Question and Answers Forum

AllQuestion and Answers: Page 78

Question Number 215395 Answers: 1 Comments: 3

Question Number 215322 Answers: 1 Comments: 0

Question Number 215315 Answers: 2 Comments: 0

Evaluate lim_(n→∞) ((1/(2n+1))+(1/(2n+2))+...+(1/(4n)))

$$\mathrm{Evaluate}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{2}}+...+\frac{\mathrm{1}}{\mathrm{4}{n}}\right) \\ $$

Question Number 215313 Answers: 2 Comments: 0

Question Number 215312 Answers: 1 Comments: 1

Question Number 215282 Answers: 1 Comments: 3

Show that x=64 and y=729 (x)^(1/3) +(y)^(1/3) =13 (√x)+(√y)=35

$${Show}\:{that}\:{x}=\mathrm{64}\:{and}\:{y}=\mathrm{729}\: \\ $$$$\sqrt[{\mathrm{3}}]{{x}}+\sqrt[{\mathrm{3}}]{{y}}=\mathrm{13} \\ $$$$\sqrt{{x}}+\sqrt{{y}}=\mathrm{35} \\ $$

Question Number 215280 Answers: 2 Comments: 2

volume of solids generated by revolving the region bounded by the curve and line the y-axis y=x ,y = x/2 ,x = 2

$${volume}\:{of}\:{solids}\:{generated}\:{by}\:{revolving} \\ $$$${the}\:{region}\:{bounded}\:{by}\:{the}\:{curve}\:{and}\:{line} \\ $$$$\:{the}\:{y}-{axis} \\ $$$$\:{y}={x}\:,{y}\:=\:{x}/\mathrm{2}\:,{x}\:=\:\mathrm{2} \\ $$

Question Number 215277 Answers: 1 Comments: 0

(1+cos x)(1+sin x) = (5/4) (1−cos x)(1−sin x) = ?

$$\:\:\:\left(\mathrm{1}+\mathrm{cos}\:\mathrm{x}\right)\left(\mathrm{1}+\mathrm{sin}\:\mathrm{x}\right)\:=\:\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$\:\:\:\left(\mathrm{1}−\mathrm{cos}\:\mathrm{x}\right)\left(\mathrm{1}−\mathrm{sin}\:\mathrm{x}\right)\:=\:? \\ $$

Question Number 215276 Answers: 1 Comments: 0

I produced the drawing shown below (Q. 215275) using this great app, but unfortunately it refused to get saved after clicking “Save” button several times. Also, the “Export As Image” button is not working. Does anyone know the cause of this?

$$\mathrm{I}\:\mathrm{produced}\:\mathrm{the}\:\mathrm{drawing}\: \\ $$$$\mathrm{shown}\:\mathrm{below}\:\left(\mathrm{Q}.\:\mathrm{215275}\right)\: \\ $$$$\mathrm{using}\:\mathrm{this}\:\mathrm{great}\:\mathrm{app},\:\mathrm{but}\: \\ $$$$\mathrm{unfortunately}\:\mathrm{it}\:\mathrm{refused}\:\mathrm{to}\: \\ $$$$\mathrm{get}\:\mathrm{saved}\:\mathrm{after}\:\mathrm{clicking}\: \\ $$$$``\mathrm{Save}''\:\mathrm{button}\:\mathrm{several}\:\mathrm{times}.\: \\ $$$$\mathrm{Also},\:\mathrm{the}\:``\mathrm{Export}\:\mathrm{As}\:\mathrm{Image}''\: \\ $$$$\mathrm{button}\:\mathrm{is}\:\mathrm{not}\:\mathrm{working}.\:\mathrm{Does}\: \\ $$$$\mathrm{anyone}\:\mathrm{know}\:\mathrm{the}\:\mathrm{cause}\:\mathrm{of}\: \\ $$$$\mathrm{this}? \\ $$

Question Number 215272 Answers: 1 Comments: 0

{ ((abac^(−) =(dc^(−) )^2 )),((d=((ab^(−) )/c))),((c^2 =ac^(−) )) :} abac^(−) =?

