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Question Number 58145 Answers: 1 Comments: 0

how to factorize a^3 b^2 +a^2 b^3

$${how}\:{to}\:{factorize} \\ $$$${a}^{\mathrm{3}} {b}^{\mathrm{2}} +{a}^{\mathrm{2}} {b}^{\mathrm{3}} \: \\ $$

Question Number 58135 Answers: 1 Comments: 0

Question Number 58132 Answers: 2 Comments: 1

Question Number 58113 Answers: 0 Comments: 1

once sgain: it′s boring to solve questions of minor complexity. we don′t have to, we do it to help unexperienced people to grow. you could at least type “thanks”. otherwise you might be ignored after a while...

$$\mathrm{once}\:\mathrm{sgain}:\:\mathrm{it}'\mathrm{s}\:\mathrm{boring}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{questions}\:\mathrm{of} \\ $$$$\mathrm{minor}\:\mathrm{complexity}.\:\mathrm{we}\:\mathrm{don}'\mathrm{t}\:\mathrm{have}\:\mathrm{to},\:\mathrm{we}\:\mathrm{do} \\ $$$$\mathrm{it}\:\mathrm{to}\:\mathrm{help}\:\mathrm{unexperienced}\:\mathrm{people}\:\mathrm{to}\:\mathrm{grow}. \\ $$$$\mathrm{you}\:\mathrm{could}\:\mathrm{at}\:\mathrm{least}\:\mathrm{type}\:``\mathrm{thanks}''.\:\mathrm{otherwise} \\ $$$$\mathrm{you}\:\mathrm{might}\:\mathrm{be}\:\mathrm{ignored}\:\mathrm{after}\:\mathrm{a}\:\mathrm{while}... \\ $$

Question Number 58103 Answers: 1 Comments: 0

f(x)=−x^6 +3 x^4 + 4x^2 find the zeros

$${f}\left({x}\right)=−{x}^{\mathrm{6}} +\mathrm{3}\:{x}^{\mathrm{4}} \:+\:\mathrm{4}{x}^{\mathrm{2}} {find}\:{the}\:{zeros} \\ $$

Question Number 58097 Answers: 2 Comments: 0

Find the angle between the curves: 1)x^2 y=1−y and x^3 =2−2y. 2) x^2 +y^2 =a^2 (√2) and x^2 −y^2 =a^2 .

$${Find}\:{the}\:{angle}\:{between}\:{the}\:{curves}: \\ $$$$\left.\mathrm{1}\right){x}^{\mathrm{2}} {y}=\mathrm{1}−{y}\:{and}\:{x}^{\mathrm{3}} =\mathrm{2}−\mathrm{2}{y}. \\ $$$$\left.\mathrm{2}\right)\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={a}^{\mathrm{2}} \sqrt{\mathrm{2}}\:{and}\:{x}^{\mathrm{2}} −{y}^{\mathrm{2}} ={a}^{\mathrm{2}} . \\ $$

Question Number 58092 Answers: 3 Comments: 0

6x^3 +5x^2 −6x−5=0

$$\mathrm{6}{x}^{\mathrm{3}} +\mathrm{5}{x}^{\mathrm{2}} −\mathrm{6}{x}−\mathrm{5}=\mathrm{0} \\ $$

Question Number 58091 Answers: 2 Comments: 0

27x^(3−1=0)

$$\mathrm{27}{x}^{\mathrm{3}−\mathrm{1}=\mathrm{0}} \\ $$

Question Number 58090 Answers: 1 Comments: 0

(x^4 −x^3 −38x^2 −31x+45)÷(x+5)

$$\left({x}^{\mathrm{4}} −{x}^{\mathrm{3}} −\mathrm{38}{x}^{\mathrm{2}} −\mathrm{31}{x}+\mathrm{45}\right)\boldsymbol{\div}\left({x}+\mathrm{5}\right) \\ $$

Question Number 58084 Answers: 0 Comments: 3

Question Number 58079 Answers: 1 Comments: 0

f(x)=2^3 +x^2 −5x+2;x+2

$${f}\left({x}\right)=\mathrm{2}^{\mathrm{3}} +{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{2};{x}+\mathrm{2} \\ $$

Question Number 58077 Answers: 1 Comments: 0

(3/(10))×2

$$\frac{\mathrm{3}}{\mathrm{10}}×\mathrm{2} \\ $$

Question Number 58076 Answers: 2 Comments: 1

a^x =m ⇒log_a m = x So is following true i^2 =−1 log_i (−1)=2

$$\:{a}^{{x}} ={m} \\ $$$$\Rightarrow\mathrm{log}_{{a}} {m}\:=\:{x} \\ $$$${So}\:{is}\:{following}\:{true} \\ $$$${i}^{\mathrm{2}} =−\mathrm{1} \\ $$$$\mathrm{log}_{{i}} \left(−\mathrm{1}\right)=\mathrm{2}\: \\ $$

Question Number 58060 Answers: 0 Comments: 3

Having given log 2 = 0.30103, find the position of the first significant figure in 2^(−37) .

$${Having}\:{given}\:\mathrm{log}\:\mathrm{2}\:=\:\mathrm{0}.\mathrm{30103},\:{find}\:{the}\:{position} \\ $$$${of}\:{the}\:{first}\:{significant}\:{figure}\:{in}\:\mathrm{2}^{−\mathrm{37}} . \\ $$

Question Number 58056 Answers: 1 Comments: 0

If the constant forces 2i−5j+6k and −i+2j−k act on a particle due to which it is displaced from a point A(4,−3,−2) to a point B(6, 1,−3), then the work done by the forces is

$$\mathrm{If}\:\mathrm{the}\:\mathrm{constant}\:\mathrm{forces}\:\mathrm{2}\boldsymbol{\mathrm{i}}−\mathrm{5}\boldsymbol{\mathrm{j}}+\mathrm{6}\boldsymbol{\mathrm{k}}\:\mathrm{and} \\ $$$$−\boldsymbol{\mathrm{i}}+\mathrm{2}\boldsymbol{\mathrm{j}}−\boldsymbol{\mathrm{k}}\:\mathrm{act}\:\mathrm{on}\:\mathrm{a}\:\mathrm{particle}\:\mathrm{due}\:\mathrm{to}\:\mathrm{which} \\ $$$$\mathrm{it}\:\mathrm{is}\:\mathrm{displaced}\:\mathrm{from}\:\mathrm{a}\:\mathrm{point}\:{A}\left(\mathrm{4},−\mathrm{3},−\mathrm{2}\right) \\ $$$$\mathrm{to}\:\mathrm{a}\:\mathrm{point}\:{B}\left(\mathrm{6},\:\mathrm{1},−\mathrm{3}\right),\:\mathrm{then}\:\mathrm{the}\:\mathrm{work}\: \\ $$$$\mathrm{done}\:\mathrm{by}\:\mathrm{the}\:\mathrm{forces}\:\mathrm{is} \\ $$

Question Number 58055 Answers: 1 Comments: 0

If the constant forces 2i−5j+6k and −i+2j−k act on a particle due to which it is displaced from a point A(4,−3,−2) to a point B(6, 1,−3), then the work done by the forces is

Question Number 58051 Answers: 1 Comments: 0