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AlgebraQuestion and Answers: Page 67

Question Number 191394    Answers: 1   Comments: 0

Question Number 191393    Answers: 0   Comments: 0

Question Number 191389    Answers: 1   Comments: 0

Factorize: x^3 +(1/x^3 )+((11)/8)

$$\mathrm{Factorize}: \\ $$$${x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+\frac{\mathrm{11}}{\mathrm{8}} \\ $$

Question Number 191343    Answers: 1   Comments: 0

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

$$ \\ $$$$\:\:\:\:\:\:\mathrm{solve}\:\:{in}\:\:\mathbb{R} \\ $$$$ \\ $$$$\:\:\:\:\:\:\lfloor\:\frac{\mathrm{1}}{{x}}\:\rfloor\:+\:\lfloor\:\frac{\mathrm{2}}{{x}}\:\rfloor\:+\:\lfloor\:\frac{\mathrm{3}}{{x}}\:\rfloor\:=\:\mathrm{1} \\ $$$$ \\ $$

Question Number 191301    Answers: 2   Comments: 0

Question Number 191300    Answers: 1   Comments: 0

Question Number 191304    Answers: 2   Comments: 0

Question Number 191250    Answers: 0   Comments: 4

Question Number 191283    Answers: 1   Comments: 0

Question Number 191222    Answers: 0   Comments: 0

Question Number 191221    Answers: 0   Comments: 0

Question Number 191430    Answers: 3   Comments: 0

factorise 4x^4 +81

$${factorise}\:\mathrm{4}{x}^{\mathrm{4}} +\mathrm{81} \\ $$

Question Number 191191    Answers: 3   Comments: 0

Question Number 191169    Answers: 1   Comments: 0

Question Number 191148    Answers: 0   Comments: 0

Question Number 191147    Answers: 0   Comments: 0

Question Number 191146    Answers: 0   Comments: 0

Question Number 191138    Answers: 0   Comments: 0

Question Number 191127    Answers: 0   Comments: 0

Question Number 191111    Answers: 0   Comments: 2

bb55bb

$${bb}\mathrm{55}{bb} \\ $$

Question Number 191103    Answers: 1   Comments: 0

Question Number 191077    Answers: 3   Comments: 2

a^→ ∙ b^(→) = 4 b^→ ∙ c^(→) = 5 7a^(→) = 4b^(→) + 2c^(→) Find: ∣c^(→) ∣ = ?

$$\overset{\rightarrow} {\mathrm{a}}\:\centerdot\:\overset{\rightarrow} {\mathrm{b}}\:=\:\mathrm{4} \\ $$$$\overset{\rightarrow} {\mathrm{b}}\:\centerdot\:\overset{\rightarrow} {\mathrm{c}}\:=\:\mathrm{5} \\ $$$$\overset{\rightarrow} {\mathrm{7a}}\:=\:\overset{\rightarrow} {\mathrm{4b}}\:+\:\overset{\rightarrow} {\mathrm{2c}}\:\: \\ $$$$\mathrm{Find}:\:\:\:\mid\overset{\rightarrow} {\mathrm{c}}\:\mid\:=\:? \\ $$

Question Number 191061    Answers: 0   Comments: 0

Question Number 191027    Answers: 0   Comments: 0

Question Number 191024    Answers: 1   Comments: 0

Question Number 191022    Answers: 0   Comments: 1

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