \\ $$

Question Number 23884 Answers: 1 Comments: 0

If C_r stands for^n C_r = ((n!)/(r! n − r!)) and Σ_(r=1) ^n r.C_r ^2 = λ for n ≥ 2, then λ is divisible by (1) 3 (n − 1) (2) n + 1 (3) n (2n − 1) (4) n^2 + 1

$$\mathrm{If}\:{C}_{{r}} \:\mathrm{stands}\:\mathrm{for}\:^{{n}} {C}_{{r}} \:=\:\frac{{n}!}{{r}!\:{n}\:−\:{r}!}\:\mathrm{and} \\ $$$$\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{r}.{C}_{{r}} ^{\mathrm{2}} \:=\:\lambda\:\mathrm{for}\:{n}\:\geqslant\:\mathrm{2},\:\mathrm{then}\:\lambda\:\mathrm{is}\:\mathrm{divisible} \\ $$$$\mathrm{by} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{3}\:\left({n}\:−\:\mathrm{1}\right) \\ $$$$\left(\mathrm{2}\right)\:{n}\:+\:\mathrm{1} \\ $$$$\left(\mathrm{3}\right)\:{n}\:\left(\mathrm{2}{n}\:−\:\mathrm{1}\right) \\ $$$$\left(\mathrm{4}\right)\:{n}^{\mathrm{2}} \:+\:\mathrm{1} \\ $$

Question Number 23865 Answers: 0 Comments: 7

A man of mass m, standing at the bottom of the staircase of height L climbs it and stands at its top. (1) Work done by all forces on man is equal to the rise in potential energy mgL. (2) Work done by all forces on man is zero. (3) Work done by the gravitational force on man is mgL. (4) The reaction force from a step does not do work because the point of application of the force does not move while the force exists.

$$\mathrm{A}\:\mathrm{man}\:\mathrm{of}\:\mathrm{mass}\:{m},\:\mathrm{standing}\:\mathrm{at}\:\mathrm{the} \\ $$$$\mathrm{bottom}\:\mathrm{of}\:\mathrm{the}\:\mathrm{staircase}\:\mathrm{of}\:\mathrm{height}\:{L} \\ $$$$\mathrm{climbs}\:\mathrm{it}\:\mathrm{and}\:\mathrm{stands}\:\mathrm{at}\:\mathrm{its}\:\mathrm{top}. \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Work}\:\mathrm{done}\:\mathrm{by}\:\mathrm{all}\:\mathrm{forces}\:\mathrm{on}\:\mathrm{man}\:\mathrm{is} \\ $$$$\mathrm{equal}\:\mathrm{to}\:\mathrm{the}\:\mathrm{rise}\:\mathrm{in}\:\mathrm{potential}\:\mathrm{energy} \\ $$$${mgL}. \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Work}\:\mathrm{done}\:\mathrm{by}\:{all}\:\mathrm{forces}\:\mathrm{on}\:\mathrm{man}\:\mathrm{is} \\ $$$$\mathrm{zero}. \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Work}\:\mathrm{done}\:\mathrm{by}\:\mathrm{the}\:\mathrm{gravitational} \\ $$$$\mathrm{force}\:\mathrm{on}\:\mathrm{man}\:\mathrm{is}\:{mgL}. \\ $$$$\left(\mathrm{4}\right)\:\mathrm{The}\:\mathrm{reaction}\:\mathrm{force}\:\mathrm{from}\:\mathrm{a}\:\mathrm{step}\:\mathrm{does} \\ $$$$\mathrm{not}\:\mathrm{do}\:\mathrm{work}\:\mathrm{because}\:\mathrm{the}\:\mathrm{point}\:\mathrm{of} \\ $$$$\mathrm{application}\:\mathrm{of}\:\mathrm{the}\:\mathrm{force}\:\mathrm{does}\:\mathrm{not}\:\mathrm{move} \\ $$$$\mathrm{while}\:\mathrm{the}\:\mathrm{force}\:\mathrm{exists}. \\ $$

Question Number 23863 Answers: 0 Comments: 0

From the following data, calculate the enthalpy change for the combustion of cyclopropane at 298 K. The enthalpy of formation of CO_2 (g), H_2 O (l) and propene (g) are −393.5, −285.8 and 20.42 kJ mol^(−1) respectively. The enthalpy of isomerisation of cyclopropane to propene is −33 kJ mol^(−1) .

