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AlgebraQuestion and Answers: Page 67

Question Number 195253 Answers: 2 Comments: 1

If 10sin^4 x+15cos^4 x=6. find the value of 27cosec^6 x+8sec^6 x

$${If}\:\mathrm{10sin}\:^{\mathrm{4}} {x}+\mathrm{15cos}\:^{\mathrm{4}} {x}=\mathrm{6}. \\ $$$${find}\:{the}\:{value}\:{of} \\ $$$$\mathrm{27cosec}\:^{\mathrm{6}} {x}+\mathrm{8sec}\:^{\mathrm{6}} {x} \\ $$$$ \\ $$

Question Number 195252 Answers: 1 Comments: 0

∫_(spillover) (dx/( (√e^(5x) ) (√((e^(2x) +e^(−2x) )^3 ))))

$$\int_{\boldsymbol{{spillover}}} \:\:\:\:\:\:\frac{{dx}}{\:\sqrt{{e}^{\mathrm{5}{x}} }\:\sqrt{\left({e}^{\mathrm{2}{x}} +{e}^{−\mathrm{2}{x}} \right)^{\mathrm{3}} }} \\ $$

Question Number 195251 Answers: 0 Comments: 1

If x^([16(log _5 x)^3 −68log _5 x]) =5^(−16) then Find the the product of x

$${If}\:\:{x}^{\left[\mathrm{16}\left(\mathrm{log}\:_{\mathrm{5}} {x}\right)^{\mathrm{3}} −\mathrm{68log}\:_{\mathrm{5}} {x}\right]} =\mathrm{5}^{−\mathrm{16}} \: \\ $$$$\:{then}\:{Find}\:{the}\:{the}\:{product}\:{of}\:{x} \\ $$$$ \\ $$

Question Number 195248 Answers: 0 Comments: 3

Question Number 195229 Answers: 0 Comments: 9

below equestion is show elips and hypharabollah (x^2 /(cos3))+(y^2 /(sin3))=1

$${below}\:{equestion}\:{is}\:{show}\:\:{elips}\:{and} \\ $$$${hypharabollah} \\ $$$$\frac{{x}^{\mathrm{2}} }{{cos}\mathrm{3}}+\frac{{y}^{\mathrm{2}} }{{sin}\mathrm{3}}=\mathrm{1} \\ $$

Question Number 195227 Answers: 0 Comments: 0

α_1 ^3 [((Π_(i=2) ^n (x−α_i ))/(Π_(i=2) ^n (α_1 −α_i )))]+Σ_(j=2) ^n (α_j ^3 [((Π_(i=1) ^(j−1) (x−α_i )Π_(i=j+1) ^n (x−α_j ))/(Π_(i=1) ^(j−1) (α_j −α_i )Π_(i=j+1) ^n (α_j −α_i )))]+α_n ^3 [((Π_(i=1) ^(n−1) (x−α_i ))/(Π_(i=1) ^(n−1) (α_n −α_i )))]−x^3 =0 solve for x . [ where n≥5 ]

$$ \\ $$$$\alpha_{\mathrm{1}} ^{\mathrm{3}} \left[\frac{\underset{{i}=\mathrm{2}} {\overset{{n}} {\prod}}\left({x}−\alpha_{{i}} \right)}{\underset{{i}=\mathrm{2}} {\overset{{n}} {\prod}}\left(\alpha_{\mathrm{1}} −\alpha_{{i}} \right)}\right]+\underset{{j}=\mathrm{2}} {\overset{{n}} {\sum}}\left(\alpha_{{j}} ^{\mathrm{3}} \left[\frac{\underset{{i}=\mathrm{1}} {\overset{{j}−\mathrm{1}} {\prod}}\left({x}−\alpha_{{i}} \right)\underset{{i}={j}+\mathrm{1}} {\overset{{n}} {\prod}}\left({x}−\alpha_{{j}} \right)}{\underset{{i}=\mathrm{1}} {\overset{{j}−\mathrm{1}} {\prod}}\left(\alpha_{{j}} −\alpha_{{i}} \right)\underset{{i}={j}+\mathrm{1}} {\overset{{n}} {\prod}}\left(\alpha_{{j}} −\alpha_{{i}} \right)}\right]+\alpha_{{n}} ^{\mathrm{3}} \left[\frac{\underset{{i}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\prod}}\left({x}−\alpha_{{i}} \right)}{\underset{{i}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\alpha_{{n}} −\alpha_{{i}} \right)}\right]−{x}^{\mathrm{3}} =\mathrm{0}\right. \\ $$$${solve}\:{for}\:{x}\:.\:\:\:\:\:\:\:\:\:\:\:\:\left[\:{where}\:{n}\geqslant\mathrm{5}\:\right] \\ $$

