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Question Number 95342 Answers: 2 Comments: 0

determine the value of k such that the point A(4,−2,6) B(0,1,0) C(1,0,−5) and D(1,k,−2) lie on the same plane

$$\mathrm{determine}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{k}\:\mathrm{such}\:\mathrm{that}\: \\ $$$$\mathrm{the}\:\mathrm{point}\:\mathrm{A}\left(\mathrm{4},−\mathrm{2},\mathrm{6}\right)\:\mathrm{B}\left(\mathrm{0},\mathrm{1},\mathrm{0}\right)\:\mathrm{C}\left(\mathrm{1},\mathrm{0},−\mathrm{5}\right) \\ $$$$\mathrm{and}\:\mathrm{D}\left(\mathrm{1},\mathrm{k},−\mathrm{2}\right)\:\mathrm{lie}\:\mathrm{on}\:\mathrm{the}\:\mathrm{same} \\ $$$$\mathrm{plane}\: \\ $$

Question Number 94572 Answers: 0 Comments: 0

Question Number 94314 Answers: 0 Comments: 0

Question Number 93874 Answers: 1 Comments: 4

find k if the vector (1^ −2,k) in R^3 be a linear combination of the vectors (3,0,2) &(2,−1,−5)

$$\mathrm{find}\:\mathrm{k}\:\mathrm{if}\:\mathrm{the}\:\mathrm{vector}\:\left(\bar {\mathrm{1}}−\mathrm{2},\mathrm{k}\right)\:\mathrm{in}\:\mathbb{R}^{\mathrm{3}} \\ $$$$\mathrm{be}\:\mathrm{a}\:\mathrm{linear}\:\mathrm{combination}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{vectors}\:\left(\mathrm{3},\mathrm{0},\mathrm{2}\right)\:\&\left(\mathrm{2},−\mathrm{1},−\mathrm{5}\right)\: \\ $$

Question Number 93299 Answers: 0 Comments: 2

find a reduction formulae for I_n = ∫_(−1) ^0 x^n (1 + x)^2 dx

$$\mathrm{find}\:\mathrm{a}\:\mathrm{reduction}\:\mathrm{formulae}\:\mathrm{for}\:{I}_{{n}} \:=\:\int_{−\mathrm{1}} ^{\mathrm{0}} {x}^{{n}} \left(\mathrm{1}\:+\:{x}\right)^{\mathrm{2}} \:{dx} \\ $$

Question Number 93249 Answers: 0 Comments: 4

calculate∫_0 ^((Π/2) ) (dx/(2+cos2x))

$${calculate}\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}\:} \frac{{dx}}{\mathrm{2}+{cos}\mathrm{2}{x}} \\ $$

Question Number 92235 Answers: 0 Comments: 0

let 0<p<1 and x>0 prove that x^2 ≤ (1−p)( ^((1−p)) (√x) ) +p (^p (√x))

$${let}\:\:\mathrm{0}<{p}<\mathrm{1}\:\:{and}\:\:{x}>\mathrm{0} \\ $$$${prove}\:{that}\:\:\:\:{x}^{\mathrm{2}} \leqslant\:\left(\mathrm{1}−{p}\right)\left(\:\:\:^{\left(\mathrm{1}−{p}\right)} \sqrt{{x}}\:\right)\:+{p}\:\left(\:^{{p}} \sqrt{{x}}\right) \\ $$$$ \\ $$$$ \\ $$

Question Number 92521 Answers: 0 Comments: 10

Question Number 90468 Answers: 0 Comments: 1

Question Number 89033 Answers: 1 Comments: 0

Question Number 88860 Answers: 0 Comments: 3

Question Number 88756 Answers: 0 Comments: 0

E is reported in (i^→ ;j^→ ) base. e_1 ^→ =2i^→ +3j^→ ; e_2 ^→ =i^→ −2j^→ and e_3 ^→ =4i^→ −5j^→ belong to E. 1)Determinate the cordonnates of e_3 ^→ in the base B(e_1 ^→ ;e_2 ^→ ).

Question Number 88214 Answers: 2 Comments: 0

Question Number 87759 Answers: 0 Comments: 2

Question Number 87682 Answers: 0 Comments: 0

what is ▽^2 ((1/r^→ )) if ▽^→ = i^ (∂/∂x)+j^ (∂/∂y)+k^ (∂/∂z) and r^→ = i^ x + j^ y + k^ z

$$\mathrm{what}\:\mathrm{is}\:\bigtriangledown^{\mathrm{2}} \left(\frac{\mathrm{1}}{\overset{\rightarrow} {\mathrm{r}}}\right)\:\mathrm{if}\: \\ $$$$\overset{\rightarrow} {\bigtriangledown}\:=\:\hat {\mathrm{i}}\:\frac{\partial}{\partial\mathrm{x}}+\hat {\mathrm{j}}\frac{\partial}{\partial\mathrm{y}}+\hat {\mathrm{k}}\:\frac{\partial}{\partial\mathrm{z}} \\ $$$$\mathrm{and}\:\overset{\rightarrow} {\mathrm{r}}\:=\:\hat {\mathrm{i}x}\:+\:\hat {\mathrm{j}y}\:+\:\hat {\mathrm{k}z}\: \\ $$

