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AllQuestion and Answers: Page 348

Question Number 186361    Answers: 1   Comments: 1

Question Number 186359    Answers: 2   Comments: 0

Question Number 186348    Answers: 0   Comments: 2

QUIZ: If the sum of the numbers 1 to 100 can be calculated with Carl Gauss′s method, then How do we calculate the sum of the numbers starting from 2 and going up to 100 by twos in an easy way with the same method? Or 2+4+6+8+10+...+100=?

$$\mathrm{Q}{UIZ}: \\ $$$${If}\:{the}\:{sum}\:{of}\:{the}\:{numbers}\:\mathrm{1}\:{to}\:\mathrm{100}\:{can} \\ $$$${be}\:{calculated}\:{with}\:{Carl}\:{Gauss}'{s}\:{method}, \\ $$$${then}\:{How}\:{do}\:{we}\:{calculate}\:{the}\:{sum}\:{of}\:{the} \\ $$$${numbers}\:{starting}\:{from}\:\mathrm{2}\:{and}\:{going}\:{up}\:{to} \\ $$$$\mathrm{100}\:{by}\:{twos}\:{in}\:{an}\:{easy}\:{way}\:{with}\:{the}\:{same} \\ $$$${method}? \\ $$$${Or}\:\mathrm{2}+\mathrm{4}+\mathrm{6}+\mathrm{8}+\mathrm{10}+...+\mathrm{100}=? \\ $$

Question Number 186347    Answers: 1   Comments: 1

∫_(−2) ^2 ((x^5 − 1 + 2)/(x^4 + x −2)) dx

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\underset{−\mathrm{2}} {\overset{\mathrm{2}} {\int}}\:\:\:\:\frac{\boldsymbol{{x}}^{\mathrm{5}} \:−\:\:\mathrm{1}\:\:+\:\:\mathrm{2}}{\boldsymbol{{x}}^{\mathrm{4}} \:\:+\:\:\boldsymbol{{x}}\:\:−\mathrm{2}}\:\:\:\:\boldsymbol{{dx}}\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 186346    Answers: 0   Comments: 0

if S_a =cos(a)+sin(x+a) then ∫(S_1 /S_2 )−((x+S_1 )/(x−S_3 ))=?

$${if}\:{S}_{{a}} =\mathrm{cos}\left({a}\right)+\mathrm{sin}\left({x}+{a}\right) \\ $$$${then}\:\int\frac{{S}_{\mathrm{1}} }{{S}_{\mathrm{2}} }−\frac{{x}+{S}_{\mathrm{1}} }{{x}−{S}_{\mathrm{3}} }=? \\ $$

Question Number 186345    Answers: 1   Comments: 0

Question Number 186337    Answers: 1   Comments: 1

Question Number 186331    Answers: 0   Comments: 5

How many possibilities are there for picking random numbers that ranges from 1-3 that sum up to 7?

$$ \\ $$How many possibilities are there for picking random numbers that ranges from 1-3 that sum up to 7?

Question Number 186330    Answers: 0   Comments: 1

Question Number 186329    Answers: 1   Comments: 0

Question Number 186327    Answers: 0   Comments: 0

lim_(x→0) (((2/3)x^4 +x^2 −tan^2 x)/x^6 ) =?

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\mathrm{2}}{\mathrm{3}}{x}^{\mathrm{4}} +{x}^{\mathrm{2}} −\mathrm{tan}\:^{\mathrm{2}} {x}}{{x}^{\mathrm{6}} }\:=? \\ $$

Question Number 186323    Answers: 1   Comments: 1

Question Number 186322    Answers: 1   Comments: 0

Question Number 186321    Answers: 3   Comments: 0

Question Number 186320    Answers: 1   Comments: 0

Question Number 186310    Answers: 1   Comments: 0

((∫x(x^2 +5)^(1/2) dx − 3∫x(x^2 +5)^(−1/2) dx)/(∫ ((x[(x^2 +5)−3])/( (√(x^2 +5 )))) dx)) =??

$$ \\ $$$$\:\:\:\frac{\int\boldsymbol{{x}}\left(\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{5}\right)^{\mathrm{1}/\mathrm{2}} \boldsymbol{{dx}}\:−\:\mathrm{3}\int\boldsymbol{{x}}\left(\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{5}\right)^{−\mathrm{1}/\mathrm{2}} \:\boldsymbol{{dx}}}{\int\:\:\frac{\boldsymbol{{x}}\left[\left(\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{5}\right)−\mathrm{3}\right]}{\:\sqrt{\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{5}\:\:}}\:\boldsymbol{{dx}}}\:=??\:\:\:\: \\ $$$$ \\ $$

Question Number 186306    Answers: 0   Comments: 2

Evaluate ∫((ln(sin x))/(ln(tan x)+1)) dx

$$\mathrm{Evaluate}\:\int\frac{\mathrm{ln}\left(\mathrm{sin}\:{x}\right)}{\mathrm{ln}\left(\mathrm{tan}\:{x}\right)+\mathrm{1}}\:{dx} \\ $$

Question Number 186305    Answers: 0   Comments: 3

log (((3.2^ )/(3.1^ ))) find Characteristic?