Question Number 195194 Answers: 2 Comments: 0

Question Number 195175 Answers: 1 Comments: 0

Question Number 195124 Answers: 1 Comments: 0

Question Number 195122 Answers: 0 Comments: 0

Question Number 195116 Answers: 0 Comments: 2

hi. please represntation (2)^(1/3) on number′s axis with ruler and compass and pen. thank you

$${hi}.\:{please}\:{represntation}\:\sqrt[{\mathrm{3}}]{\mathrm{2}}\:{on}\:{number}'{s}\:{axis} \\ $$$${with}\:{ruler}\:{and}\:{compass}\:{and}\:{pen}. \\ $$$${thank}\:{you} \\ $$

Question Number 195121 Answers: 2 Comments: 0

x^2 −x−1=0 x^8 +2x^7 −47x=?

$${x}^{\mathrm{2}} −{x}−\mathrm{1}=\mathrm{0} \\ $$$${x}^{\mathrm{8}} +\mathrm{2}{x}^{\mathrm{7}} −\mathrm{47}{x}=? \\ $$

Question Number 195107 Answers: 1 Comments: 0

Question Number 195101 Answers: 2 Comments: 0

(√(ln2)) >^? ln2

$$\sqrt{{ln}\mathrm{2}}\:\:\overset{?} {>}{ln}\mathrm{2} \\ $$

Question Number 195027 Answers: 1 Comments: 2

a_3 x^3 −x^2 +a_1 x−7=0 is a cubic polynomial in x whose Roots are α , β , γ positive real numbers satisfying ((225α^2 )/(α^2 +7))=((144β^2 )/(β^2 +7))=((100γ^2 )/(γ^2 +7)) find (a_1 )

$${a}_{\mathrm{3}} {x}^{\mathrm{3}} −{x}^{\mathrm{2}} +{a}_{\mathrm{1}} {x}−\mathrm{7}=\mathrm{0}\:{is}\:{a}\:{cubic}\:{polynomial}\:{in}\:{x} \\ $$$${whose}\:{Roots}\:{are}\:\alpha\:,\:\beta\:,\:\gamma\:{positive}\:{real}\:{numbers} \\ $$$${satisfying} \\ $$$$\frac{\mathrm{225}\alpha^{\mathrm{2}} }{\alpha^{\mathrm{2}} +\mathrm{7}}=\frac{\mathrm{144}\beta^{\mathrm{2}} }{\beta^{\mathrm{2}} +\mathrm{7}}=\frac{\mathrm{100}\gamma^{\mathrm{2}} }{\gamma^{\mathrm{2}} +\mathrm{7}} \\ $$$${find}\:\left({a}_{\mathrm{1}} \right) \\ $$

Question Number 195017 Answers: 3 Comments: 0

(x+1)^3 =1 x=?

$$\left({x}+\mathrm{1}\right)^{\mathrm{3}} =\mathrm{1}\:\:\:\:\:\:\:\:\:\:{x}=? \\ $$

Question Number 195002 Answers: 1 Comments: 0

x^(√x) =((√x))^x x=?

