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Question Number 185490 Answers: 1 Comments: 1

Question Number 185486 Answers: 2 Comments: 0

Question Number 185484 Answers: 1 Comments: 0

Question Number 185482 Answers: 1 Comments: 1

Question Number 185481 Answers: 0 Comments: 0

Question Number 185472 Answers: 4 Comments: 2

if ω^7 =1 (1/(ω+ω^6 ))+(1/(ω^2 +ω^5 ))+(1/(ω^3 +ω^4 ))=?

$${if}\:\omega^{\mathrm{7}} =\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\omega+\omega^{\mathrm{6}} }+\frac{\mathrm{1}}{\omega^{\mathrm{2}} +\omega^{\mathrm{5}} }+\frac{\mathrm{1}}{\omega^{\mathrm{3}} +\omega^{\mathrm{4}} }=? \\ $$

Question Number 185471 Answers: 2 Comments: 0

∫_( 0) ^( r) (1/( (√(r^2 −x^2 ))))dx=?

$$\int_{\:\mathrm{0}} ^{\:{r}} \frac{\mathrm{1}}{\:\sqrt{{r}^{\mathrm{2}} −{x}^{\mathrm{2}} }}{dx}=? \\ $$

Question Number 185470 Answers: 0 Comments: 3

Determine whether the series U_n =((1+2n^2 )/(1+n^2 )) is convergent or not M.m

$$\mathrm{Determine}\:\mathrm{whether}\:\mathrm{the}\:\mathrm{series} \\ $$$$\mathrm{U}_{\mathrm{n}} =\frac{\mathrm{1}+\mathrm{2n}^{\mathrm{2}} }{\mathrm{1}+\mathrm{n}^{\mathrm{2}} }\:\mathrm{is}\:\mathrm{convergent}\:\mathrm{or}\:\mathrm{not} \\ $$$$ \\ $$$$\mathrm{M}.\mathrm{m} \\ $$

Question Number 185455 Answers: 1 Comments: 0

∫ ((x^3 .e^x^2 )/((x^2 +1)^2 )) dx =?

$$\:\int\:\frac{{x}^{\mathrm{3}} .{e}^{{x}^{\mathrm{2}} } }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\:{dx}\:=? \\ $$

Question Number 185453 Answers: 1 Comments: 9

Find the largest possible area of trapezoid that can be drawn under the x−axis so that one of its bases is on the x−axis and the other two vertices are on the curve y=x^2 −9

$$\:{Find}\:{the}\:{largest}\:{possible}\:{area} \\ $$$$\:{of}\:{trapezoid}\:{that}\:{can}\:{be}\:{drawn}\: \\ $$$$\:{under}\:{the}\:{x}−{axis}\:{so}\:{that}\:{one}\: \\ $$$$\:{of}\:{its}\:{bases}\:{is}\:{on}\:{the}\:{x}−{axis}\: \\ $$$$\:{and}\:{the}\:{other}\:{two}\:{vertices}\:{are} \\ $$$$\:{on}\:{the}\:{curve}\:{y}={x}^{\mathrm{2}} −\mathrm{9}\: \\ $$

Question Number 185451 Answers: 0 Comments: 2

Question Number 185450 Answers: 0 Comments: 1

Write down the expansion of (a) cosx (b) (1/(l+x)), and hence show that ((cosx)/(1+x)) = 1−x+(x^2 /2)−(x^3 /2)+((13x^4 )/(24))+... M.m

$$\mathrm{Write}\:\mathrm{down}\:\mathrm{the}\:\mathrm{expansion}\:\mathrm{of}\: \\ $$$$\left(\mathrm{a}\right)\:\mathrm{cosx}\:\left(\mathrm{b}\right)\:\frac{\mathrm{1}}{\mathrm{l}+\mathrm{x}},\:\mathrm{and}\:\mathrm{hence}\:\mathrm{show} \\ $$$$\mathrm{that}\:\frac{\mathrm{cosx}}{\mathrm{1}+\mathrm{x}}\:=\:\mathrm{1}−\mathrm{x}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{2}}+\frac{\mathrm{13x}^{\mathrm{4}} }{\mathrm{24}}+... \\ $$$$ \\ $$$$ \\ $$$$\mathrm{M}.\mathrm{m} \\ $$

Question Number 185449 Answers: 1 Comments: 0

Find the first four terms in the expansion of ((x−3)/((1−x^2 )^2 (2+x^2 ))) in ascending power of x. M.m

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{first}\:\mathrm{four}\:\mathrm{terms}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{expansion}\:\mathrm{of}\:\frac{\mathrm{x}−\mathrm{3}}{\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} \left(\mathrm{2}+\mathrm{x}^{\mathrm{2}} \right)}\:\mathrm{in} \\ $$$$\mathrm{ascending}\:\mathrm{power}\:\mathrm{of}\:\mathrm{x}. \\ $$$$ \\ $$$$ \\ $$$$\mathrm{M}.\mathrm{m} \\ $$

Question Number 185446 Answers: 0 Comments: 0

Question Number 185438 Answers: 1 Comments: 0

lim_(x→0) ((a^x sin (ax)−b^x sin (bx))/(c^x sin (cx)−d^x sin dx)))= ?

