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Question Number 224095 Answers: 1 Comments: 2

how to prove that x + 9 = x is not has solution because (x + 9)^2 = x^2 x^2 + 18x + 81 = x^2 18x = −81 x = − ((81)/(18)) = −(9/2)

$${how}\:{to}\:{prove}\:{that}\:\:{x}\:+\:\mathrm{9}\:=\:{x}\:{is}\:{not}\:{has}\:{solution} \\ $$$${because}\: \\ $$$$\left({x}\:+\:\mathrm{9}\right)^{\mathrm{2}} \:=\:{x}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} \:+\:\mathrm{18}{x}\:+\:\mathrm{81}\:=\:{x}^{\mathrm{2}} \\ $$$$\mathrm{18}{x}\:=\:−\mathrm{81} \\ $$$${x}\:=\:−\:\frac{\mathrm{81}}{\mathrm{18}}\:=\:−\frac{\mathrm{9}}{\mathrm{2}} \\ $$

Question Number 224092 Answers: 0 Comments: 3

what′s the matter? yesterday the app was not accessible for many hours. now it seems to work normally again. but actually it doesn′t, at least with me. when i tip “view older” to scroll through the old posts, the app crashes always at some point, showing a message “Math Editor isn′t responding. × Close it ⊝ Wait” are you also experiencing the same problem? what has happened since yesterday?

$${what}'{s}\:{the}\:{matter}? \\ $$$${yesterday}\:{the}\:{app}\:{was}\:{not}\:{accessible} \\ $$$${for}\:{many}\:{hours}.\:{now}\:{it}\:{seems}\:{to} \\ $$$${work}\:{normally}\:{again}.\:{but}\:{actually} \\ $$$${it}\:{doesn}'{t},\:{at}\:{least}\:{with}\:{me}.\:{when} \\ $$$${i}\:{tip}\:``\boldsymbol{{view}}\:\boldsymbol{{older}}''\:{to}\:{scroll}\:{through} \\ $$$${the}\:{old}\:{posts},\:{the}\:{app}\:{crashes}\:{always} \\ $$$${at}\:{some}\:{point},\:{showing}\:{a}\:{message} \\ $$$$``{Math}\:{Editor}\:{isn}'{t}\:{responding}. \\ $$$$×\:{Close}\:{it} \\ $$$$\circleddash\:{Wait}'' \\ $$$${are}\:{you}\:{also}\:{experiencing}\:{the}\:{same} \\ $$$${problem}?\: \\ $$$${what}\:{has}\:{happened}\:{since}\:{yesterday}? \\ $$

Question Number 224085 Answers: 2 Comments: 0

If f(x)=4x^3 +3x^2 +x, Then solve for a and b: max_(x∈R) {∫_x ^2 f(t)dt}=a where x=b

$$\mathrm{If}\:{f}\left({x}\right)=\mathrm{4}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} +{x},\:\mathrm{Then}\:\mathrm{solve}\:\mathrm{for}\:{a}\:\mathrm{and}\:{b}: \\ $$$$\underset{{x}\in\mathbb{R}} {\mathrm{max}}\left\{\int_{{x}} ^{\mathrm{2}} {f}\left({t}\right){dt}\right\}={a}\:\mathrm{where}\:{x}={b} \\ $$

Question Number 224080 Answers: 0 Comments: 0

Use choleski′s method to solve the following system of equation 4x_1 −2x_2 +2x_3 =6 4x_1 −3x_2 −2x_3 =−8 2x_1 +3x_2 −x_3 =5

