Question and Answers Forum

All Questions   Topic List

AllQuestion and Answers: Page 68

Question Number 216645    Answers: 0   Comments: 0

Question Number 216665    Answers: 1   Comments: 1

Question Number 216638    Answers: 1   Comments: 1

without using LHopital rule evalute lim_(x→0) ((ln(1−x)−sin(x) )/(1−cox^2 (x)))

$$\:\:\boldsymbol{{without}}\:\boldsymbol{{using}}\:\boldsymbol{{LHopital}} \\ $$$$\:\:\:\boldsymbol{{rule}}\:\boldsymbol{{evalute}}\: \\ $$$$\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\boldsymbol{{ln}}\left(\mathrm{1}−{x}\right)−\boldsymbol{{sin}}\left(\boldsymbol{{x}}\right)\:}{\mathrm{1}−\boldsymbol{{cox}}^{\mathrm{2}} \left(\boldsymbol{{x}}\right)} \\ $$

Question Number 216630    Answers: 2   Comments: 0

the circles x² + y² -4x -2y +3 =0 and x² +y² + 2x +4y -3 =0 touches each other Find the coordinates of the point of contact

the circles x² + y² -4x -2y +3 =0 and x² +y² + 2x +4y -3 =0 touches each other Find the coordinates of the point of contact

Question Number 216621    Answers: 2   Comments: 0

Question Number 216618    Answers: 2   Comments: 0

∫_( 0) ^( 2π) (√(1 − cos^2 x)) dx Is the answer 0 or 4????

$$\int_{\:\mathrm{0}} ^{\:\mathrm{2}\pi} \:\sqrt{\mathrm{1}\:\:−\:\:\mathrm{cos}^{\mathrm{2}} \mathrm{x}}\:\:\mathrm{dx} \\ $$$$\mathrm{Is}\:\mathrm{the}\:\mathrm{answer}\:\:\mathrm{0}\:\:\:\mathrm{or}\:\:\:\:\mathrm{4}???? \\ $$

Question Number 216615    Answers: 2   Comments: 0

∫_0 ^1 sin nπx J_0 (j_(0m) x)dx

$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{sin}\:{n}\pi{x}\:{J}_{\mathrm{0}} \left({j}_{\mathrm{0}{m}} {x}\right){dx} \\ $$

Question Number 216613    Answers: 0   Comments: 1

∫_a ^( x) ((1−bln (x/a))/( (√(1−(1−bln (x/a))^2 )))) dx

$$\int_{{a}} ^{\:{x}} \frac{\mathrm{1}−{b}\mathrm{ln}\:\frac{{x}}{{a}}}{\:\sqrt{\mathrm{1}−\left(\mathrm{1}−{b}\mathrm{ln}\:\frac{{x}}{{a}}\right)^{\mathrm{2}} }}\:{dx} \\ $$

Question Number 216607    Answers: 1   Comments: 0

Question Number 216596    Answers: 1   Comments: 0

∫∫_(D) ((sin(x^2 +y^2 )+tan(x^2 +y^2 ))/(cos(x^2 +y^2 )+tan(x^2 +y^2 )))dxdy D=[0,(π/4)]×[0,(π/4)]

$$\underset{\mathcal{D}} {\int\int}\:\:\:\frac{\mathrm{sin}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)+\mathrm{tan}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)}{\mathrm{cos}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)+\mathrm{tan}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)}\mathrm{d}{x}\mathrm{d}{y} \\ $$$$\mathcal{D}=\left[\mathrm{0},\frac{\pi}{\mathrm{4}}\right]×\left[\mathrm{0},\frac{\pi}{\mathrm{4}}\right] \\ $$

Question Number 216593    Answers: 2   Comments: 0

Question Number 216592    Answers: 0   Comments: 0

if we have the following system: ((tanx)/(tany))=a x±y=α we have the general candition: −1≤(((a−1)/(a+1)))sinα≤1 if you apply the general candition by the following system it does not give us the reality, despite this system have the solution. ((tanx)/(tany))=−3 x−y=(π/2)

$${if}\:{we}\:{have}\:{the}\:{following}\:{system}: \\ $$$$\frac{{tanx}}{{tany}}={a} \\ $$$${x}\pm{y}=\alpha \\ $$$${we}\:{have}\:{the}\:{general}\:{candition}: \\ $$$$−\mathrm{1}\leqslant\left(\frac{{a}−\mathrm{1}}{{a}+\mathrm{1}}\right){sin}\alpha\leqslant\mathrm{1} \\ $$$${if}\:{you}\:{apply}\:{the}\:{general}\:{candition}\:{by} \\ $$$${the}\:{following}\:{system}\:{it}\:{does}\:{not}\:{give} \\ $$$${us}\:{the}\:{reality},\:{despite}\:{this}\:{system}\:{have} \\ $$$${the}\:{solution}. \\ $$$$\frac{{tanx}}{{tany}}=−\mathrm{3} \\ $$$${x}−{y}=\frac{\pi}{\mathrm{2}} \\ $$

