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AlgebraQuestion and Answers: Page 237

Question Number 104037 Answers: 4 Comments: 0

(1) { ((x^3 +y^6 = 91)),((x+y^2 = 7 )) :} find x−y^6 . (2) 2a+(2/a) = 8 ⇒ ((a^6 +1)/a^3 ) ?

$$\left(\mathrm{1}\right)\begin{cases}{{x}^{\mathrm{3}} +{y}^{\mathrm{6}} \:=\:\mathrm{91}}\\{{x}+{y}^{\mathrm{2}} \:=\:\mathrm{7}\:}\end{cases} \\ $$$${find}\:{x}−{y}^{\mathrm{6}} \:. \\ $$$$\left(\mathrm{2}\right)\:\mathrm{2}{a}+\frac{\mathrm{2}}{{a}}\:=\:\mathrm{8}\:\Rightarrow\:\frac{{a}^{\mathrm{6}} +\mathrm{1}}{{a}^{\mathrm{3}} }\:? \\ $$

Question Number 104023 Answers: 0 Comments: 1

Question Number 103983 Answers: 0 Comments: 0

If α and β are two unequal angle, which satisfy the equation, a cos(α) + b sin(β) = c, show that (i) sin(((α + β)/2)) sec(((α − β)/2)) = (b/c) (ii) tan((α/2)) tan((β/2)) = ((c − a)/(c + a))

$$\mathrm{If}\:\:\alpha\:\:\mathrm{and}\:\:\beta\:\:\mathrm{are}\:\mathrm{two}\:\mathrm{unequal}\:\mathrm{angle},\:\mathrm{which}\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{equation}, \\ $$$$\:\:\:\:\mathrm{a}\:\mathrm{cos}\left(\alpha\right)\:\:+\:\:\mathrm{b}\:\mathrm{sin}\left(\beta\right)\:\:=\:\:\mathrm{c},\:\:\:\:\mathrm{show}\:\mathrm{that} \\ $$$$\left(\mathrm{i}\right)\:\:\:\:\mathrm{sin}\left(\frac{\alpha\:\:+\:\beta}{\mathrm{2}}\right)\:\mathrm{sec}\left(\frac{\alpha\:\:−\:\:\beta}{\mathrm{2}}\right)\:\:=\:\:\frac{\mathrm{b}}{\mathrm{c}} \\ $$$$\left(\mathrm{ii}\right)\:\:\:\:\:\mathrm{tan}\left(\frac{\alpha}{\mathrm{2}}\right)\:\mathrm{tan}\left(\frac{\beta}{\mathrm{2}}\right)\:\:=\:\:\frac{\mathrm{c}\:\:−\:\:\mathrm{a}}{\mathrm{c}\:\:+\:\:\mathrm{a}} \\ $$

Question Number 103921 Answers: 1 Comments: 0

Prove that ∀ x ∈ R^ , ∣ cos x ∣ ≤ 1 − sin^2 x

$$\mathrm{Prove}\:\mathrm{that}\:\forall\:{x}\:\in\:\bar {\mathbb{R}}\:,\:\mid\:\mathrm{cos}\:{x}\:\mid\:\leqslant\:\mathrm{1}\:−\:\mathrm{sin}^{\mathrm{2}} \:{x} \\ $$

Question Number 103914 Answers: 2 Comments: 2

(2+(√3))^x^2 + (2−(√3))^x^2 = 4

$$\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{{x}^{\mathrm{2}} } \:+\:\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{{x}^{\mathrm{2}} } \:=\:\mathrm{4}\: \\ $$

Question Number 103888 Answers: 2 Comments: 0

find all such numbers: if we make its last digit, say k, as its first digit, the number becomes k times large as before. (□□...□k)→(k□□...□)=k×(□□...□k)

$${find}\:{all}\:{such}\:{numbers}: \\ $$$${if}\:{we}\:{make}\:{its}\:{last}\:{digit},\:{say}\:{k},\:{as}\:{its} \\ $$$${first}\:{digit},\:{the}\:{number}\:{becomes}\:{k} \\ $$$${times}\:{large}\:{as}\:{before}. \\ $$$$\left(\Box\Box...\Box{k}\right)\rightarrow\left({k}\Box\Box...\Box\right)={k}×\left(\Box\Box...\Box{k}\right) \\ $$

Question Number 103804 Answers: 0 Comments: 0

Solve for r ((h−p^2 )/(p−r))=(1/(2p)) , ((h−q)/(q^2 −r))= 2q (h−p^2 )^2 +(p−r)^2 =r^2 (h−q)^2 +(q^2 −r)^2 =r^2

$$\:\:\:\boldsymbol{{S}}{olve}\:{for}\:\boldsymbol{{r}} \\ $$$$\:\:\frac{\boldsymbol{{h}}−\boldsymbol{{p}}^{\mathrm{2}} }{\boldsymbol{{p}}−\boldsymbol{{r}}}=\frac{\mathrm{1}}{\mathrm{2}\boldsymbol{{p}}}\:\:\:\:,\:\:\:\:\frac{\boldsymbol{{h}}−\boldsymbol{{q}}}{\boldsymbol{{q}}^{\mathrm{2}} −\boldsymbol{{r}}}=\:\mathrm{2}\boldsymbol{{q}} \\ $$$$\left(\boldsymbol{{h}}−\boldsymbol{{p}}^{\mathrm{2}} \right)^{\mathrm{2}} +\left(\boldsymbol{{p}}−\boldsymbol{{r}}\right)^{\mathrm{2}} =\boldsymbol{{r}}^{\mathrm{2}} \\ $$$$\:\left(\boldsymbol{{h}}−\boldsymbol{{q}}\right)^{\mathrm{2}} +\left(\boldsymbol{{q}}^{\mathrm{2}} −\boldsymbol{{r}}\right)^{\mathrm{2}} =\boldsymbol{{r}}^{\mathrm{2}} \\ $$

Question Number 103736 Answers: 0 Comments: 0

Solve: a^2 + c^2 = 196 ... (i) b^2 + (c − a)^2 = 169 ... (ii) c^2 + (b − c)^2 = 225 ... (iii)

$$\mathrm{Solve}:\:\:\:\:\:\:\mathrm{a}^{\mathrm{2}} \:\:+\:\:\mathrm{c}^{\mathrm{2}} \:\:=\:\:\:\mathrm{196}\:\:\:\:\:\:...\:\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{b}^{\mathrm{2}} \:\:+\:\:\left(\mathrm{c}\:\:−\:\:\mathrm{a}\right)^{\mathrm{2}} \:\:=\:\:\mathrm{169}\:\:\:\:\:...\:\left(\mathrm{ii}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{c}^{\mathrm{2}} \:\:+\:\:\left(\mathrm{b}\:\:−\:\:\mathrm{c}\right)^{\mathrm{2}} \:\:=\:\:\mathrm{225}\:\:\:\:\:...\:\left(\mathrm{iii}\right) \\ $$

Question Number 103685 Answers: 1 Comments: 0