Question and Answers Forum

AllQuestion and Answers: Page 1000

Question Number 112209 Answers: 0 Comments: 6

A triangle ABC has the following properties BC=1, AB=BC and that the angle bisector from vertex B is also a median. Find all possible triangle(s) with its/their side−lengths and angles.

$$\mathrm{A}\:\mathrm{triangle}\:\mathrm{ABC}\:\mathrm{has}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{properties}\:\mathrm{BC}=\mathrm{1},\:\mathrm{AB}=\mathrm{BC}\:\mathrm{and}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{angle}\:\mathrm{bisector}\:\mathrm{from}\:\mathrm{vertex}\:\mathrm{B}\:\mathrm{is} \\ $$$$\mathrm{also}\:\mathrm{a}\:\mathrm{median}.\:\mathrm{Find}\:\mathrm{all}\:\mathrm{possible} \\ $$$$\mathrm{triangle}\left(\mathrm{s}\right)\:\mathrm{with}\:\mathrm{its}/\mathrm{their} \\ $$$$\mathrm{side}−\mathrm{lengths}\:\mathrm{and}\:\mathrm{angles}. \\ $$

Question Number 112208 Answers: 1 Comments: 3

Question Number 112206 Answers: 0 Comments: 0

∫ln(sin(x)) dx

$$\int{ln}\left({sin}\left({x}\right)\right)\:{dx} \\ $$

Question Number 112199 Answers: 0 Comments: 0

Let Ω denote the circumcircle of ABC. The tangent to Ω at A meets BC at X. Let the angle bisectors of ∠AXB meet AC and AB at E and F respectively. D is the foot of the angle bisector from ∠BAC on BC. Let AD intersect EF at K and Ω again at L(other than A). Prove that AEDF is a rhombus and further prove that the circle defined by triangle KLX passes through the midpoint of line segment BC.

$$\mathrm{Let}\:\Omega\:\mathrm{denote}\:\mathrm{the}\:\mathrm{circumcircle}\:\mathrm{of}\:\mathrm{ABC}. \\ $$$$\mathrm{The}\:\mathrm{tangent}\:\mathrm{to}\:\Omega\:\mathrm{at}\:\mathrm{A}\:\mathrm{meets}\:\mathrm{BC}\:\mathrm{at}\:\mathrm{X}. \\ $$$$\mathrm{Let}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{bisectors}\:\mathrm{of}\:\angle\mathrm{AXB}\:\mathrm{meet} \\ $$$$\mathrm{AC}\:\mathrm{and}\:\mathrm{AB}\:\mathrm{at}\:\mathrm{E}\:\mathrm{and}\:\mathrm{F} \\ $$$$\mathrm{respectively}.\:\mathrm{D}\:\mathrm{is}\:\mathrm{the}\:\mathrm{foot}\:\mathrm{of}\:\mathrm{the}\:\mathrm{angle} \\ $$$$\mathrm{bisector}\:\mathrm{from}\:\angle\mathrm{BAC}\:\mathrm{on}\:\mathrm{BC}.\:\mathrm{Let}\:\mathrm{AD} \\ $$$$\mathrm{intersect}\:\mathrm{EF}\:\mathrm{at}\:\mathrm{K}\:\mathrm{and}\:\Omega\:\mathrm{again}\:\mathrm{at} \\ $$$$\mathrm{L}\left(\mathrm{other}\:\mathrm{than}\:\mathrm{A}\right).\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{AEDF}\:\mathrm{is} \\ $$$$\mathrm{a}\:\mathrm{rhombus}\:\mathrm{and}\:\mathrm{further}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{the} \\ $$$$\mathrm{circle}\:\mathrm{defined}\:\mathrm{by}\:\mathrm{triangle}\:\mathrm{KLX}\:\mathrm{passes} \\ $$$$\mathrm{through}\:\mathrm{the}\:\mathrm{midpoint}\:\mathrm{of}\:\mathrm{line}\:\mathrm{segment} \\ $$$$\mathrm{BC}. \\ $$

Question Number 112195 Answers: 1 Comments: 0

Σ_(n=1 ) ^(11) (((−1)^(n+1) (4n+2))/(4n(n+1))) please help

$$\underset{\mathrm{n}=\mathrm{1}\:} {\overset{\mathrm{11}} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} \left(\mathrm{4n}+\mathrm{2}\right)}{\mathrm{4n}\left(\mathrm{n}+\mathrm{1}\right)} \\ $$$$\mathrm{please}\:\mathrm{help} \\ $$