Question Number 216587    Answers: 1   Comments: 4

Question Number 216585    Answers: 0   Comments: 3

is this right? i had let θ= { ((tan^(−1) ((d/c))),((c>0,d>0))),((π−tan^(−1) ((d/c))),((c<0,d>0))),((−π+tan^(−1) ((d/c))),((c<0,d<0))),((−tan^(−1) ((d/c))),((c>0,d<0))) :} before i calculated below (a+bi)^(c+di) =∣a+di∣^(c+di) e^(i(c+di)θ) =∣a+bi∣^c ∣a+bi∣^di e^(icθ) e^(−dθ) =((√(a^2 +b^2 )))^c ((√(a^2 +b^2 )))^di e^(icθ) e^(−dθ) =((√(a^2 +b^2 )))^c e^((1/2)idln(a^2 +b^2 )) e^(icθ) e^(−dθ) =((√(a^2 +b^2 )))^c e^(−dθ) e^(i(dln(a^2 +b^2 )+cθ)) =((√(a^2 +b^2 )))^c e^(−dθ) (cos(dln(a^2 +b^2 )+cθ)+isin(dln(a^2 +b^2 )+cθ))

$$\mathrm{is}\:\mathrm{this}\:\mathrm{right}? \\ $$$$\mathrm{i}\:\mathrm{had}\:\mathrm{let}\:\theta=\begin{cases}{\mathrm{tan}^{−\mathrm{1}} \left(\frac{{d}}{{c}}\right)}&{\left({c}>\mathrm{0},{d}>\mathrm{0}\right)}\\{\pi−\mathrm{tan}^{−\mathrm{1}} \left(\frac{{d}}{{c}}\right)}&{\left({c}<\mathrm{0},{d}>\mathrm{0}\right)}\\{−\pi+\mathrm{tan}^{−\mathrm{1}} \left(\frac{{d}}{{c}}\right)}&{\left({c}<\mathrm{0},{d}<\mathrm{0}\right)}\\{−\mathrm{tan}^{−\mathrm{1}} \left(\frac{{d}}{{c}}\right)}&{\left({c}>\mathrm{0},{d}<\mathrm{0}\right)}\end{cases}\:\mathrm{before}\:\mathrm{i}\:\mathrm{calculated}\:\mathrm{below} \\ $$$$\left({a}+{bi}\right)^{{c}+{di}} =\mid{a}+{di}\mid^{{c}+{di}} {e}^{{i}\left({c}+{di}\right)\theta} \\ $$$$=\mid{a}+{bi}\mid^{{c}} \mid{a}+{bi}\mid^{{di}} {e}^{{ic}\theta} {e}^{−{d}\theta} \\ $$$$=\left(\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)^{{c}} \left(\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)^{{di}} {e}^{{ic}\theta} {e}^{−{d}\theta} \\ $$$$=\left(\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)^{{c}} {e}^{\frac{\mathrm{1}}{\mathrm{2}}{id}\mathrm{ln}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)} {e}^{{ic}\theta} {e}^{−{d}\theta} \\ $$$$=\left(\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)^{{c}} {e}^{−{d}\theta} {e}^{{i}\left({d}\mathrm{ln}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)+{c}\theta\right)} \\ $$$$=\left(\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)^{{c}} {e}^{−{d}\theta} \left(\mathrm{cos}\left({d}\mathrm{ln}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)+{c}\theta\right)+{i}\mathrm{sin}\left({d}\mathrm{ln}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)+{c}\theta\right)\right) \\ $$

Question Number 216582    Answers: 0   Comments: 3

using first principle solve y=((x+2)/( (√x)+2)) is it possible with first principle

$$ \\ $$$$\mathrm{using}\:\mathrm{first}\:\mathrm{principle}\:\mathrm{solve} \\ $$$$\mathrm{y}=\frac{\mathrm{x}+\mathrm{2}}{\:\sqrt{\mathrm{x}}+\mathrm{2}} \\ $$$$\mathrm{is}\:\mathrm{it}\:\mathrm{possible}\:\mathrm{with}\:\mathrm{first}\:\mathrm{principle} \\ $$