$${x}^{\sqrt{{x}}} =\left(\sqrt{{x}}\right)^{{x}} \:\:\:\:\:{x}=? \\ $$

Question Number 194991 Answers: 0 Comments: 4

a_3 x^3 −x^2 +a_1 x−7=0 is a cubic polynomial in x whose Roots are α , β , γ positive real numbers satisfying ((225α^2 )/(α^2 +7))=((144β^2 )/(β^2 +7))=((100γ^2 )/(γ^2 +7)) Find (a_1 )

$$ \\ $$$${a}_{\mathrm{3}} {x}^{\mathrm{3}} −{x}^{\mathrm{2}} +{a}_{\mathrm{1}} {x}−\mathrm{7}=\mathrm{0}\:{is}\:{a}\:{cubic}\:{polynomial}\:{in}\:{x} \\ $$$${whose}\:{Roots}\:{are}\:\alpha\:,\:\beta\:,\:\gamma\:{positive}\:{real}\:{numbers} \\ $$$${satisfying} \\ $$$$\frac{\mathrm{225}\alpha^{\mathrm{2}} }{\alpha^{\mathrm{2}} +\mathrm{7}}=\frac{\mathrm{144}\beta^{\mathrm{2}} }{\beta^{\mathrm{2}} +\mathrm{7}}=\frac{\mathrm{100}\gamma^{\mathrm{2}} }{\gamma^{\mathrm{2}} +\mathrm{7}} \\ $$$${Find}\:\left({a}_{\mathrm{1}} \right) \\ $$

Question Number 194953 Answers: 1 Comments: 0

Let P(x)= x^2 +(x/2)+b and Q(x)=x^2 +cx+d be two polynomial with real coefficients such that P(x)Q(x)= Q(P(x)) for all real x . Find all the real roots of P(Q(x))=0

$$\:{Let}\:{P}\left({x}\right)=\:{x}^{\mathrm{2}} +\frac{{x}}{\mathrm{2}}+{b}\:{and} \\ $$$$\:\:{Q}\left({x}\right)={x}^{\mathrm{2}} +{cx}+{d}\:{be}\:{two}\: \\ $$$$\:\:{polynomial}\:{with}\:{real}\:{coefficients} \\ $$$$\:\:{such}\:{that}\:{P}\left({x}\right){Q}\left({x}\right)=\:{Q}\left({P}\left({x}\right)\right) \\ $$$$\:{for}\:{all}\:{real}\:{x}\:. \\ $$$$\:\:{Find}\:{all}\:{the}\:{real}\:{roots}\:{of}\: \\ $$$$\:\:{P}\left({Q}\left({x}\right)\right)=\mathrm{0}\: \\ $$

Question Number 194952 Answers: 1 Comments: 0

x + (√(17−x^2 )) + x(√(17−x^2 )) =9 find the possible value of X

$$\boldsymbol{\mathrm{x}}\:+\:\sqrt{\mathrm{17}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\:\:+\:\boldsymbol{\mathrm{x}}\sqrt{\mathrm{17}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} \:}\:=\mathrm{9} \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{possible}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\:\:\boldsymbol{\mathrm{X}} \\ $$

Question Number 194942 Answers: 2 Comments: 2

= ((√5)+1)y + ((√5)+1)x x ≠ y and x+y ≠ 0 ((x/y))^(2025) +((y/x))^(2025)

$$\:\:\: =\: \left(\sqrt{\mathrm{5}}+\mathrm{1}\right)\mathrm{y} \\ $$$$\:\:\:\:\:\:\:\: +\:\left(\sqrt{\mathrm{5}}+\mathrm{1}\right)\mathrm{x} \\ $$$$\:\:\:\:\:\:\:\mathrm{x}\:\neq\:\mathrm{y}\:\mathrm{and}\:\mathrm{x}+\mathrm{y}\:\neq\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\: \left(\frac{\mathrm{x}}{\mathrm{y}}\right)^{\mathrm{2025}} +\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{\mathrm{2025}} \\ $$

Question Number 194914 Answers: 2 Comments: 0