Question and Answers Forum

AllQuestion and Answers: Page 503

Question Number 168707 Answers: 1 Comments: 1

Question Number 168698 Answers: 3 Comments: 0

3f(x)+2f(((x+59)/(x−1)))=10x+30 for all rael x=1 faind volue of f(7)=?

$$\mathrm{3}{f}\left({x}\right)+\mathrm{2}{f}\left(\frac{{x}+\mathrm{59}}{{x}−\mathrm{1}}\right)=\mathrm{10}{x}+\mathrm{30}\:{for}\:{all} \\ $$$${rael}\:\:{x}\cancel{=}\mathrm{1}\:{faind}\:{volue}\:{of} \\ $$$${f}\left(\mathrm{7}\right)=? \\ $$$$ \\ $$

Question Number 168696 Answers: 1 Comments: 0

Calculate :: lim_(x→0) ((∫_(2x−1) ^(2x+1) e^t^2 dt−∫_(−1) ^1 e^t^2 dt)/x^2 )=8e ,Don′t use L′Hospital′s rule.

$$\mathrm{Calculate}\:\:\:::\:\:\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\int_{\mathrm{2x}−\mathrm{1}} ^{\mathrm{2x}+\mathrm{1}} \mathrm{e}^{\mathrm{t}^{\mathrm{2}} } \mathrm{dt}−\int_{−\mathrm{1}} ^{\mathrm{1}} \mathrm{e}^{\mathrm{t}^{\mathrm{2}} } \mathrm{dt}}{\mathrm{x}^{\mathrm{2}} }=\mathrm{8e}\:\:\:,\mathrm{Don}'\mathrm{t}\:\mathrm{use}\:\mathrm{L}'\mathrm{Hospital}'\mathrm{s}\:\mathrm{rule}. \\ $$

Question Number 168686 Answers: 0 Comments: 0

Question Number 168679 Answers: 2 Comments: 0

Question Number 168677 Answers: 2 Comments: 1

Question Number 168680 Answers: 0 Comments: 1

Question Number 168669 Answers: 0 Comments: 1

convert the intigeral ∫_1 ^( 2) ∫_0 ^( 1) 4xy^3 dxdy to polar cordinaite and it solve ?

$$\boldsymbol{{convert}}\:\boldsymbol{{the}}\:\boldsymbol{{intigeral}}\:\int_{\mathrm{1}} ^{\:\mathrm{2}} \:\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{4}\boldsymbol{{xy}}^{\mathrm{3}} \boldsymbol{{dxdy}}\:\boldsymbol{{to}}\:\boldsymbol{{polar}}\:\boldsymbol{{cordinaite}}\:\boldsymbol{{and}}\:\boldsymbol{{it}}\:\boldsymbol{{solve}}\:? \\ $$

Question Number 168665 Answers: 1 Comments: 1

Question Number 168664 Answers: 1 Comments: 0

{ ((8cos x−8sin x=3)),((55tan x+((55)/(tan x)) =?)) :}

$$\:\:\:\:\:\begin{cases}{\mathrm{8cos}\:{x}−\mathrm{8sin}\:{x}=\mathrm{3}}\\{\mathrm{55tan}\:{x}+\frac{\mathrm{55}}{\mathrm{tan}\:{x}}\:=?}\end{cases} \\ $$

Question Number 168663 Answers: 0 Comments: 1

how to solve? ((cos(30°)∙sin(45°)−cos(30°+dx)∙sin(45°+dx))/dx)=? dx→0

$${how}\:{to}\:{solve}? \\ $$$$\frac{{cos}\left(\mathrm{30}°\right)\centerdot{sin}\left(\mathrm{45}°\right)−{cos}\left(\mathrm{30}°+{dx}\right)\centerdot{sin}\left(\mathrm{45}°+{dx}\right)}{{dx}}=? \\ $$$${dx}\rightarrow\mathrm{0} \\ $$

Question Number 168662 Answers: 0 Comments: 0

∫sin^3 (x)cos^4 (5x)dx=?

