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Question Number 29286 Answers: 0 Comments: 1

To Developer Tinku Tara: Since the recent updates, it is not more possible to change an existing image via Edit Post”. This was possible in earlier versions of the app. Can you please fix this problem? Thank you!

$${To}\:{Developer}\:{Tinku}\:{Tara}: \\ $$$${Since}\:{the}\:{recent}\:{updates},\:{it}\:{is}\:{not}\:{more} \\ $$$${possible}\:{to}\:{change}\:{an}\:{existing}\:{image}\:{via} \\ $$$${Edit}\:{Post}''.\:{This}\:{was}\:{possible}\:{in}\:{earlier} \\ $$$${versions}\:{of}\:{the}\:{app}.\:{Can}\:{you}\:{please} \\ $$$${fix}\:{this}\:{problem}?\:{Thank}\:{you}! \\ $$

Question Number 29276 Answers: 1 Comments: 0

if the (x/a)+(y/b)=1 passes through the point lf intersection of the lines x+y=3 and 2x−3y=1 and is parallel to the line y=x−6 then find the value of a and b.

$${if}\:{the}\:\frac{{x}}{{a}}+\frac{{y}}{{b}}=\mathrm{1}\:{passes}\:{through}\:{the}\:{point}\:{lf}\:{intersection}\:{of}\:{the}\:{lines}\:{x}+{y}=\mathrm{3}\:{and}\:\mathrm{2}{x}−\mathrm{3}{y}=\mathrm{1}\:{and}\:{is}\:{parallel}\:{to}\:{the}\:{line}\:{y}={x}−\mathrm{6}\:{then}\:{find}\:\:{the}\:{value}\:{of}\:{a}\:\:{and}\:{b}.\: \\ $$

Question Number 29275 Answers: 0 Comments: 0

x(1−2xln x)y^((2)) +(1−4x^2 ln x)y^((1)) −(2+4x)y=0 y_1 =ln x the public answer?

$${x}\left(\mathrm{1}−\mathrm{2}{x}\mathrm{ln}\:{x}\right){y}^{\left(\mathrm{2}\right)} +\left(\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} \mathrm{ln}\:{x}\right){y}^{\left(\mathrm{1}\right)} −\left(\mathrm{2}+\mathrm{4}{x}\right){y}=\mathrm{0} \\ $$$${y}_{\mathrm{1}} =\mathrm{ln}\:{x} \\ $$$${the}\:{public}\:{answer}? \\ $$

Question Number 29268 Answers: 0 Comments: 2

A man moves 20m North , then 12m East and finally 15m South.His displacement from the starting point is now (a) 13m (b) 27m (c) 47m (d) 23m

$$\mathrm{A}\:\mathrm{man}\:\mathrm{moves}\:\:\mathrm{20m}\:\:\mathrm{North}\:,\:\:\mathrm{then}\:\:\mathrm{12m}\:\mathrm{East}\:\:\mathrm{and}\:\:\mathrm{finally}\:\:\mathrm{15m}\:\:\mathrm{South}.\mathrm{His} \\ $$$$\mathrm{displacement}\:\mathrm{from}\:\mathrm{the}\:\mathrm{starting}\:\mathrm{point}\:\mathrm{is}\:\mathrm{now} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{13m}\:\:\left(\mathrm{b}\right)\:\:\mathrm{27m}\:\:\left(\mathrm{c}\right)\:\:\mathrm{47m}\:\:\left(\mathrm{d}\right)\:\:\mathrm{23m} \\ $$

Question Number 29272 Answers: 0 Comments: 0

(2x−3x^3 )y^((2)) +4y^((1)) +6xy=0 this equation has an answer in the form of several sentences. get the public answer.

$$\left(\mathrm{2}{x}−\mathrm{3}{x}^{\mathrm{3}} \right){y}^{\left(\mathrm{2}\right)} +\mathrm{4}{y}^{\left(\mathrm{1}\right)} +\mathrm{6}{xy}=\mathrm{0} \\ $$$${this}\:{equation}\:{has}\:{an}\:{answer}\:{in}\:{the}\:{form}\:{of}\:{several}\:{sentences}. \\ $$$${get}\:{the}\:{public}\:{answer}. \\ $$