$$\boldsymbol{{Use}}\:\boldsymbol{{choleski}}'\boldsymbol{{s}}\:\boldsymbol{{method}}\:\boldsymbol{{to}}\:\boldsymbol{{solve}}\:\boldsymbol{{the}}\:\boldsymbol{{following}}\:\boldsymbol{{system}} \\ $$$$\boldsymbol{{of}}\:\boldsymbol{{equation}} \\ $$$$\mathrm{4}\boldsymbol{{x}}_{\mathrm{1}} −\mathrm{2}\boldsymbol{{x}}_{\mathrm{2}} +\mathrm{2}\boldsymbol{{x}}_{\mathrm{3}} =\mathrm{6} \\ $$$$\mathrm{4}\boldsymbol{{x}}_{\mathrm{1}} −\mathrm{3}\boldsymbol{{x}}_{\mathrm{2}} −\mathrm{2}\boldsymbol{{x}}_{\mathrm{3}} =−\mathrm{8} \\ $$$$\mathrm{2}\boldsymbol{{x}}_{\mathrm{1}} +\mathrm{3}\boldsymbol{{x}}_{\mathrm{2}} −\boldsymbol{{x}}_{\mathrm{3}} =\mathrm{5} \\ $$

Question Number 224079 Answers: 0 Comments: 0

For the given function f(x),let x_0 =0,x_1 =0.6 and x_2 =0.9. construct the lagrange interpolating polynomials of degree. (1) at most 1 (2)at most 2 to approximate f(0.45) if (a) f(x)=cosx (b) f(x)=(√(1+x)) (c) f(x)=In(1+x) (d) f(x)=tanx

$$\boldsymbol{{For}}\:\boldsymbol{{the}}\:\boldsymbol{{given}}\:\boldsymbol{{function}}\:\boldsymbol{{f}}\left(\boldsymbol{{x}}\right),\boldsymbol{{let}}\:\boldsymbol{{x}}_{\mathrm{0}} =\mathrm{0},\boldsymbol{{x}}_{\mathrm{1}} =\mathrm{0}.\mathrm{6} \\ $$$$\boldsymbol{{and}}\:\boldsymbol{{x}}_{\mathrm{2}} =\mathrm{0}.\mathrm{9}.\:\boldsymbol{{construct}}\:\boldsymbol{{the}}\:\boldsymbol{{lagrange}}\:\boldsymbol{{interpolating}} \\ $$$$\boldsymbol{{polynomials}}\:\boldsymbol{{of}}\:\boldsymbol{{degree}}.\:\left(\mathrm{1}\right)\:\boldsymbol{{at}}\:\boldsymbol{{most}}\:\mathrm{1}\:\left(\mathrm{2}\right)\boldsymbol{{at}}\:\boldsymbol{{most}}\:\mathrm{2} \\ $$$$\boldsymbol{{to}}\:\boldsymbol{{approximate}}\:\boldsymbol{{f}}\left(\mathrm{0}.\mathrm{45}\right)\:\boldsymbol{{if}}\: \\ $$$$\left(\boldsymbol{{a}}\right)\:\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\boldsymbol{{cosx}}\:\:\left(\boldsymbol{{b}}\right)\:\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\sqrt{\mathrm{1}+\boldsymbol{{x}}}\:\left(\boldsymbol{{c}}\right)\:\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\boldsymbol{{In}}\left(\mathrm{1}+\boldsymbol{{x}}\right) \\ $$$$\left(\boldsymbol{{d}}\right)\:\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\boldsymbol{{tanx}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Question Number 224078 Answers: 0 Comments: 0

Evaluate ∫^(𝛑/2) _0 sinxdx with h=(𝛑/(12)),correct to 5 decimal places,using (1)Trapezoidal rule (2)Newton−Cotes formula for n=4 (3)Simpson 3/8 −rule then find the truncation error in each case.

$$\boldsymbol{{Evaluate}}\:\underset{\mathrm{0}} {\int}^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \boldsymbol{{sinxdx}}\:\boldsymbol{{with}}\:\boldsymbol{{h}}=\frac{\boldsymbol{\pi}}{\mathrm{12}},\boldsymbol{{correct}}\:\boldsymbol{{to}} \\ $$$$\mathrm{5}\:\boldsymbol{{decimal}}\:\boldsymbol{{places}},\boldsymbol{{using}} \\ $$$$\left(\mathrm{1}\right)\boldsymbol{{Trapezoidal}}\:\boldsymbol{{rule}} \\ $$$$\left(\mathrm{2}\right)\boldsymbol{{Newton}}−\boldsymbol{{Cotes}}\:\boldsymbol{{formula}}\:\boldsymbol{{for}}\:\boldsymbol{{n}}=\mathrm{4} \\ $$$$\left(\mathrm{3}\right)\boldsymbol{{Simpson}}\:\mathrm{3}/\mathrm{8}\:−\boldsymbol{{rule}} \\ $$$$\boldsymbol{{then}}\:\boldsymbol{{find}}\:\boldsymbol{{the}}\:\boldsymbol{{truncation}}\:\boldsymbol{{error}}\:\boldsymbol{{in}}\:\boldsymbol{{each}}\:\boldsymbol{{case}}. \\ $$