Question Number 112189 Answers: 0 Comments: 0

prove that ∫_0 ^∞ (1/(cos(x)+sinh(x)))dx=1.4917.

$${prove}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{cos}\left({x}\right)+{sinh}\left({x}\right)}{dx}=\mathrm{1}.\mathrm{4917}. \\ $$

Question Number 112179 Answers: 1 Comments: 1

Question Number 112173 Answers: 1 Comments: 0

find the solution of equation (√(cos 2x−sin^3 x+3)) = sin x

$$\:\mathrm{find}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{equation}\: \\ $$$$\sqrt{\mathrm{cos}\:\mathrm{2x}−\mathrm{sin}\:^{\mathrm{3}} \mathrm{x}+\mathrm{3}}\:=\:\mathrm{sin}\:\mathrm{x}\: \\ $$

Question Number 112169 Answers: 2 Comments: 0

∫ tan^3 x sec^3 x dx ?

$$\:\int\:\mathrm{tan}\:^{\mathrm{3}} {x}\:\mathrm{sec}\:^{\mathrm{3}} {x}\:{dx}\:? \\ $$

Question Number 112163 Answers: 1 Comments: 0

In a class of 30 pupils , 12 are girls and 2 of them are short−sighted. Among 18 boys , 6 are short−sighted . If a pupil is selected at random, what is the probability that the pupil chosen will be { (((a) a girls)),(((b) short−sighted)),(((c) a short−sighted boy)) :}

$$\mathrm{In}\:\mathrm{a}\:\mathrm{class}\:\mathrm{of}\:\mathrm{30}\:\mathrm{pupils}\:,\:\mathrm{12}\:\mathrm{are}\:\mathrm{girls}\:\mathrm{and}\: \\ $$$$\mathrm{2}\:\mathrm{of}\:\mathrm{them}\:\mathrm{are}\:\mathrm{short}−\mathrm{sighted}.\:\mathrm{Among}\: \\ $$$$\mathrm{18}\:\mathrm{boys}\:,\:\mathrm{6}\:\mathrm{are}\:\mathrm{short}−\mathrm{sighted}\:.\:\mathrm{If}\:\mathrm{a}\:\mathrm{pupil} \\ $$$$\mathrm{is}\:\mathrm{selected}\:\mathrm{at}\:\mathrm{random},\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{probability} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{pupil}\:\mathrm{chosen}\:\mathrm{will}\:\mathrm{be}\: \\ $$$$\begin{cases}{\left(\mathrm{a}\right)\:\mathrm{a}\:\mathrm{girls}}\\{\left(\mathrm{b}\right)\:\mathrm{short}−\mathrm{sighted}}\\{\left(\mathrm{c}\right)\:\mathrm{a}\:\mathrm{short}−\mathrm{sighted}\:\mathrm{boy}}\end{cases} \\ $$

Question Number 112162 Answers: 1 Comments: 3

Question Number 112159 Answers: 1 Comments: 0

(1)Find the equation of hyperbola with centre point at (1,−2) and coordinates of foci is (6,−2) and (−4,−2) (2) If hyperbola ((x^2 −2nx+n^2 )/(25))−((y^2 −2my+m^2 )/(16))=1 have a asympyotes passes through at (0,1), then 5m−4n =

$$\left(\mathrm{1}\right)\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{hyperbola}\:\mathrm{with}\: \\ $$$$\mathrm{centre}\:\mathrm{point}\:\mathrm{at}\:\left(\mathrm{1},−\mathrm{2}\right)\:\mathrm{and}\:\mathrm{coordinates} \\ $$$$\mathrm{of}\:\mathrm{foci}\:\mathrm{is}\:\left(\mathrm{6},−\mathrm{2}\right)\:\mathrm{and}\:\left(−\mathrm{4},−\mathrm{2}\right) \\ $$$$\left(\mathrm{2}\right)\:\mathrm{If}\:\mathrm{hyperbola}\:\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{2nx}+\mathrm{n}^{\mathrm{2}} }{\mathrm{25}}−\frac{\mathrm{y}^{\mathrm{2}} −\mathrm{2my}+\mathrm{m}^{\mathrm{2}} }{\mathrm{16}}=\mathrm{1} \\ $$$$\mathrm{have}\:\mathrm{a}\:\mathrm{asympyotes}\:\mathrm{passes}\:\mathrm{through}\:\mathrm{at}\: \\ $$$$\left(\mathrm{0},\mathrm{1}\right),\:\mathrm{then}\:\mathrm{5m}−\mathrm{4n}\:=\: \\ $$

