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Question Number 161482 Answers: 3 Comments: 1

Question Number 161476 Answers: 0 Comments: 2

Between (3/6) and −(4/5) How do i list two rational numbers please?

$$\:{Between}\:\frac{\mathrm{3}}{\mathrm{6}}\:\:{and}\:−\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\:{How}\:{do}\:{i}\:{list}\:{two}\:{rational}\: \\ $$$$\:{numbers}\:{please}? \\ $$

Question Number 161538 Answers: 2 Comments: 0

lim_( x → −2 ) (((2+ 3x + 3x^( 2) + x^( 3) )/( sin ( ((πx)/2) ))) )=? −−−−

$$ \\ $$$${lim}_{\:{x}\:\rightarrow\:−\mathrm{2}\:\:} \left(\frac{\mathrm{2}+\:\mathrm{3}{x}\:+\:\mathrm{3}{x}^{\:\mathrm{2}} \:+\:{x}^{\:\mathrm{3}} }{\:{sin}\:\left(\:\frac{\pi{x}}{\mathrm{2}}\:\right)}\:\right)=? \\ $$$$\:\:\:\:−−−− \\ $$

Question Number 161491 Answers: 0 Comments: 0

Question Number 161464 Answers: 0 Comments: 1

Given that in ΔABC, (sin A+sin B):(sin B+sin C):(sin C+sin A)= 6: 4: 5 Find the angle A.

$$\mathrm{Given}\:\mathrm{that}\:\mathrm{in}\:\Delta\mathrm{ABC}, \\ $$$$\left(\mathrm{sin}\:\mathrm{A}+\mathrm{sin}\:\mathrm{B}\right):\left(\mathrm{sin}\:\mathrm{B}+\mathrm{sin}\:\mathrm{C}\right):\left(\mathrm{sin}\:\mathrm{C}+\mathrm{sin}\:\mathrm{A}\right)=\:\mathrm{6}:\:\mathrm{4}:\:\mathrm{5} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{A}. \\ $$

Question Number 161463 Answers: 2 Comments: 0

Let f(x)= x+∣x∣−1. Find lim_(x→0^+ ) ((f(x)−f(0))/(x−0)) and lim_(x→0^− ) ((f(x)−f(0))/(x−0)). Hence, determine whether f(x) is differentiable at x=0.

$$\mathrm{Let}\:{f}\left({x}\right)=\:{x}+\mid{x}\mid−\mathrm{1}.\: \\ $$$$\mathrm{Find}\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{{f}\left({x}\right)−{f}\left(\mathrm{0}\right)}{{x}−\mathrm{0}}\:\mathrm{and}\:\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\:\frac{{f}\left({x}\right)−{f}\left(\mathrm{0}\right)}{{x}−\mathrm{0}}. \\ $$$$\mathrm{Hence},\:\mathrm{determine}\:\mathrm{whether}\:{f}\left({x}\right)\:\mathrm{is}\: \\ $$$$\mathrm{differentiable}\:\mathrm{at}\:{x}=\mathrm{0}. \\ $$

Question Number 161462 Answers: 1 Comments: 0

Find the particular solution to the differential equation 2y′′+5y′+2y=0 subject to the initial conditions y(0)=2y , y′(0)=1 .

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{particular}\:\mathrm{solution}\:\mathrm{to}\:\mathrm{the}\:\mathrm{differential} \\ $$$$\mathrm{equation}\:\mathrm{2}{y}''+\mathrm{5}{y}'+\mathrm{2}{y}=\mathrm{0}\:\mathrm{subject}\:\mathrm{to}\:\mathrm{the}\:\mathrm{initial} \\ $$$$\mathrm{conditions}\:\:{y}\left(\mathrm{0}\right)=\mathrm{2}{y}\:,\:\:{y}'\left(\mathrm{0}\right)=\mathrm{1}\:. \\ $$

Question Number 161461 Answers: 0 Comments: 3

Let the region bounded by the curve y=x(1−x) and the x-axis be R. The line y=mx divides R into two parts, find the value of (1−m)^3 .

$$\mathrm{Let}\:\mathrm{the}\:\mathrm{region}\:\mathrm{bounded}\:\mathrm{by}\:\mathrm{the}\:\mathrm{curve}\: \\ $$$${y}={x}\left(\mathrm{1}−{x}\right)\:\mathrm{and}\:\mathrm{the}\:{x}-\mathrm{axis}\:\mathrm{be}\:\mathrm{R}. \\ $$$$\mathrm{The}\:\mathrm{line}\:{y}={mx}\:\mathrm{divides}\:\mathrm{R}\:\mathrm{into}\:\mathrm{two}\:\mathrm{parts}, \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\left(\mathrm{1}−{m}\right)^{\mathrm{3}} . \\ $$

Question Number 161456 Answers: 0 Comments: 0

Question Number 161450 Answers: 0 Comments: 0

A ector field is given by v= (x^2 −y^2 +x)i−(2xy+y)j. Show that vector v is irrotational hence find the scalar potential

$$\mathrm{A}\: \mathrm{ector}\:\mathrm{field}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by}\:\mathrm{v}= \\ $$$$\left(\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} +\mathrm{x}\right)\mathrm{i}−\left(\mathrm{2xy}+\mathrm{y}\right)\mathrm{j}.\:\mathrm{Show}\:\mathrm{that} \\ $$$$\mathrm{vector}\:\mathrm{v}\:\mathrm{is}\:\mathrm{irrotational}\:\mathrm{hence}\:\mathrm{find} \\ $$$$\:\mathrm{the}\:\mathrm{scalar}\:\mathrm{potential} \\ $$