$$\mathrm{From}\:\mathrm{the}\:\mathrm{following}\:\mathrm{data},\:\mathrm{calculate}\:\mathrm{the} \\ $$$$\mathrm{enthalpy}\:\mathrm{change}\:\mathrm{for}\:\mathrm{the}\:\mathrm{combustion}\:\mathrm{of} \\ $$$$\mathrm{cyclopropane}\:\mathrm{at}\:\mathrm{298}\:\mathrm{K}.\:\mathrm{The}\:\mathrm{enthalpy} \\ $$$$\mathrm{of}\:\mathrm{formation}\:\mathrm{of}\:\mathrm{CO}_{\mathrm{2}} \:\left({g}\right),\:\mathrm{H}_{\mathrm{2}} \mathrm{O}\:\left({l}\right)\:\mathrm{and} \\ $$$$\mathrm{propene}\:\left({g}\right)\:\mathrm{are}\:−\mathrm{393}.\mathrm{5},\:−\mathrm{285}.\mathrm{8}\:\mathrm{and} \\ $$$$\mathrm{20}.\mathrm{42}\:\mathrm{kJ}\:\mathrm{mol}^{−\mathrm{1}} \:\mathrm{respectively}.\:\mathrm{The} \\ $$$$\mathrm{enthalpy}\:\mathrm{of}\:\mathrm{isomerisation}\:\mathrm{of}\:\mathrm{cyclopropane} \\ $$$$\mathrm{to}\:\mathrm{propene}\:\mathrm{is}\:−\mathrm{33}\:\mathrm{kJ}\:\mathrm{mol}^{−\mathrm{1}} . \\ $$

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 23864 Answers: 1 Comments: 0

One mole of steam is condensed at 100°C, the water is cooled to 0°C and frozen to ice. Which of the following statements are correct, given heat of vapourization and fusion are 540 cal/ gm and 80 cal/gm? (average heat capacity of liquid water = 1 cal gm^(−1) degree^(−1) ) (1) Entropy change during the condensation of steam is −26.06 cal/°C (2) Entropy change during cooling of water from 100°C to 0°C is −5.62 cal/°C (3) Entropy change during freezing of water at 0°C is −5.27 cal/°C (4) Total entropy change is −36.95 cal/°C

$$\mathrm{One}\:\mathrm{mole}\:\mathrm{of}\:\mathrm{steam}\:\mathrm{is}\:\mathrm{condensed}\:\mathrm{at} \\ $$$$\mathrm{100}°\mathrm{C},\:\mathrm{the}\:\mathrm{water}\:\mathrm{is}\:\mathrm{cooled}\:\mathrm{to}\:\mathrm{0}°\mathrm{C}\:\mathrm{and} \\ $$$$\mathrm{frozen}\:\mathrm{to}\:\mathrm{ice}.\:\mathrm{Which}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{statements}\:\mathrm{are}\:\mathrm{correct},\:\mathrm{given}\:\mathrm{heat}\:\mathrm{of} \\ $$$$\mathrm{vapourization}\:\mathrm{and}\:\mathrm{fusion}\:\mathrm{are}\:\mathrm{540}\:\mathrm{cal}/ \\ $$$$\mathrm{gm}\:\mathrm{and}\:\mathrm{80}\:\mathrm{cal}/\mathrm{gm}?\:\left(\mathrm{average}\:\mathrm{heat}\right. \\ $$$$\mathrm{capacity}\:\mathrm{of}\:\mathrm{liquid}\:\mathrm{water}\:=\:\mathrm{1}\:\mathrm{cal}\:\mathrm{gm}^{−\mathrm{1}} \\ $$$$\left.\mathrm{degree}^{−\mathrm{1}} \right) \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Entropy}\:\mathrm{change}\:\mathrm{during}\:\mathrm{the} \\ $$$$\mathrm{condensation}\:\mathrm{of}\:\mathrm{steam}\:\mathrm{is}\:−\mathrm{26}.\mathrm{06}\:\mathrm{cal}/°\mathrm{C} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Entropy}\:\mathrm{change}\:\mathrm{during}\:\mathrm{cooling}\:\mathrm{of} \\ $$$$\mathrm{water}\:\mathrm{from}\:\mathrm{100}°\mathrm{C}\:\mathrm{to}\:\mathrm{0}°\mathrm{C}\:\mathrm{is}\:−\mathrm{5}.\mathrm{62}\:\mathrm{cal}/°\mathrm{C} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Entropy}\:\mathrm{change}\:\mathrm{during}\:\mathrm{freezing}\:\mathrm{of} \\ $$$$\mathrm{water}\:\mathrm{at}\:\mathrm{0}°\mathrm{C}\:\mathrm{is}\:−\mathrm{5}.\mathrm{27}\:\mathrm{cal}/°\mathrm{C} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{Total}\:\mathrm{entropy}\:\mathrm{change}\:\mathrm{is}\:−\mathrm{36}.\mathrm{95} \\ $$$$\mathrm{cal}/°\mathrm{C} \\ $$