Question Number 112145 Answers: 1 Comments: 0

4^(tan^2 x) + 2^(1/(cos^2 x)) − 80 = 0

$$\mathrm{4}^{\mathrm{tan}\:^{\mathrm{2}} {x}} \:+\:\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{cos}\:^{\mathrm{2}} {x}}} \:−\:\mathrm{80}\:=\:\mathrm{0} \\ $$

Question Number 112136 Answers: 2 Comments: 0

find (5/(6+7sin 2β)) , if tan β = 0.2

$${find}\:\:\:\frac{\mathrm{5}}{\mathrm{6}+\mathrm{7sin}\:\mathrm{2}\beta}\:,\:\:\:{if}\:\:\:\mathrm{tan}\:\beta\:\:=\:\:\mathrm{0}.\mathrm{2} \\ $$

Question Number 112133 Answers: 1 Comments: 0

Question Number 112128 Answers: 5 Comments: 2

Question Number 112318 Answers: 2 Comments: 2

A triangle ABC has the following properties BC=1, AB=AC and that the angle bisector from vertex B is also a median. Find all possible triangle(s) with its/their side−lengths and angles.

$$\mathrm{A}\:\mathrm{triangle}\:\mathrm{ABC}\:\mathrm{has}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{properties}\:\mathrm{BC}=\mathrm{1},\:\boldsymbol{\mathrm{AB}}=\boldsymbol{\mathrm{AC}}\:\mathrm{and}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{angle}\:\mathrm{bisector}\:\mathrm{from}\:\mathrm{vertex}\:\mathrm{B}\:\mathrm{is} \\ $$$$\mathrm{also}\:\mathrm{a}\:\mathrm{median}.\:\mathrm{Find}\:\mathrm{all}\:\mathrm{possible} \\ $$$$\mathrm{triangle}\left(\mathrm{s}\right)\:\mathrm{with}\:\mathrm{its}/\mathrm{their} \\ $$$$\mathrm{side}−\mathrm{lengths}\:\mathrm{and}\:\mathrm{angles}. \\ $$

Question Number 112135 Answers: 3 Comments: 0

solve ∫_(−1) ^1 ∣3^x −2^x ∣dx

$${solve} \\ $$$$\int_{−\mathrm{1}} ^{\mathrm{1}} \mid\mathrm{3}^{{x}} −\mathrm{2}^{{x}} \mid{dx} \\ $$

Question Number 112119 Answers: 1 Comments: 0

....calculus.... prove that::: if Ω =∫_(0 ) ^( 1) ln(ln(1−(√x) ))dx then Re(Ω) := −γ + ln(2).... m.n. july 1970#

$$\:\:\:\:\:\:\:\:\:\:\:\:\:....{calculus}.... \\ $$$${prove}\:{that}::: \\ $$$${if}\:\:\:\Omega\:=\int_{\mathrm{0}\:\:} ^{\:\mathrm{1}} {ln}\left({ln}\left(\mathrm{1}−\sqrt{{x}}\:\right)\right){dx} \\ $$$${then} \\ $$$$\mathscr{R}{e}\left(\Omega\right)\::=\:−\gamma\:+\:{ln}\left(\mathrm{2}\right).... \\ $$$$ \\ $$$${m}.{n}.\:{july}\:\mathrm{1970}# \\ $$

Question Number 112114 Answers: 0 Comments: 0

solution of Φ=∫_0 ^1 xH_x dx =^(γ+ψ(x+1)) ∫_0 ^( 1) x(γ+ψ(x+1))dx =^(ψ(x+1)=(1/x)+ψ(x)) ∫_0 ^( 1) x(γ+(1/x)+ψ(x))dx =(γ/2)+1+ ∫_0 ^( 1) x.(d/dx)(ln(Γ(x))) =(γ/2)+1+[xln(Γ(x))]_0 ^1 −∫_0 ^( 1) ln(Γ(x))dx we know (why?) ∫_0 ^( 1) ln(Γ(x))dx=ln((√(2π)) ) and lim_(x→0^+ ) (xln(Γ(x)))=0 (why??) finally Φ =∫_0 ^( 1) xH_x dx=(γ/2)+1−ln((√(2π))) ✓ m.n.july 1970....