$$\int{sin}^{\mathrm{3}} \left({x}\right){cos}^{\mathrm{4}} \left(\mathrm{5}{x}\right){dx}=? \\ $$

Question Number 168672 Answers: 0 Comments: 3

2^x =y^2 +7 x,y∈N x=? y=?

$$\mathrm{2}^{{x}} ={y}^{\mathrm{2}} +\mathrm{7}\:\:\:\:\:\:\:\:\:{x},{y}\in{N} \\ $$$${x}=?\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{y}=? \\ $$

Question Number 168649 Answers: 0 Comments: 3

Help!!!

$$\boldsymbol{{Help}}!!! \\ $$

Question Number 168652 Answers: 0 Comments: 0

Question Number 168643 Answers: 0 Comments: 3

Question Number 168642 Answers: 0 Comments: 0

Let △ABC be a triangle with ∠ABC=60° and ∠ACB=50°. IABI=a^2 −2 , IBCI=a in this instance , prove that IACI=(√3).

$$\mathrm{Let}\:\:\bigtriangleup\mathrm{ABC}\:\:\mathrm{be}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{with} \\ $$$$\angle\mathrm{ABC}=\mathrm{60}°\:\:\mathrm{and}\:\:\angle\mathrm{ACB}=\mathrm{50}°. \\ $$$$\mathrm{IABI}=\mathrm{a}^{\mathrm{2}} −\mathrm{2}\:,\:\mathrm{IBCI}=\mathrm{a}\:\:\mathrm{in}\:\mathrm{this} \\ $$$$\mathrm{instance}\:,\:\mathrm{prove}\:\mathrm{that}\:\:\mathrm{IACI}=\sqrt{\mathrm{3}}. \\ $$

Question Number 168628 Answers: 1 Comments: 0

montrer que d(x,y)=((∣u−v∣)/(1+∣u−v∣)) une distance sur R

$${montrer}\:{que} \\ $$$${d}\left({x},{y}\right)=\frac{\mid{u}−{v}\mid}{\mathrm{1}+\mid{u}−{v}\mid}\: \\ $$$${une}\:{distance}\:{sur}\:{R} \\ $$

Question Number 168617 Answers: 0 Comments: 2

Question Number 168616 Answers: 1 Comments: 0

Question Number 168613 Answers: 3 Comments: 1

Resolve 1) x(dy/dx)−y=y^3 2) (x−y)ydx−x^2 dy=0 3) (2x−y)dx+(4x−2y+3)dy=0

$${Resolve}\: \\ $$$$\left.\mathrm{1}\right)\:{x}\frac{{dy}}{{dx}}−{y}={y}^{\mathrm{3}} \\ $$$$\left.\mathrm{2}\right)\:\left({x}−{y}\right){ydx}−{x}^{\mathrm{2}} {dy}=\mathrm{0} \\ $$$$\left.\mathrm{3}\right)\:\left(\mathrm{2}{x}−{y}\right){dx}+\left(\mathrm{4}{x}−\mathrm{2}{y}+\mathrm{3}\right){dy}=\mathrm{0} \\ $$

Question Number 168609 Answers: 2 Comments: 1

Question Number 168608 Answers: 2 Comments: 0

Question Number 168606 Answers: 1 Comments: 6

Question Number 168605 Answers: 0 Comments: 4

((cos^2 10°+sin^2 25°−cos^2 15°)/(sin^2 10°+sin^2 25°−sin^2 15°))=?

$$\:\:\:\:\:\frac{\mathrm{cos}\:^{\mathrm{2}} \mathrm{10}°+\mathrm{sin}\:^{\mathrm{2}} \mathrm{25}°−\mathrm{cos}\:^{\mathrm{2}} \mathrm{15}°}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{10}°+\mathrm{sin}\:^{\mathrm{2}} \mathrm{25}°−\mathrm{sin}\:^{\mathrm{2}} \mathrm{15}°}=? \\ $$

Question Number 168603 Answers: 1 Comments: 1

Pg 498 Pg 499 Pg 500 Pg 501 Pg 502 Pg 503 Pg 504 Pg 505 Pg 506 Pg 507

Contact: info@tinkutara.com