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Question Number 8150    Answers: 0   Comments: 1

show that the plane x+2y−3z+d=0 is perpendiculr to each of the plane is 2x+5y+4z+1=0 and 4x+7y+3z+2=0 .

$${show}\:{that}\:{the}\:{plane}\:\:\:{x}+\mathrm{2}{y}−\mathrm{3}{z}+{d}=\mathrm{0}\:\:{is}\:{perpendiculr}\:{to}\:{each}\:{of} \\ $$$${the}\:{plane}\:{is}\:\:\:\mathrm{2}{x}+\mathrm{5}{y}+\mathrm{4}{z}+\mathrm{1}=\mathrm{0}\:\:{and}\:\:\:\mathrm{4}{x}+\mathrm{7}{y}+\mathrm{3}{z}+\mathrm{2}=\mathrm{0}\:.\: \\ $$

Question Number 8151    Answers: 1   Comments: 0

what is rectangular paralleopiped?

$${what}\:{is}\:\:{rectangular}\:\:\:{paralleopiped}? \\ $$

Question Number 8139    Answers: 1   Comments: 4

is it correct? when S_n ={ 1×2+1×3+1×4+1×5+.......+1×n +2×3+2×4+2×5+.......+2×n +3×4+3×5+.......+3×n +4×5+.......+4×n .... +(n−1)×n } find S_n . /////////////// S_n +S_n +(1^2 +2^2 +3^2 +4^2 +...+n^2 )={ 1×1+1×2+1×3+1×4+.......+1×n+ 2×1+2×2+2×3+2×4+.......+2×n+ 3×1+3×2+3×3+3×4+.......+3×n+ 4×1+4×1+4×3+4×4+.......+4×n+ ... n×1+n×2+n×3+n×4+......+n×n } ⇔ =(1+2+3+4+...+n)(1+2+3+4+...+n) ⇔ 2S_n +((n(n+1)(2n+1))/6)={((n(n+1))/2)}^2 2S_n =((n(n+1))/2){((n(n+1))/2)−((2n+1)/3)} =((n(n+1))/2)×((3n^2 −n−2)/6) S_n =((n(n+1))/4)×(((3n+2)(n−1))/6) S_n =(((n−1)n(n+1)(3n+2))/(24))

$${is}\:{it}\:{correct}? \\ $$$${when} \\ $$$${S}_{{n}} =\left\{\right. \\ $$$$\mathrm{1}×\mathrm{2}+\mathrm{1}×\mathrm{3}+\mathrm{1}×\mathrm{4}+\mathrm{1}×\mathrm{5}+.......+\mathrm{1}×{n} \\ $$$$\:\:\:\:\:\:\:\:\:\:+\mathrm{2}×\mathrm{3}+\mathrm{2}×\mathrm{4}+\mathrm{2}×\mathrm{5}+.......+\mathrm{2}×{n} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{3}×\mathrm{4}+\mathrm{3}×\mathrm{5}+.......+\mathrm{3}×{n} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{4}×\mathrm{5}+.......+\mathrm{4}×{n} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:.... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left({n}−\mathrm{1}\right)×{n} \\ $$$$\left.\right\} \\ $$$${find}\:{S}_{{n}} \:. \\ $$$$/////////////// \\ $$$${S}_{{n}} +{S}_{{n}} +\left(\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} +...+{n}^{\mathrm{2}} \right)=\left\{\right. \\ $$$$\mathrm{1}×\mathrm{1}+\mathrm{1}×\mathrm{2}+\mathrm{1}×\mathrm{3}+\mathrm{1}×\mathrm{4}+.......+\mathrm{1}×{n}+ \\ $$$$\mathrm{2}×\mathrm{1}+\mathrm{2}×\mathrm{2}+\mathrm{2}×\mathrm{3}+\mathrm{2}×\mathrm{4}+.......+\mathrm{2}×{n}+ \\ $$$$\mathrm{3}×\mathrm{1}+\mathrm{3}×\mathrm{2}+\mathrm{3}×\mathrm{3}+\mathrm{3}×\mathrm{4}+.......+\mathrm{3}×{n}+ \\ $$$$\mathrm{4}×\mathrm{1}+\mathrm{4}×\mathrm{1}+\mathrm{4}×\mathrm{3}+\mathrm{4}×\mathrm{4}+.......+\mathrm{4}×{n}+ \\ $$$$... \\ $$$$\left.{n}×\mathrm{1}+{n}×\mathrm{2}+{n}×\mathrm{3}+{n}×\mathrm{4}+......+{n}×{n}\:\right\} \\ $$$$\Leftrightarrow \\ $$$$=\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+...+{n}\right)\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+...+{n}\right) \\ $$$$\Leftrightarrow \\ $$$$\mathrm{2}{S}_{{n}} +\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}=\left\{\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\right\}^{\mathrm{2}} \\ $$$$\mathrm{2}{S}_{{n}} =\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\left\{\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}−\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{3}}\right\} \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}×\frac{\mathrm{3}{n}^{\mathrm{2}} −{n}−\mathrm{2}}{\mathrm{6}} \\ $$$${S}_{{n}} =\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{4}}×\frac{\left(\mathrm{3}{n}+\mathrm{2}\right)\left({n}−\mathrm{1}\right)}{\mathrm{6}} \\ $$$${S}_{{n}} =\frac{\left({n}−\mathrm{1}\right){n}\left({n}+\mathrm{1}\right)\left(\mathrm{3}{n}+\mathrm{2}\right)}{\mathrm{24}} \\ $$$$ \\ $$