Question Number 216579    Answers: 0   Comments: 0

∫_0 ^(π/4) ∫_0 ^(π/4) ((tan(x^2 +y^2 )+sin(x^2 +y^2 ))/(tan(x^2 +y^2 )+cos(x^2 +y^2 )))dxdy

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{tan}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)+\mathrm{sin}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)}{\mathrm{tan}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)+\mathrm{cos}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)}{dxdy} \\ $$

Question Number 216560    Answers: 1   Comments: 0

An object of mass M, initially at rest at the coordinate origin, explodes into three parts. Fragment A has a mass M/2, and fragments B and C have a mass M/4 each. After the explosion, fragment A moves in the +X direction at 10 m/s and fragment B moves in the +Y direction at 8 m/s. Find the direction and speed of fragment C

An object of mass M, initially at rest at the coordinate origin, explodes into three parts. Fragment A has a mass M/2, and fragments B and C have a mass M/4 each. After the explosion, fragment A moves in the +X direction at 10 m/s and fragment B moves in the +Y direction at 8 m/s. Find the direction and speed of fragment C

Question Number 216542    Answers: 0   Comments: 1

if y=cosu then prove that y′=sinu∙u′ by newton′s formula.

$${if}\:{y}={cosu}\:{then}\:{prove}\:{that}\:{y}'={sinu}\centerdot{u}' \\ $$$${by}\:{newton}'{s}\:{formula}. \\ $$

Question Number 216538    Answers: 0   Comments: 4

Find all integer solutions of 3^m =2n^2 +1. I only found m=1, 2, 5 by computer from m=1 to m=30000. Is there any greater solutions?

$$\mathrm{Find}\:\mathrm{all}\:\mathrm{integer}\:\mathrm{solutions}\:\mathrm{of} \\ $$$$\mathrm{3}^{{m}} =\mathrm{2}{n}^{\mathrm{2}} +\mathrm{1}. \\ $$$$ \\ $$$${I}\:{only}\:{found}\:{m}=\mathrm{1},\:\mathrm{2},\:\mathrm{5}\:{by}\:{computer} \\ $$$${from}\:{m}=\mathrm{1}\:{to}\:{m}=\mathrm{30000}. \\ $$$${Is}\:{there}\:{any}\:{greater}\:{solutions}? \\ $$

Question Number 216537    Answers: 0   Comments: 0

using first principle solve y=((x+2)/( (√x)+2)) is it possible with first principle

$$\mathrm{using}\:\mathrm{first}\:\mathrm{principle}\:\mathrm{solve} \\ $$$$\mathrm{y}=\frac{\mathrm{x}+\mathrm{2}}{\:\sqrt{\mathrm{x}}+\mathrm{2}} \\ $$$$\mathrm{is}\:\mathrm{it}\:\mathrm{possible}\:\mathrm{with}\:\mathrm{first}\:\mathrm{principle} \\ $$

Question Number 216534    Answers: 0   Comments: 24

Question Number 216572    Answers: 1   Comments: 0

Question Number 216532    Answers: 2   Comments: 0

Let f :R_+ →R such as f(xy)=f(x)+f(y) 1) Prove that f is derivable iff f is derivable at x=1. 2) Prove that if so, f(x)=Log_a x) where a is positive value to precise

$${Let}\:{f}\::\mathbb{R}_{+} \rightarrow\mathbb{R}\:{such}\:{as}\:{f}\left({xy}\right)={f}\left({x}\right)+{f}\left({y}\right) \\ $$$$\left.\mathrm{1}\right)\:{Prove}\:{that}\:\:{f}\:{is}\:{derivable}\:{iff}\:\: \\ $$$${f}\:{is}\:{derivable}\:{at}\:{x}=\mathrm{1}. \\ $$$$\left.\mathrm{2}\left.\right)\:{Prove}\:{that}\:{if}\:{so},\:{f}\left({x}\right)={Log}_{{a}} {x}\right)\: \\ $$$${where}\:{a}\:{is}\:{positive}\:{value}\:{to}\:{precise} \\ $$

Question Number 216526    Answers: 1   Comments: 0

Question Number 216525    Answers: 2   Comments: 0

Question Number 216515    Answers: 2   Comments: 0

  Pg 63      Pg 64      Pg 65      Pg 66      Pg 67      Pg 68      Pg 69      Pg 70      Pg 71      Pg 72   

Terms of Service

Privacy Policy

Contact: info@tinkutara.com