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{a}^{{x}} \mathrm{sin}\:\left({ax}\right)−{b}^{{x}} \mathrm{sin}\:\left({bx}\right)}{\left.{c}^{{x}} \mathrm{sin}\:\left({cx}\right)−{d}^{{x}} \mathrm{sin}\:{dx}\right)}=\:? \\ $$

Question Number 185419 Answers: 0 Comments: 0

What color do I write in this app, it does not change from black to other colors, what is the reason

$$ \\ $$What color do I write in this app, it does not change from black to other colors, what is the reason

Question Number 185414 Answers: 0 Comments: 1

2^x =a 3^y =b : 6^(xy) =? ⇒6^(xy) =(2×3)^(xy) ⇒2^(xy) ×3^(xy) ⇒(2^x )^y ×(3^y )^x ⇒a^y ×b^x

$$\mathrm{2}^{{x}} ={a} \\ $$$$\mathrm{3}^{{y}} ={b}\:\:\:\:\:\::\:\mathrm{6}^{{xy}} =? \\ $$$$\Rightarrow\mathrm{6}^{{xy}} =\left(\mathrm{2}×\mathrm{3}\right)^{{xy}} \Rightarrow\mathrm{2}^{{xy}} ×\mathrm{3}^{{xy}} \Rightarrow\left(\mathrm{2}^{{x}} \right)^{{y}} ×\left(\mathrm{3}^{{y}} \right)^{{x}} \Rightarrow{a}^{{y}} ×{b}^{{x}} \\ $$

Question Number 185413 Answers: 1 Comments: 0

2^x =a

$$\mathrm{2}^{{x}} ={a} \\ $$$$ \\ $$

Question Number 185406 Answers: 1 Comments: 0

If a,b,c and d are constants such that lim_(x→0) ((ax^2 +sin bx+sin cx+sin dx)/(3x^2 +5x^4 +7x^6 ))=8 find the value of the sum a+b+c+d

$${If}\:{a},{b},{c}\:{and}\:{d}\:{are}\:{constants}\:{such}\:{that} \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{ax}^{\mathrm{2}} +\mathrm{sin}\:{bx}+\mathrm{sin}\:{cx}+\mathrm{sin}\:{dx}}{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{5}{x}^{\mathrm{4}} +\mathrm{7}{x}^{\mathrm{6}} }=\mathrm{8} \\ $$$${find}\:{the}\:{value}\:{of}\:{the}\:{sum}\:{a}+{b}+{c}+{d} \\ $$

Question Number 185405 Answers: 1 Comments: 0

lim_(x→∞) (((x+2)^(1/x) −x^(1/x) )/((x+3)^(1/x) −x^(1/x) )) =?

$$\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left({x}+\mathrm{2}\right)^{\mathrm{1}/{x}} −{x}^{\mathrm{1}/{x}} }{\left({x}+\mathrm{3}\right)^{\mathrm{1}/{x}} −{x}^{\mathrm{1}/{x}} }\:=? \\ $$

Question Number 185403 Answers: 2 Comments: 0

Σ_(k=1) ^(16) ((√k)−(√(k−1)))=?

$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{16}} {\sum}}\left(\sqrt{{k}}−\sqrt{{k}−\mathrm{1}}\right)=? \\ $$

Question Number 185402 Answers: 0 Comments: 0

Question Number 185401 Answers: 0 Comments: 1

Show that ((1+z)/(1−z)) + ((1+z^ )/(1−z^ )) = ((2(1−∣z∣^2 ))/(∣1−z∣^2 )) Where z is a complex number Help!

$$\mathrm{Show}\:\mathrm{that}\:\frac{\mathrm{1}+\mathrm{z}}{\mathrm{1}−\mathrm{z}}\:+\:\frac{\mathrm{1}+\bar {\mathrm{z}}}{\mathrm{1}−\bar {\mathrm{z}}}\:=\:\frac{\mathrm{2}\left(\mathrm{1}−\mid\mathrm{z}\mid^{\mathrm{2}} \right)}{\mid\mathrm{1}−\mathrm{z}\mid^{\mathrm{2}} } \\ $$$$\mathrm{Where}\:\mathrm{z}\:\mathrm{is}\:\mathrm{a}\:\mathrm{complex}\:\mathrm{number} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Help}! \\ $$

Question Number 185395 Answers: 1 Comments: 0

Question Number 185394 Answers: 1 Comments: 0

Question Number 185387 Answers: 2 Comments: 0

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