Question Number 8134    Answers: 1   Comments: 0

Question Number 8132    Answers: 1   Comments: 1

The coefficient of x^(n−2) in the polynomial (x−1)(x−2)....(x−n) is

$$\mathrm{The}\:\:\mathrm{coefficient}\:\mathrm{of}\:{x}^{{n}−\mathrm{2}} \:\mathrm{in}\:\mathrm{the}\:\mathrm{polynomial} \\ $$$$\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)....\left({x}−{n}\right)\:\:\mathrm{is} \\ $$

Question Number 8130    Answers: 0   Comments: 0

It is desired to set the operating point at 3.5V, 2.2mA by biasing a silicon transistor with feedback resistor). if Beta = 100, find the value of R_b and R_c taking V_(be) = 0.7V

$${It}\:{is}\:{desired}\:{to}\:{set}\:{the}\:{operating}\:{point}\:{at}\:\mathrm{3}.\mathrm{5}{V},\: \\ $$$$\mathrm{2}.\mathrm{2}{mA}\:{by}\:{biasing}\:{a}\:{silicon}\:{transistor}\:{with}\: \\ $$$$\left.{feedback}\:{resistor}\right).\:{if}\:\:{Beta}\:=\:\mathrm{100},\:{find}\:{the}\:{value} \\ $$$${of}\:{R}_{{b}} \:{and}\:{R}_{{c}} \:\:{taking}\:\:{V}_{{be}} \:=\:\mathrm{0}.\mathrm{7}{V} \\ $$

Question Number 8129    Answers: 0   Comments: 0

An ellipse having focii at (3 3)and (−4 4) and passing through origin has e??

$${An}\:{ellipse}\:{having}\:{focii}\:{at}\:\left(\mathrm{3}\:\mathrm{3}\right){and}\:\left(−\mathrm{4}\:\mathrm{4}\right)\:{and}\:{passing}\:{through}\:{origin}\:{has}\:{e}?? \\ $$

Question Number 8127    Answers: 1   Comments: 3

For ∣x∣<1, we have that (1+x)^(1/2) =1+(1/2)x+((((1/2))((1/2)−1))/(2!))x^2 +(((1/2)((1/2)−1)((1/2)−2))/(3!))x^3 +... (1+x)^(1/2) =1+Σ_(r=1) ^∞ ((Π_(k=0) ^(r−1) (0.5−k))/(r!))x^r . Let g(r)=Π_(k=0) ^(r−1) (0.5−k). Is it true that for x=(1/2)i⇒∣x∣=0.5<1 (1+(1/2)i)^(1/2) =1+Σ_(r=1) ^∞ ((g(r))/(r!))×(1/2^r )i^r ? (i=(√(−1)))

$${For}\:\mid{x}\mid<\mathrm{1},\:{we}\:{have}\:{that} \\ $$$$\left(\mathrm{1}+{x}\right)^{\mathrm{1}/\mathrm{2}} =\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}{x}+\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right)}{\mathrm{2}!}{x}^{\mathrm{2}} +\frac{\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{2}\right)}{\mathrm{3}!}{x}^{\mathrm{3}} +... \\ $$$$\left(\mathrm{1}+{x}\right)^{\mathrm{1}/\mathrm{2}} =\mathrm{1}+\underset{{r}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\underset{{k}=\mathrm{0}} {\overset{{r}−\mathrm{1}} {\prod}}\left(\mathrm{0}.\mathrm{5}−{k}\right)}{{r}!}{x}^{{r}} . \\ $$$${Let}\:{g}\left({r}\right)=\underset{{k}=\mathrm{0}} {\overset{{r}−\mathrm{1}} {\prod}}\left(\mathrm{0}.\mathrm{5}−{k}\right). \\ $$$${Is}\:{it}\:{true}\:{that}\:{for}\:{x}=\frac{\mathrm{1}}{\mathrm{2}}{i}\Rightarrow\mid{x}\mid=\mathrm{0}.\mathrm{5}<\mathrm{1} \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}{i}\right)^{\mathrm{1}/\mathrm{2}} =\mathrm{1}+\underset{{r}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{g}\left({r}\right)}{{r}!}×\frac{\mathrm{1}}{\mathrm{2}^{{r}} }{i}^{{r}} \:\:? \\ $$$$\left({i}=\sqrt{−\mathrm{1}}\right) \\ $$