$$\begin{cases}{\overline {{abac}}=\left(\overline {{dc}}\right)^{\mathrm{2}} }\\{{d}=\frac{\overline {{ab}}}{{c}}}\\{{c}^{\mathrm{2}} =\overline {{ac}}}\end{cases} \\ $$$$\overline {{abac}}=? \\ $$

Question Number 215275 Answers: 0 Comments: 0

Question Number 215256 Answers: 1 Comments: 2

Question Number 215255 Answers: 0 Comments: 0

a,b,c are pythagorean triples For the following values: determinant ((a,b,( c)),(5,(12),(13))) a^2 =b+c _(⇒5^2 =12+13=25) determinant ((a,b,( c)),((45),(1012),(1013)))a^2 =b+c_( ⇒45^2 =1012+1013=2025) ⇒45^2 =1012+1013=2025 determinant ((( _ determinant (((HAPPY NEW YEAR!)))^ )))

$${a},{b},{c}\:{are}\:{pythagorean}\:{triples} \\ $$$${For}\:{the}\:{following}\:{values}: \\ $$$$\begin{array}{|c|c|}{{a}}&\hline{{b}}&\hline{\:{c}}\\{\mathrm{5}}&\hline{\mathrm{12}}&\hline{\mathrm{13}}\\\hline\end{array}\:\:\underset{\Rightarrow\mathrm{5}^{\mathrm{2}} =\mathrm{12}+\mathrm{13}=\mathrm{25}} {{a}^{\mathrm{2}} ={b}+{c}\:\:} \\ $$$$\begin{array}{|c|c|}{{a}}&\hline{{b}}&\hline{\:{c}}\\{\mathrm{45}}&\hline{\mathrm{1012}}&\hline{\mathrm{1013}}\\\hline\end{array}\underset{\:\:\:\:\:\:\:\Rightarrow\mathrm{45}^{\mathrm{2}} =\mathrm{1012}+\mathrm{1013}=\mathrm{2025}} {{a}^{\mathrm{2}} ={b}+{c}} \\ $$$$\Rightarrow\mathrm{45}^{\mathrm{2}} =\mathrm{1012}+\mathrm{1013}=\mathrm{2025} \\ $$$$\: \\ $$$$\begin{array}{|c|}{\:_{\:} \begin{array}{|c|}{\mathcal{HAPPY}\:\:\mathcal{NEW}\:\:\mathcal{YEAR}!}\\\hline\end{array}^{\:} \:}\\\hline\end{array} \\ $$

Question Number 215248 Answers: 2 Comments: 1

((10)/7)÷(9/4)×((21)/8)

$$\frac{\mathrm{10}}{\mathrm{7}}\boldsymbol{\div}\frac{\mathrm{9}}{\mathrm{4}}×\frac{\mathrm{21}}{\mathrm{8}} \\ $$

Question Number 215234 Answers: 0 Comments: 0

Question Number 215224 Answers: 0 Comments: 1

I have just discovered that it is only alphabetic letters that can be made bold while typing with the keyboard of this app. Numbers 0 - 9 can′t be made bold, after being selected. Am I right, or is it only my phone that is having the issue? If it is general, admin team should kindly see to it in year 2025. Kudos to them for their hard work!