Question Number 29265 Answers: 0 Comments: 0

An electric pump with efficiency of 70% raises water to a height of 15m . If water is delivered at the rate of 350 dm^3 per minute. (i) what is the power rating of the pump ? (mass of 1 dm^3 = 1 kg) (ii) what is the energy lost by the pump ? (g = 10 m/s^2 ) (Answer: 1250W. 22.5 KJ)

$$\mathrm{An}\:\mathrm{electric}\:\mathrm{pump}\:\mathrm{with}\:\mathrm{efficiency}\:\mathrm{of}\:\:\:\mathrm{70\%}\:\:\mathrm{raises}\:\mathrm{water}\:\mathrm{to}\:\mathrm{a}\:\mathrm{height}\:\mathrm{of}\:\:\mathrm{15m} \\ $$$$.\:\mathrm{If}\:\mathrm{water}\:\mathrm{is}\:\mathrm{delivered}\:\mathrm{at}\:\mathrm{the}\:\mathrm{rate}\:\mathrm{of}\:\:\mathrm{350}\:\mathrm{dm}^{\mathrm{3}} \:\mathrm{per}\:\mathrm{minute}.\:\: \\ $$$$\left(\mathrm{i}\right)\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{power}\:\mathrm{rating}\:\mathrm{of}\:\mathrm{the}\:\mathrm{pump}\:?\:\:\:\left(\mathrm{mass}\:\mathrm{of}\:\mathrm{1}\:\mathrm{dm}^{\mathrm{3}} \:=\:\mathrm{1}\:\mathrm{kg}\right) \\ $$$$\left(\mathrm{ii}\right)\:\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{energy}\:\mathrm{lost}\:\mathrm{by}\:\mathrm{the}\:\mathrm{pump}\:?\:\:\:\left(\mathrm{g}\:=\:\mathrm{10}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \right) \\ $$$$\left(\mathrm{Answer}:\:\:\:\mathrm{1250W}.\:\:\:\:\mathrm{22}.\mathrm{5}\:\mathrm{KJ}\right) \\ $$

Question Number 29264 Answers: 1 Comments: 0

there are 25 persons in a conical tent every person needs an area of 4 sq m on the ground under the tent. if height if the tent is 18m.find the volume of the tent.

$$\mathrm{there}\:\mathrm{are}\:\mathrm{25}\:\mathrm{persons}\:\mathrm{in}\:\mathrm{a}\:\mathrm{conical}\:\mathrm{tent} \\ $$$$\mathrm{every}\:\mathrm{person}\:\mathrm{needs}\:\mathrm{an}\:\mathrm{area}\:\mathrm{of}\:\mathrm{4}\:\mathrm{sq}\:\mathrm{m}\: \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{ground}\:\mathrm{under}\:\mathrm{the}\:\mathrm{tent}.\:\mathrm{if} \\ $$$$\mathrm{height}\:\mathrm{if}\:\mathrm{the}\:\mathrm{tent}\:\mathrm{is}\:\mathrm{18m}.\mathrm{find}\:\mathrm{the}\:\mathrm{volume} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{tent}. \\ $$

Question Number 29273 Answers: 0 Comments: 4

Question Number 29260 Answers: 0 Comments: 0

The eriodic time of a simple pendulum is given by T=2π(√(L/g)). if L=100±0.1cm(s.e) and the time s for 10 ossilations are: 48.2,48.2,48.7,48.5,48.4,48.2,48.4, 48.6,48.5,48.2. calcukate g and the standard error involved.

Question Number 29259 Answers: 1 Comments: 0

solve the equation 2x+3y=5 3x+4y=4

$${solve}\:{the}\:{equation} \\ $$$$\mathrm{2}{x}+\mathrm{3}{y}=\mathrm{5} \\ $$$$\mathrm{3}{x}+\mathrm{4}{y}=\mathrm{4}\: \\ $$

Question Number 29308 Answers: 2 Comments: 0

Solve: w^3 = − 16

$$\mathrm{Solve}:\:\:\:\mathrm{w}^{\mathrm{3}} \:=\:−\:\mathrm{16} \\ $$

Question Number 29249 Answers: 1 Comments: 1

Question Number 29241 Answers: 1 Comments: 0

Question Number 29242 Answers: 0 Comments: 0

Question Number 29254 Answers: 1 Comments: 1

Question Number 29222 Answers: 1 Comments: 0

3x−4y=12, xy=2

$$\mathrm{3}{x}−\mathrm{4}{y}=\mathrm{12},\:{xy}=\mathrm{2} \\ $$

Question Number 29217 Answers: 1 Comments: 0

The velocity of a physical system is given by V={(√((p+1/n)))}/x where p=pressure. Find the dimensions of n and x

$${The}\:{velocity}\:{of}\:{a}\:{physical}\:{system} \\ $$$${is}\:{given}\:{by}\:{V}=\left\{\sqrt{\left({p}+\mathrm{1}/{n}\right)}\right\}/{x} \\ $$$${where}\:{p}={pressure}.\:{Find}\:{the} \\ $$$${dimensions}\:{of}\:{n}\:{and}\:{x} \\ $$