Question Number 8115    Answers: 2   Comments: 0

if xy+y^2 =1. Find (d^2 y/dx) at (0,1)

$${if}\:{xy}+{y}^{\mathrm{2}} =\mathrm{1}.\:{Find}\:\frac{{d}^{\mathrm{2}} {y}}{{dx}}\:{at}\:\left(\mathrm{0},\mathrm{1}\right) \\ $$

Question Number 8114    Answers: 2   Comments: 0

Evaluate ∫cos^6 x

$${Evaluate}\:\int\mathrm{cos}\:^{\mathrm{6}} {x} \\ $$

Question Number 8113    Answers: 1   Comments: 0

Find an equation of the tangent line to the curve y=tan^2 x at the point((x/3) , 0)

$${Find}\:{an}\:{equation}\:{of}\:{the}\:{tangent} \\ $$$${line}\:{to}\:{the}\:{curve}\:{y}={tan}^{\mathrm{2}} {x}\:{at}\:{the}\: \\ $$$${point}\left(\frac{{x}}{\mathrm{3}}\:,\:\mathrm{0}\right) \\ $$

Question Number 8107    Answers: 1   Comments: 0

The general term in the expansion of (1−2x)^(3/4) is

$$\mathrm{The}\:\mathrm{general}\:\mathrm{term}\:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion}\:\mathrm{of} \\ $$$$\left(\mathrm{1}−\mathrm{2}{x}\right)^{\mathrm{3}/\mathrm{4}} \:\:\mathrm{is} \\ $$

Question Number 8105    Answers: 1   Comments: 1

Prove that ,for any acute angle α secα cosecα +tanα +cotα ≥4.

$${Prove}\:{that}\:,{for}\:{any}\:{acute}\:{angle}\:\alpha\: \\ $$$${sec}\alpha\:{cosec}\alpha\:+{tan}\alpha\:+{cot}\alpha\:\geqslant\mathrm{4}. \\ $$

Question Number 8093    Answers: 0   Comments: 10

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$$\mathrm{App}\:\mathrm{has}\:\mathrm{been}\:\mathrm{updated}\:\left(\mathrm{ver}\:\mathrm{1}.\mathrm{48}\right) \\ $$$$\bullet\:\mathrm{Customizable}\:\mathrm{key}\:\mathrm{screen}\:\bigstar \\ $$$$\:\:\:\:\mathrm{long}\:\mathrm{press}\:\mathrm{on}\:\mathrm{any}\:\mathrm{key}\:\mathrm{on}\:\mathrm{that}\:\mathrm{screen} \\ $$$$\:\:\:\:\mathrm{to}\:\mathrm{change}\:\mathrm{the}\:\mathrm{symbol}\:\mathrm{on}\:\mathrm{that}\:\mathrm{location} \\ $$$$\bullet\:\:\mathrm{Choose}\:\mathrm{default}\:\mathrm{font}\:\mathrm{size}\:\mathrm{and}\:\mathrm{style} \\ $$$$\bullet\:\:\mathrm{option}\:\mathrm{menu}\:\mathrm{for}\:\mathrm{quicker}\:\mathrm{access}\:\mathrm{to}\:\mathrm{long} \\ $$$$\:\:\:\:\:\mathrm{press}\:\mathrm{menu} \\ $$$$\bullet\:\:\:\mathrm{new}\:\mathrm{symbols}\:\mathrm{and}\:\mathrm{right}\:\mathrm{curly}\:\mathrm{brace}\:\mathrm{matrix} \\ $$$$\bullet\:\:\:\mathrm{Ability}\:\mathrm{to}\:\mathrm{change}\:\mathrm{selected}\:\mathrm{text}\:\mathrm{to}\:\mathrm{bold}/\mathrm{italic} \\ $$$$\mathrm{Please}\:\mathrm{update}\:\mathrm{to}\:\mathrm{version}\:\mathrm{1}.\mathrm{48}. \\ $$$$\mathrm{For}\:\mathrm{any}\:\mathrm{issues}/\mathrm{questions}.\:\mathrm{Please}\:\mathrm{contact}\:\mathrm{us} \\ $$$$\mathrm{at}\:\mathrm{infoattinkutara}.\mathrm{com} \\ $$