Question Number 23851 Answers: 0 Comments: 0

Solve sin(83) − cos(17) in surd form

$$\mathrm{Solve}\:\:\:\:\:\:\mathrm{sin}\left(\mathrm{83}\right)\:−\:\mathrm{cos}\left(\mathrm{17}\right)\:\:\:\:\:\:\mathrm{in}\:\mathrm{surd}\:\mathrm{form} \\ $$

Question Number 23849 Answers: 2 Comments: 0

∫((x^6 +1)/(x^4 +1))dx

$$\int\frac{{x}^{\mathrm{6}} +\mathrm{1}}{{x}^{\mathrm{4}} +\mathrm{1}}{dx} \\ $$

Question Number 23833 Answers: 1 Comments: 3

f(x^2 + x) + 2f(x^2 − 3x + 2) = 9x^2 − 15x then what is the value of f(10) ?

$${f}\left({x}^{\mathrm{2}} \:+\:{x}\right)\:+\:\mathrm{2}{f}\left({x}^{\mathrm{2}} \:−\:\mathrm{3}{x}\:+\:\mathrm{2}\right)\:=\:\mathrm{9}{x}^{\mathrm{2}} \:−\:\mathrm{15}{x} \\ $$$$\mathrm{then}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{f}\left(\mathrm{10}\right)\:? \\ $$

Question Number 23827 Answers: 0 Comments: 0

∫ (√(cosecx)) dx

$$\int\:\sqrt{\mathrm{cosecx}}\:\:\:\mathrm{dx} \\ $$

Question Number 23823 Answers: 1 Comments: 0

The period of a simple pendulum A is 6 secs. Find the period of a simple pendulum B, if A makes 30 oscillation in the time it takes B to make 70 oscillation.

$$\mathrm{The}\:\mathrm{period}\:\mathrm{of}\:\mathrm{a}\:\mathrm{simple}\:\mathrm{pendulum}\:\mathrm{A}\:\mathrm{is}\:\mathrm{6}\:\mathrm{secs}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{period}\:\mathrm{of}\:\mathrm{a}\:\mathrm{simple} \\ $$$$\mathrm{pendulum}\:\mathrm{B},\:\mathrm{if}\:\mathrm{A}\:\mathrm{makes}\:\mathrm{30}\:\mathrm{oscillation}\:\mathrm{in}\:\mathrm{the}\:\mathrm{time}\:\mathrm{it}\:\mathrm{takes}\:\mathrm{B}\:\mathrm{to}\:\mathrm{make}\:\mathrm{70} \\ $$$$\mathrm{oscillation}. \\ $$

Question Number 23814 Answers: 1 Comments: 1

Question Number 23781 Answers: 0 Comments: 0

P(X=299), μ=318, standard deviation=42 P(z=((299−318)/(42))) P(z=−0.45) Is it P(z=−0.45)=P(−0.45<z<−0.44)? If not, how to find P(z=−0.45)?

$${P}\left({X}=\mathrm{299}\right),\:\mu=\mathrm{318},\:{standard}\:{deviation}=\mathrm{42} \\ $$$${P}\left({z}=\frac{\mathrm{299}−\mathrm{318}}{\mathrm{42}}\right) \\ $$$${P}\left({z}=−\mathrm{0}.\mathrm{45}\right) \\ $$$$ \\ $$$${Is}\:{it}\:{P}\left({z}=−\mathrm{0}.\mathrm{45}\right)={P}\left(−\mathrm{0}.\mathrm{45}<{z}<−\mathrm{0}.\mathrm{44}\right)? \\ $$$$ \\ $$$${If}\:{not},\:{how}\:{to}\:{find}\:{P}\left({z}=−\mathrm{0}.\mathrm{45}\right)? \\ $$