Question Number 29216 Answers: 1 Comments: 0

A horizontal force of 0.8N is required to pull a 5kg block across a table top at a constant speed. With the block initially at rest,a 20g bullet fired horizontally into the block to slide 1.5m before coming to rest again.Determine the speed v of the bullet,where the bullet is assumed to be embedded in the block.

$${A}\:{horizontal}\:{force}\:{of}\:\mathrm{0}.\mathrm{8}{N}\:{is} \\ $$$${required}\:{to}\:{pull}\:{a}\:\mathrm{5}{kg}\:{block}\:{across} \\ $$$${a}\:{table}\:{top}\:{at}\:{a}\:{constant}\:{speed}. \\ $$$${With}\:{the}\:{block}\:{initially}\:{at}\:{rest},{a} \\ $$$$\mathrm{20}{g}\:{bullet}\:{fired}\:{horizontally}\:{into} \\ $$$${the}\:{block}\:{to}\:{slide}\:\mathrm{1}.\mathrm{5}{m}\:{before} \\ $$$${coming}\:{to}\:{rest}\:{again}.{Determine} \\ $$$${the}\:{speed}\:{v}\:{of}\:{the}\:{bullet},{where} \\ $$$${the}\:{bullet}\:{is}\:{assumed}\:{to}\:{be} \\ $$$${embedded}\:{in}\:{the}\:{block}. \\ $$

Question Number 29213 Answers: 0 Comments: 1

Two masses m and 2m,approach each l along a path at right angle to each other and move off at 2m/s at angle 37° to the original direction of mass m. What where the initial speeds of the two particles?

$${Two}\:{masses}\:{m}\:{and}\:\mathrm{2}{m},{approach} \\ $$$${each}\:{l}\:{along}\:{a}\:{path}\:{at}\:{right} \\ $$$${angle}\:{to}\:{each}\:{other}\:{and}\:{move}\:{off} \\ $$$${at}\:\mathrm{2}{m}/{s}\:{at}\:{angle}\:\mathrm{37}°\:{to}\:{the} \\ $$$${original}\:{direction}\:{of}\:{mass}\:{m}. \\ $$$${What}\:{where}\:{the}\:{initial}\:{speeds}\:{of} \\ $$$${the}\:{two}\:{particles}? \\ $$

Question Number 29212 Answers: 1 Comments: 1

A block of wood of mass 0.6kg is balanced on top of vertical port 2m high.A 10g bullet is fired horizontally into the block and the embedded bullet land at a 4m from the base of the port.Find the initial velocity of the bullet.

$${A}\:{block}\:{of}\:{wood}\:{of}\:{mass}\:\mathrm{0}.\mathrm{6}{kg}\:{is} \\ $$$${balanced}\:{on}\:{top}\:{of}\:{vertical}\:{port} \\ $$$$\mathrm{2}{m}\:{high}.{A}\:\mathrm{10}{g}\:{bullet}\:{is}\:{fired} \\ $$$${horizontally}\:{into}\:{the}\:{block}\:{and} \\ $$$${the}\:{embedded}\:{bullet}\:{land}\:{at}\:{a}\:\mathrm{4}{m} \\ $$$${from}\:{the}\:{base}\:{of}\:{the}\:{port}.{Find} \\ $$$${the}\:{initial}\:{velocity}\:{of}\:{the}\:{bullet}. \\ $$

Question Number 29209 Answers: 1 Comments: 4

Question Number 29202 Answers: 1 Comments: 0

Find area between by y=1 and y=((1−x^2 )/(1+x^2 )) .

$${Find}\:{area}\:{between}\:{by}\:{y}=\mathrm{1}\:\:{and} \\ $$$${y}=\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }\:. \\ $$

Question Number 29201 Answers: 2 Comments: 0

Question Number 29198 Answers: 1 Comments: 1

Question Number 29196 Answers: 0 Comments: 0

Let s = n_c_1 − (1+(1/2))n_c_2 +(1+(1/2)+(1/3))n_c_3 +.......+(−1)^(n−1) (1+(1/2)+(1/3)+....+(1/n))n_c_n then prove that s×n =1.

$$\mathrm{Let}\:\mathrm{s}\:=\:\mathrm{n}_{\mathrm{c}_{\mathrm{1}} } \:−\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)\mathrm{n}_{\mathrm{c}_{\mathrm{2}} } \:+\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}\right)\mathrm{n}_{\mathrm{c}_{\mathrm{3}} } \\ $$$$+.......+\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+....+\frac{\mathrm{1}}{\mathrm{n}}\right)\mathrm{n}_{\mathrm{c}_{\mathrm{n}} } \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{s}×\mathrm{n}\:=\mathrm{1}. \\ $$

Question Number 29231 Answers: 1 Comments: 0

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