Question and Answers Forum

All Questions   Topic List

AllQuestion and Answers: Page 493

Question Number 169906    Answers: 0   Comments: 0

Evaluate ∫∫e^(2x+3y) dxdy over the triangle bounded by the lines x = 0, y = 0, x+y = 1.

$$\mathrm{Evaluate}\:\int\int{e}^{\mathrm{2}{x}+\mathrm{3}{y}} {dxdy}\:\mathrm{over}\:\mathrm{the}\:\mathrm{triangle} \\ $$$$\mathrm{bounded}\:\mathrm{by}\:\mathrm{the}\:\mathrm{lines}\:{x}\:=\:\mathrm{0},\:{y}\:=\:\mathrm{0},\:{x}+{y}\:=\:\mathrm{1}. \\ $$

Question Number 170010    Answers: 1   Comments: 1

Question Number 170012    Answers: 1   Comments: 0

Question Number 169901    Answers: 0   Comments: 0

Question Number 169900    Answers: 0   Comments: 0

montrer que ∀z∈C/ ∣z∣=2, ∣(1/(z^4 −5z+1))∣≤(1/5)

$$\boldsymbol{{montrer}}\:\boldsymbol{{que}}\:\forall\boldsymbol{{z}}\in\mathbb{C}/\:\mid\boldsymbol{{z}}\mid=\mathrm{2}, \\ $$$$\mid\frac{\mathrm{1}}{\boldsymbol{{z}}^{\mathrm{4}} −\mathrm{5}\boldsymbol{{z}}+\mathrm{1}}\mid\leqslant\frac{\mathrm{1}}{\mathrm{5}} \\ $$

Question Number 169899    Answers: 1   Comments: 2

Question Number 169897    Answers: 0   Comments: 1

Question Number 169896    Answers: 1   Comments: 0

Question Number 169892    Answers: 1   Comments: 0

(dy/dx) = 2xe^(−y) , y(1) = 0

$$\frac{{dy}}{{dx}}\:=\:\mathrm{2}{xe}^{−{y}} \:,\:\:{y}\left(\mathrm{1}\right)\:=\:\mathrm{0} \\ $$

Question Number 169885    Answers: 1   Comments: 0

∣a^→ ∣=13 ∣b^→ ∣=19 ∣a^→ +b^→ ∣=24 ∣a^→ −b^→ ∣=?

$$\mid\overset{\rightarrow} {{a}}\mid=\mathrm{13} \\ $$$$\mid\overset{\rightarrow} {{b}}\mid=\mathrm{19} \\ $$$$\mid\overset{\rightarrow} {{a}}+\overset{\rightarrow} {{b}}\mid=\mathrm{24} \\ $$$$\mid\overset{\rightarrow} {{a}}−\overset{\rightarrow} {{b}}\mid=? \\ $$

Question Number 169878    Answers: 2   Comments: 0

Question Number 169874    Answers: 1   Comments: 0

Σ_(d∣6) d=?

$$\underset{{d}\mid\mathrm{6}} {\sum}{d}=? \\ $$

Question Number 169873    Answers: 2   Comments: 0

∫_0 ^1 ((ln(1+x))/x)dx=?

$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}{dx}=? \\ $$

Question Number 169870    Answers: 0   Comments: 0

In an artistic design by a contractor of the students’ centre in a university, four pillars were erected to form two triangular figures with the vertices at the first floor of the building and two of the pillars crossing each other. Proof that the spaces occupied by these two triangular figures are equal.

$$ \\ $$In an artistic design by a contractor of the students’ centre in a university, four pillars were erected to form two triangular figures with the vertices at the first floor of the building and two of the pillars crossing each other. Proof that the spaces occupied by these two triangular figures are equal.

Question Number 169865    Answers: 3   Comments: 2

x+y=−2 xy=4 find x^8 +8y^5 =?

$${x}+{y}=−\mathrm{2} \\ $$$${xy}=\mathrm{4} \\ $$$${find}\:{x}^{\mathrm{8}} +\mathrm{8}{y}^{\mathrm{5}} =? \\ $$

Question Number 169864    Answers: 1   Comments: 0

Question Number 169863    Answers: 1   Comments: 0

∫_o ^1 xln∣x^2 −2x∣dx

$$\int_{{o}} ^{\mathrm{1}} {xln}\mid{x}^{\mathrm{2}} −\mathrm{2}{x}\mid{dx} \\ $$

Question Number 169860    Answers: 1   Comments: 0

y = x^2 + 1 y = 0 x = - 1 x = 2 find the volume of the object obtained by rotating the figure bounded by lines around the abscissa axis

$$\mathrm{y}\:=\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{1} \\ $$$$\mathrm{y}\:=\:\mathrm{0} \\ $$$$\mathrm{x}\:=\:-\:\mathrm{1} \\ $$$$\mathrm{x}\:=\:\mathrm{2} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{the}\:\mathrm{object}\:\mathrm{obtained} \\ $$$$\mathrm{by}\:\mathrm{rotating}\:\mathrm{the}\:\mathrm{figure}\:\mathrm{bounded}\:\mathrm{by}\:\mathrm{lines} \\ $$$$\mathrm{around}\:\mathrm{the}\:\mathrm{abscissa}\:\mathrm{axis} \\ $$

Question Number 169859    Answers: 0   Comments: 0

Question Number 169855    Answers: 1   Comments: 0

Question Number 169852    Answers: 1   Comments: 0

y = x^2 + 2 y = - x x = 0 x = 1 find the area of the figure bounded by lines

$$\mathrm{y}\:=\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{2} \\ $$$$\mathrm{y}\:=\:-\:\mathrm{x} \\ $$$$\mathrm{x}\:=\:\mathrm{0} \\ $$$$\mathrm{x}\:=\:\mathrm{1} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{figure}\:\mathrm{bounded}\:\mathrm{by} \\ $$$$\mathrm{lines} \\ $$

Question Number 169850    Answers: 0   Comments: 1

Question Number 169849    Answers: 1   Comments: 0

Question Number 169842    Answers: 1   Comments: 0

Question Number 169837    Answers: 0   Comments: 0

$$ \\ $$

Question Number 169836    Answers: 1   Comments: 0

  Pg 488      Pg 489      Pg 490      Pg 491      Pg 492      Pg 493      Pg 494      Pg 495      Pg 496      Pg 497   

Terms of Service

Privacy Policy

Contact: info@tinkutara.com