$$\mathrm{I}\:\mathrm{have}\:\mathrm{just}\:\mathrm{discovered}\:\mathrm{that}\:\mathrm{it}\: \\ $$$$\mathrm{is}\:\mathrm{only}\:\mathrm{alphabetic}\:\mathrm{letters}\:\mathrm{that} \\ $$$$\mathrm{can}\:\mathrm{be}\:\mathrm{made}\:\mathrm{bold}\:\mathrm{while}\: \\ $$$$\mathrm{typing}\:\mathrm{with}\:\mathrm{the}\:\mathrm{keyboard}\:\mathrm{of}\: \\ $$$$\mathrm{this}\:\mathrm{app}.\:\mathrm{Numbers}\:\mathrm{0}\:-\:\mathrm{9}\:\mathrm{can}'\mathrm{t} \\ $$$$\mathrm{be}\:\mathrm{made}\:\mathrm{bold},\:\mathrm{after}\:\mathrm{being}\: \\ $$$$\mathrm{selected}.\: \\ $$$$ \\ $$$$\mathrm{Am}\:\mathrm{I}\:\mathrm{right},\:\mathrm{or}\:\mathrm{is}\:\mathrm{it}\:\mathrm{only}\:\mathrm{my}\: \\ $$$$\mathrm{phone}\:\mathrm{that}\:\mathrm{is}\:\mathrm{having}\:\mathrm{the}\: \\ $$$$\mathrm{issue}? \\ $$$$ \\ $$$$\mathrm{If}\:\mathrm{it}\:\mathrm{is}\:\mathrm{general},\:\mathrm{admin}\:\mathrm{team}\: \\ $$$$\mathrm{should}\:\mathrm{kindly}\:\mathrm{see}\:\mathrm{to}\:\mathrm{it}\:\mathrm{in}\:\mathrm{year}\: \\ $$$$\mathrm{2025}.\:\mathrm{Kudos}\:\mathrm{to}\:\mathrm{them}\:\mathrm{for}\:\mathrm{their} \\ $$$$\mathrm{hard}\:\mathrm{work}! \\ $$

Question Number 215227 Answers: 2 Comments: 0

(20+25)^2 =2025 only 3 other 4 digit numbers: (00+01)^2 =0001 (30+25)^2 =3025 (98+01)^2 =9801

$$\left(\mathrm{20}+\mathrm{25}\right)^{\mathrm{2}} =\mathrm{2025} \\ $$$$\mathrm{only}\:\mathrm{3}\:\mathrm{other}\:\mathrm{4}\:\mathrm{digit}\:\mathrm{numbers}: \\ $$$$\left(\mathrm{00}+\mathrm{01}\right)^{\mathrm{2}} =\mathrm{0001} \\ $$$$\left(\mathrm{30}+\mathrm{25}\right)^{\mathrm{2}} =\mathrm{3025} \\ $$$$\left(\mathrm{98}+\mathrm{01}\right)^{\mathrm{2}} =\mathrm{9801} \\ $$

Question Number 215200 Answers: 4 Comments: 1

HAPPY NEW YEAR𝚺_(n=1) ^9 n^3 !

$$\boldsymbol{\mathcal{HAPPY}}\:\:\boldsymbol{\mathcal{NEW}}\:\:\boldsymbol{\mathcal{YEAR}}\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\mathrm{9}} {\boldsymbol{\sum}}{n}}^{\mathrm{3}} \:! \\ $$

Question Number 215199 Answers: 0 Comments: 0

F(x ,y) = ln (√(x^2 +y^2 )) where x(r,s) = r e^s and y(r,s) = r e^(−s ) Find (a) (∂F/∂r) (b) (∂F/∂s)

$$\:\:\:\: \mathrm{F}\left(\mathrm{x}\:,\mathrm{y}\right)\:=\:\mathrm{ln}\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }\: \\ $$$$\:\:\:\mathrm{where}\:\mathrm{x}\left(\mathrm{r},\mathrm{s}\right)\:=\:\mathrm{r}\:\mathrm{e}^{\mathrm{s}} \:\mathrm{and}\:\mathrm{y}\left(\mathrm{r},\mathrm{s}\right)\:=\:\mathrm{r}\:\mathrm{e}^{−\mathrm{s}\:} \\ $$$$\:\:\mathrm{Find}\: \\ $$$$\:\left(\mathrm{a}\right)\:\frac{\partial\mathrm{F}}{\partial\mathrm{r}}\:\:\:\:\: \\ $$$$\:\left(\mathrm{b}\right)\:\frac{\partial\mathrm{F}}{\partial\mathrm{s}} \\ $$

Question Number 215195 Answers: 1 Comments: 1