$$\mathrm{log}\:\left(\frac{\mathrm{3}.\bar {\mathrm{2}}}{\mathrm{3}.\bar {\mathrm{1}}}\right)\:\:\:\:\:\:\:{find}\:\mathrm{Characteristic}? \\ $$

Question Number 186302    Answers: 1   Comments: 3

Question Number 186301    Answers: 1   Comments: 0

∫5x+6^x dx

$$\int\mathrm{5}{x}+\mathrm{6}^{{x}} {dx} \\ $$

Question Number 186300    Answers: 1   Comments: 0

solve in R ⌊ 2log_( 8) (x) + (1/3) ⌋ = log_( 4) (x )+ (1/2)

$$\:\:\: \\ $$$$\:\:\:\:\:\mathrm{solve}\:\:\mathrm{in}\:\:\:\mathbb{R} \\ $$$$ \\ $$$$\:\:\:\lfloor\:\:\mathrm{2log}_{\:\mathrm{8}} \left({x}\right)\:+\:\frac{\mathrm{1}}{\mathrm{3}}\:\rfloor\:=\:\mathrm{log}_{\:\mathrm{4}} \left({x}\:\right)+\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$

Question Number 186298    Answers: 1   Comments: 0

Question Number 186597    Answers: 1   Comments: 0

Question Number 186594    Answers: 5   Comments: 1

Question Number 186293    Answers: 0   Comments: 4

Question Number 186292    Answers: 0   Comments: 0

(x−p)(x^3 −x−(1/3))=0 x^4 −px^3 −x^2 +(p−(1/3))x+(p/3)=0 (x^2 +ax+h)(x^2 +bx+k)=0 a+b=−p h+k+ab=−1 bh+ak=p−(1/3) hk=(p/3) −−−−−− say ab=t −−−−−− ah+bk+p−(1/3)=p(1+t) bh+ak−(p−(1/3))=0 ⇒ (a−b)(h−k)=pt−p+(2/3) squaring (p^2 −4t){(1+t)^2 −((4p)/3)}=(pt−p+(2/3))^2 say t+1=z (p^2 +4−4z)(z^2 −((4p)/3))=(pz−2p+(2/3))^2 ⇒ −4z^3 +(p^2 +4)z^2 +((16pz)/3)−((4p)/3)(p^2 +4) =p^2 z^2 −4p(p−(1/3))z+4(p−(1/3))^2 ⇒ z^3 −z^2 −p(p+1)z+(p−(1/3))^2 +(p/3)(p^2 +4)=0 .....

$$\left({x}−{p}\right)\left({x}^{\mathrm{3}} −{x}−\frac{\mathrm{1}}{\mathrm{3}}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{4}} −{px}^{\mathrm{3}} −{x}^{\mathrm{2}} +\left({p}−\frac{\mathrm{1}}{\mathrm{3}}\right){x}+\frac{{p}}{\mathrm{3}}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} +{ax}+{h}\right)\left({x}^{\mathrm{2}} +{bx}+{k}\right)=\mathrm{0} \\ $$$${a}+{b}=−{p} \\ $$$${h}+{k}+{ab}=−\mathrm{1} \\ $$$${bh}+{ak}={p}−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${hk}=\frac{{p}}{\mathrm{3}} \\ $$$$−−−−−− \\ $$$${say}\:{ab}={t} \\ $$$$−−−−−− \\ $$$${ah}+{bk}+{p}−\frac{\mathrm{1}}{\mathrm{3}}={p}\left(\mathrm{1}+{t}\right) \\ $$$${bh}+{ak}−\left({p}−\frac{\mathrm{1}}{\mathrm{3}}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\left({a}−{b}\right)\left({h}−{k}\right)={pt}−{p}+\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${squaring} \\ $$$$\left({p}^{\mathrm{2}} −\mathrm{4}{t}\right)\left\{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} −\frac{\mathrm{4}{p}}{\mathrm{3}}\right\}=\left({pt}−{p}+\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$${say}\:\:{t}+\mathrm{1}={z} \\ $$$$\left({p}^{\mathrm{2}} +\mathrm{4}−\mathrm{4}{z}\right)\left({z}^{\mathrm{2}} −\frac{\mathrm{4}{p}}{\mathrm{3}}\right)=\left({pz}−\mathrm{2}{p}+\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$$−\mathrm{4}{z}^{\mathrm{3}} +\left({p}^{\mathrm{2}} +\mathrm{4}\right){z}^{\mathrm{2}} +\frac{\mathrm{16}{pz}}{\mathrm{3}}−\frac{\mathrm{4}{p}}{\mathrm{3}}\left({p}^{\mathrm{2}} +\mathrm{4}\right) \\ $$$$\:\:\:\:={p}^{\mathrm{2}} {z}^{\mathrm{2}} −\mathrm{4}{p}\left({p}−\frac{\mathrm{1}}{\mathrm{3}}\right){z}+\mathrm{4}\left({p}−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:{z}^{\mathrm{3}} −{z}^{\mathrm{2}} −{p}\left({p}+\mathrm{1}\right){z}+\left({p}−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:+\frac{{p}}{\mathrm{3}}\left({p}^{\mathrm{2}} +\mathrm{4}\right)=\mathrm{0} \\ $$$$.....\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$

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