$$\:\:{solution}\:{of} \\ $$$$\Phi=\int_{\mathrm{0}} ^{\mathrm{1}} {xH}_{{x}} {dx}\:\overset{\gamma+\psi\left({x}+\mathrm{1}\right)} {=}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}\left(\gamma+\psi\left({x}+\mathrm{1}\right)\right){dx}\: \\ $$$$\:\overset{\psi\left({x}+\mathrm{1}\right)=\frac{\mathrm{1}}{{x}}+\psi\left({x}\right)} {=}\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}\left(\gamma+\frac{\mathrm{1}}{{x}}+\psi\left({x}\right)\right){dx} \\ $$$$=\frac{\gamma}{\mathrm{2}}+\mathrm{1}+\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}.\frac{{d}}{{dx}}\left({ln}\left(\Gamma\left({x}\right)\right)\right) \\ $$$$=\frac{\gamma}{\mathrm{2}}+\mathrm{1}+\left[{xln}\left(\Gamma\left({x}\right)\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left(\Gamma\left({x}\right)\right){dx} \\ $$$$\:\:{we}\:\:{know}\:\left({why}?\right)\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left(\Gamma\left({x}\right)\right){dx}={ln}\left(\sqrt{\mathrm{2}\pi}\:\right) \\ $$$$\:{and}\:\:{lim}_{{x}\rightarrow\mathrm{0}^{+} } \left({xln}\left(\Gamma\left({x}\right)\right)\right)=\mathrm{0}\:\left({why}??\right) \\ $$$${finally}\:\Phi\:=\int_{\mathrm{0}} ^{\:\mathrm{1}} {xH}_{{x}} {dx}=\frac{\gamma}{\mathrm{2}}+\mathrm{1}−{ln}\left(\sqrt{\mathrm{2}\pi}\right)\:\checkmark \\ $$$$\:\:{m}.{n}.{july}\:\mathrm{1970}.... \\ $$$$\: \\ $$

Question Number 112096 Answers: 3 Comments: 0

Question Number 112097 Answers: 0 Comments: 2

∣∣x−1∣+1∣ < x

$$\:\:\:\:\:\:\mid\mid{x}−\mathrm{1}\mid+\mathrm{1}\mid\:<\:{x} \\ $$

Question Number 112087 Answers: 2 Comments: 0

a(√a) −3 = 10(√a) → (√a) + (√a^(−1) ) =?

$$\:\:\:\mathrm{a}\sqrt{\mathrm{a}}\:−\mathrm{3}\:=\:\mathrm{10}\sqrt{\mathrm{a}}\:\rightarrow\:\sqrt{\mathrm{a}}\:+\:\sqrt{\mathrm{a}^{−\mathrm{1}} }\:=? \\ $$

Question Number 112083 Answers: 2 Comments: 0

lim_(n→∞) 4^n (1−cos ((α/2^n ))) ? (√(bemath))

$$\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{4}^{{n}} \:\left(\mathrm{1}−\mathrm{cos}\:\left(\frac{\alpha}{\mathrm{2}^{{n}} }\right)\right)\:?\: \\ $$$$\:\:\sqrt{{bemath}} \\ $$

Question Number 112076 Answers: 1 Comments: 3

Question Number 112075 Answers: 0 Comments: 2

Are there infinitely many solutions to sin((β/2))=cos(90−(β/2)) ?

$$\mathrm{Are}\:\mathrm{there}\:\mathrm{infinitely}\:\mathrm{many}\:\mathrm{solutions}\:\mathrm{to} \\ $$$$\mathrm{sin}\left(\frac{\beta}{\mathrm{2}}\right)=\mathrm{cos}\left(\mathrm{90}−\frac{\beta}{\mathrm{2}}\right)\:? \\ $$

Pg 995 Pg 996 Pg 997 Pg 998 Pg 999 Pg 1000 Pg 1001 Pg 1002 Pg 1003 Pg 1004

Contact: info@tinkutara.com