Question Number 8082    Answers: 0   Comments: 0

Question Number 8080    Answers: 0   Comments: 0

Question Number 8073    Answers: 4   Comments: 6

Question Number 8071    Answers: 1   Comments: 2

Solve −2x(x+(7/2))−(x−3)^2 ≤0

$${Solve}\:−\mathrm{2}{x}\left({x}+\frac{\mathrm{7}}{\mathrm{2}}\right)−\left({x}−\mathrm{3}\right)^{\mathrm{2}} \leqslant\mathrm{0} \\ $$

Question Number 8070    Answers: 1   Comments: 0

a line L intersects (0, 0) and the curve y=x^2 at x=t. What is the equation of the line? What is the area between L and y from x=0 to x=t?

$$\mathrm{a}\:\mathrm{line}\:{L}\:\mathrm{intersec}{ts}\:\left(\mathrm{0},\:\mathrm{0}\right)\:{and}\:{the}\:{curve} \\ $$$${y}={x}^{\mathrm{2}} \:\:\:{at}\:{x}={t}. \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{line}? \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{area}\:\mathrm{between}\:{L}\:\mathrm{and}\:{y}\:\mathrm{from} \\ $$$${x}=\mathrm{0}\:\mathrm{to}\:{x}={t}? \\ $$

Question Number 8066    Answers: 1   Comments: 0

calculate lim_(n→∞) (n/2^(√n) )

$${calculate}\: \\ $$$$\:\underset{{n}\rightarrow\infty} {{lim}}\frac{{n}}{\mathrm{2}^{\sqrt{{n}}} } \\ $$$$ \\ $$

Question Number 8050    Answers: 1   Comments: 0

The value of ((18^3 +7^3 +3 ∙ 18 ∙ 7 ∙ 25)/(3^6 +6∙243∙2+15∙181∙4+20∙27∙8+15∙9∙16+6∙3∙32+64)) is

$$\mathrm{The}\:\mathrm{value}\:\mathrm{of} \\ $$$$\frac{\mathrm{18}^{\mathrm{3}} +\mathrm{7}^{\mathrm{3}} +\mathrm{3}\:\centerdot\:\mathrm{18}\:\centerdot\:\mathrm{7}\:\centerdot\:\mathrm{25}}{\mathrm{3}^{\mathrm{6}} +\mathrm{6}\centerdot\mathrm{243}\centerdot\mathrm{2}+\mathrm{15}\centerdot\mathrm{181}\centerdot\mathrm{4}+\mathrm{20}\centerdot\mathrm{27}\centerdot\mathrm{8}+\mathrm{15}\centerdot\mathrm{9}\centerdot\mathrm{16}+\mathrm{6}\centerdot\mathrm{3}\centerdot\mathrm{32}+\mathrm{64}} \\ $$$$\mathrm{is} \\ $$

Question Number 8035    Answers: 0   Comments: 2

prove >> a^n +b^n =c^n [n>2] it has no integer roots

$${prove}\:>>\:{a}^{{n}} +{b}^{{n}} ={c}^{{n}} \:\:\left[{n}>\mathrm{2}\right] \\ $$$${it}\:{has}\:{no}\:{integer}\:{roots} \\ $$$$ \\ $$

Question Number 8032    Answers: 1   Comments: 0

find the real root: 99x^3 +297x^2 +594x−7867=0

$${find}\:{the}\:{real}\:{root}: \\ $$$$\mathrm{99}{x}^{\mathrm{3}} +\mathrm{297}{x}^{\mathrm{2}} +\mathrm{594}{x}−\mathrm{7867}=\mathrm{0} \\ $$

Question Number 8031    Answers: 1   Comments: 0

(√2) ≈((19601)/(13860))

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\sqrt{\mathrm{2}}\:\approx\frac{\mathrm{19601}}{\mathrm{13860}} \\ $$

Question Number 8043    Answers: 0   Comments: 4

solve (xz+y^2 )+(yz−zx^2 )q+2xy+z^2 =0

$${solve}\:\left({xz}+{y}^{\mathrm{2}} \right)+\left(\mathrm{yz}−\mathrm{zx}^{\mathrm{2}} \right)\mathrm{q}+\mathrm{2xy}+\mathrm{z}^{\mathrm{2}} =\mathrm{0} \\ $$

Question Number 8027    Answers: 1   Comments: 1

prove→ any prime number>2 can be written into( x^2 −y^(2 ) ) where (x,y)∈N

$${prove}\rightarrow\:{any}\:{prime}\:{number}>\mathrm{2}\: \\ $$$${can}\:{be}\:{written}\:{into}\left(\:{x}^{\mathrm{2}} −{y}^{\mathrm{2}\:} \right)\:{where} \\ $$$$\left({x},{y}\right)\in{N} \\ $$

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