Question Number 23777 Answers: 0 Comments: 0

please solve question no 23764 and 23765

$${please}\:{solve}\:{question}\:{no}\:\mathrm{23764}\:{and}\:\mathrm{23765} \\ $$

Question Number 23773 Answers: 0 Comments: 2

I have a leaderboard with the data of ′runners′. Each runner has a ′place′ (1st, 2nd, etc.), and a ′time′ (in seconds). Every day I have been comparing the data. What information is important to note when looking at data such as this? Is it possible to make predictions of future data? I′d love advice for things I can do with monitoring these kinds of statistics!

$$\mathrm{I}\:\mathrm{have}\:\mathrm{a}\:\mathrm{leaderboard}\:\mathrm{with}\:\mathrm{the}\:\mathrm{data}\:\mathrm{of}\:'\mathrm{runners}'. \\ $$$$\mathrm{Each}\:\mathrm{runner}\:\mathrm{has}\:\mathrm{a}\:'\mathrm{place}'\:\left(\mathrm{1st},\:\mathrm{2nd},\:\mathrm{etc}.\right), \\ $$$$\mathrm{and}\:\mathrm{a}\:'\mathrm{time}'\:\left(\mathrm{in}\:\mathrm{seconds}\right). \\ $$$$\: \\ $$$$\mathrm{Every}\:\mathrm{day}\:\mathrm{I}\:\mathrm{have}\:\mathrm{been}\:\mathrm{comparing}\:\mathrm{the}\:\mathrm{data}. \\ $$$$\mathrm{What}\:\mathrm{information}\:\mathrm{is}\:\mathrm{important}\:\mathrm{to}\:\mathrm{note}\:\mathrm{when} \\ $$$$\mathrm{looking}\:\mathrm{at}\:\mathrm{data}\:\mathrm{such}\:\mathrm{as}\:\mathrm{this}? \\ $$$$\: \\ $$$$\mathrm{Is}\:\mathrm{it}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{make}\:\mathrm{predictions}\:\mathrm{of}\:\mathrm{future}\:\mathrm{data}? \\ $$$$\: \\ $$$$\mathrm{I}'\mathrm{d}\:\mathrm{love}\:\mathrm{advice}\:\mathrm{for}\:\mathrm{things}\:\mathrm{I}\:\mathrm{can}\:\mathrm{do}\:\mathrm{with}\:\mathrm{monitoring} \\ $$$$\mathrm{these}\:\mathrm{kinds}\:\mathrm{of}\:\mathrm{statistics}! \\ $$

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 23768 Answers: 0 Comments: 2

Find the last two digit of 20^(17) + 17^(20)

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{last}\:\mathrm{two}\:\mathrm{digit}\:\mathrm{of}\:\:\:\:\:\mathrm{20}^{\mathrm{17}} \:+\:\mathrm{17}^{\mathrm{20}} \\ $$

Question Number 23765 Answers: 0 Comments: 0

lim_(x→∝) ((cot^(−1) (x^(−a) ln _a x))/(sec^(−1) (a^x ln _x a))) (a>1)

$$\underset{{x}\rightarrow\propto} {\mathrm{lim}}\:\frac{\mathrm{cot}^{−\mathrm{1}} \left({x}^{−{a}} \mathrm{ln}\:_{{a}} \:{x}\right)}{\mathrm{sec}^{−\mathrm{1}} \left({a}^{{x}} \:\mathrm{ln}\:_{{x}} {a}\right)}\:\: \\ $$$$\left({a}>\mathrm{1}\right) \\ $$$$ \\ $$

Question Number 23764 Answers: 0 Comments: 0

lim_(x→∝) (((2^x^n )^(1/e^x ) −(3^x^n )^(1/e^x ) )/x^n ) where nεN

$$\underset{{x}\rightarrow\propto} {\mathrm{lim}}\:\frac{\left(\mathrm{2}^{{x}^{{n}} } \overset{\mathrm{1}/{e}^{{x}} } {\right)}−\left(\mathrm{3}^{{x}^{{n}} } \right)^{\mathrm{1}/{e}^{{x}} } }{{x}^{{n}} } \\ $$$${where}\:{n}\varepsilon{N} \\ $$$$ \\ $$

Question Number 23767 Answers: 0 Comments: 0

suppose that n is the number of balls in a basket.If pronability of pulling A and B together is 0.05 what is the value of n?