Question Number 161444 Answers: 0 Comments: 2

Question Number 161443 Answers: 1 Comments: 0

Question Number 161442 Answers: 1 Comments: 0

Use the binomial theorem to write the first four terms of the expansion of (√(2+3x−x^2 ))

$$\mathrm{Use}\:\mathrm{the}\:\mathrm{binomial}\:\mathrm{theorem}\:\mathrm{to}\:\mathrm{write} \\ $$$$\mathrm{the}\:\mathrm{first}\:\mathrm{four}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{the}\:\mathrm{expansion} \\ $$$$\mathrm{of}\:\sqrt{\mathrm{2}+\mathrm{3}{x}−{x}^{\mathrm{2}} } \\ $$

Question Number 161439 Answers: 0 Comments: 1

Question Number 161437 Answers: 2 Comments: 0

lim_(x→0) ((((8x^2 −4x+1))^(1/(10)) +((7x^2 +3x+1))^(1/5) −2)/x) =?

$$\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{10}}]{\mathrm{8}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{1}}\:+\sqrt[{\mathrm{5}}]{\mathrm{7}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}}−\mathrm{2}}{{x}}\:=? \\ $$

Question Number 161433 Answers: 0 Comments: 0

if x;y;z>0 such that x+y+z=3 and 𝛌≥0 then prove that: (x/(y^3 +λy^2 )) + (y/(z^3 +λz^2 )) + (z/(x^3 +λx^2 )) ≥ (3/(λ+1))

$$\mathrm{if}\:\:\mathrm{x};\mathrm{y};\mathrm{z}>\mathrm{0}\:\:\mathrm{such}\:\mathrm{that}\:\:\mathrm{x}+\mathrm{y}+\mathrm{z}=\mathrm{3} \\ $$$$\mathrm{and}\:\:\boldsymbol{\lambda}\geqslant\mathrm{0}\:\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{x}}{\mathrm{y}^{\mathrm{3}} +\lambda\mathrm{y}^{\mathrm{2}} }\:+\:\frac{\mathrm{y}}{\mathrm{z}^{\mathrm{3}} +\lambda\mathrm{z}^{\mathrm{2}} }\:+\:\frac{\mathrm{z}}{\mathrm{x}^{\mathrm{3}} +\lambda\mathrm{x}^{\mathrm{2}} }\:\geqslant\:\frac{\mathrm{3}}{\lambda+\mathrm{1}} \\ $$

Question Number 161429 Answers: 0 Comments: 1

help me ! { ((x+3y+z=2)),((−3x+4y+2z=3)),((−2x+7y+3z=5)) :} Gauss Method...

$$\mathrm{help}\:\mathrm{me}\:! \\ $$$$\begin{cases}{{x}+\mathrm{3}{y}+{z}=\mathrm{2}}\\{−\mathrm{3}{x}+\mathrm{4}{y}+\mathrm{2}{z}=\mathrm{3}}\\{−\mathrm{2}{x}+\mathrm{7}{y}+\mathrm{3}{z}=\mathrm{5}}\end{cases} \\ $$$$\boldsymbol{\mathrm{G}}\mathrm{auss}\:\mathrm{Method}... \\ $$

Question Number 161424 Answers: 1 Comments: 0

Question Number 161418 Answers: 0 Comments: 0

4^(2021 ) =a^3 +b^3 +c^3 , (a:b:c)⇒ natural numbers

$$\mathrm{4}^{\mathrm{2021}\:} ={a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} ,\:\left({a}:{b}:{c}\right)\Rightarrow\:{natural}\:{numbers} \\ $$

Question Number 161416 Answers: 0 Comments: 0

Question Number 161412 Answers: 0 Comments: 0

Question Number 161410 Answers: 0 Comments: 3

Question Number 161409 Answers: 2 Comments: 1

Calculate lim_(x→0) ((1−cos (1−cos x))/x^4 ) lim_(x→0) ((1−cos xcos 2xcos 3x)/x^2 )

$${Calculate} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:\left(\mathrm{1}−\mathrm{cos}\:{x}\right)}{{x}^{\mathrm{4}} } \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:{x}\mathrm{cos}\:\mathrm{2}{x}\mathrm{cos}\:\mathrm{3}{x}}{{x}^{\mathrm{2}} } \\ $$

Question Number 161408 Answers: 1 Comments: 1

Question Number 161407 Answers: 1 Comments: 0

Question Number 161406 Answers: 0 Comments: 0

f (x ) = cos^( 2) ( x ) + sin^( 4) ( x ) R_( f) = ? −−−solution−−− y = cos^( 2) (x ) + sin^( 2) (x) .( 1−cos^( 2) (x)) = 1 − (1/4) sin^( 2) ( 2x) we know that : 0≤ sin^( 2) ( αx) ≤1 therefore −1≤− sin^( 2) (2x) ≤ 0 1−(1/4) ≤ 1− (1/4) sin^( 2) (2x) ≤1 R_( f) = [ (3/4) , 1 ] ◂ ★ ▶

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