Question Number 87508 Answers: 0 Comments: 1

Question Number 86918 Answers: 0 Comments: 2

Question Number 85865 Answers: 1 Comments: 0

prove that curl(r^n c^→ ×r^→ )=(n+2)r^n c^→ −nr^(n−2) (r^→ .c^→ ) . where c is the constant vector.

$$\mathrm{prove}\:\mathrm{that}\: \\ $$$$\mathrm{curl}\left(\mathrm{r}^{\mathrm{n}} \overset{\rightarrow} {\mathrm{c}}×\overset{\rightarrow} {\mathrm{r}}\right)=\left(\mathrm{n}+\mathrm{2}\right)\mathrm{r}^{\mathrm{n}} \overset{\rightarrow} {\mathrm{c}}−\mathrm{nr}^{\mathrm{n}−\mathrm{2}} \left(\overset{\rightarrow} {\mathrm{r}}.\overset{\rightarrow} {\mathrm{c}}\right)\:\:. \\ $$$$\mathrm{where}\:\mathrm{c}\:\mathrm{is}\:\mathrm{the}\:\mathrm{constant}\:\mathrm{vector}. \\ $$

Question Number 84849 Answers: 0 Comments: 5

ABC is a triangle prove that sinA+sinB+sinC>sinA sinB sinC

$${ABC}\:{is}\:{a}\:{triangle}\: \\ $$$${prove}\:{that} \\ $$$${sinA}+{sinB}+{sinC}>{sinA}\:{sinB}\:{sinC} \\ $$

Question Number 84641 Answers: 1 Comments: 1

Question Number 84296 Answers: 0 Comments: 1

Question Number 84275 Answers: 1 Comments: 2

(a.c)b−(a.b)c = a×(c×b) ?

$$\left(\boldsymbol{\mathrm{a}}.\boldsymbol{\mathrm{c}}\right)\boldsymbol{\mathrm{b}}−\left(\boldsymbol{\mathrm{a}}.\boldsymbol{\mathrm{b}}\right)\boldsymbol{\mathrm{c}}\:=\:\boldsymbol{\mathrm{a}}×\left(\boldsymbol{\mathrm{c}}×\boldsymbol{\mathrm{b}}\right)\:? \\ $$

Question Number 83556 Answers: 2 Comments: 0

find the value of (b) wich makes the line y=b divide the tow funtions into tow equal parts 1) f(x)=9−x^2 , g(x)=0 2)f(x)=9−∣x∣ , g(x)=0

$${find}\:{the}\:{value}\:{of}\:\:\left({b}\right)\:{wich}\:{makes}\:{the} \\ $$$${line}\:{y}={b}\:{divide}\:{the}\:{tow}\:{funtions}\:{into} \\ $$$${tow}\:{equal}\:{parts} \\ $$$$ \\ $$$$\left.\mathrm{1}\right)\:{f}\left({x}\right)=\mathrm{9}−{x}^{\mathrm{2}} \:,\:{g}\left({x}\right)=\mathrm{0} \\ $$$$ \\ $$$$\left.\mathrm{2}\right){f}\left({x}\right)=\mathrm{9}−\mid{x}\mid\:,\:{g}\left({x}\right)=\mathrm{0} \\ $$

Question Number 83293 Answers: 0 Comments: 0

Let A (((−2)),((−1)) ) ,B ((1),(3) ) , C (((−10)),(( 5)) ) three given points in the brand (O,I,J) such as OI=OJ and (OI)⊥(OJ) D is a point such as AD=AC+2 and CD=2 Prove correctly that BD=13 .Can you find the coordinate of D?

$${Let}\:\:{A}\begin{pmatrix}{−\mathrm{2}}\\{−\mathrm{1}}\end{pmatrix}\:\:,{B}\begin{pmatrix}{\mathrm{1}}\\{\mathrm{3}}\end{pmatrix}\:,\:{C}\begin{pmatrix}{−\mathrm{10}}\\{\:\mathrm{5}}\end{pmatrix}\:{three}\:{given}\:{points}\:{in}\:{the}\:{brand}\:\left({O},{I},{J}\right)\:{such}\:{as}\:{OI}={OJ}\:{and}\:\left({OI}\right)\bot\left({OJ}\right) \\ $$$$\:{D}\:{is}\:{a}\:{point}\:{such}\:{as}\:{AD}={AC}+\mathrm{2}\:\:{and}\:\:{CD}=\mathrm{2}\: \\ $$$${Prove}\:{correctly}\:{that}\:\:{BD}=\mathrm{13}\:.{Can}\:{you}\:{find}\:{the}\:{coordinate}\:{of}\:{D}? \\ $$

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Question Number 83268 Answers: 1 Comments: 0

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