Question Number 224076 Answers: 1 Comments: 0

D=(√((m−(n^2 /4))^2 +(e^m −n)^2 ))+(n^2 /4)(m,n∈R),D_(min) =?

$$ \\ $$$${D}=\sqrt{\left({m}−\frac{{n}^{\mathrm{2}} }{\mathrm{4}}\right)^{\mathrm{2}} +\left({e}^{{m}} −{n}\right)^{\mathrm{2}} }+\frac{{n}^{\mathrm{2}} }{\mathrm{4}}\left({m},{n}\in{R}\right),{D}_{\mathrm{min}} =? \\ $$

Question Number 224071 Answers: 1 Comments: 0

Two masses of 3 kg and 6 kg are placed 2 m apart in space. Calculate the gravitational force between them. (Take G=6.673×10^(−11) Nm^2 kg^(−2) )

$$\mathrm{Two}\:\mathrm{masses}\:\mathrm{of}\:\mathrm{3}\:\mathrm{kg}\:\mathrm{and}\:\mathrm{6}\:\mathrm{kg}\:\mathrm{are}\:\mathrm{placed}\:\mathrm{2}\:\mathrm{m}\:\:\mathrm{apart}\:\mathrm{in}\:\mathrm{space}. \\ $$$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{gravitational}\:\mathrm{force}\:\mathrm{between}\:\mathrm{them}. \\ $$$$\left(\mathrm{Take}\:\mathrm{G}=\mathrm{6}.\mathrm{673}×\mathrm{10}^{−\mathrm{11}} \:\mathrm{Nm}^{\mathrm{2}} \mathrm{kg}^{−\mathrm{2}} \right) \\ $$

Question Number 224069 Answers: 1 Comments: 0

If x^(32) =2^x then solve for x.

$$\mathrm{If}\:\mathrm{x}^{\mathrm{32}} =\mathrm{2}^{\mathrm{x}} \:\mathrm{then}\:\mathrm{solve}\:\mathrm{for}\:\mathrm{x}. \\ $$

Question Number 224067 Answers: 0 Comments: 0

Find the length of the diagonal of image formed by line segment connecting the points (2,3) and (4,5) in sequence with their respective reflection point with respect to x−axis.

$$ \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{diagonal}\:\mathrm{of}\:\mathrm{image}\:\mathrm{formed}\:\mathrm{by} \\ $$$$\mathrm{line}\:\mathrm{segment}\:\mathrm{connecting}\:\mathrm{the} \\ $$$$\mathrm{points}\:\left(\mathrm{2},\mathrm{3}\right)\:\mathrm{and}\:\left(\mathrm{4},\mathrm{5}\right)\:\mathrm{in}\:\mathrm{sequence} \\ $$$$\mathrm{with}\:\mathrm{their}\:\mathrm{respective} \\ $$$$\mathrm{reflection}\:\mathrm{point}\:\mathrm{with}\:\mathrm{respect}\:\mathrm{to} \\ $$$$\:\mathrm{x}−\mathrm{axis}. \\ $$

Question Number 224065 Answers: 0 Comments: 1

Find x, (√3)x−3x(√(1−x^2 ))=1 .

$$\mathrm{Find}\:\mathrm{x},\:\sqrt{\mathrm{3}}\mathrm{x}−\mathrm{3x}\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }=\mathrm{1}\:. \\ $$

Question Number 224082 Answers: 0 Comments: 1