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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

⇃

$$\:\:\:\:\:\downharpoonleft\underline{\:} \\ $$

Question Number 215195 Answers: 1 Comments: 1

Question Number 215193 Answers: 2 Comments: 1

Question Number 215190 Answers: 1 Comments: 0

∫(x^(p−1) /(1+x^n ))lnln(1/x)dx,p,n>0

$$\int\frac{{x}^{{p}−\mathrm{1}} }{\mathrm{1}+{x}^{{n}} }\mathrm{lnln}\frac{\mathrm{1}}{{x}}{dx},{p},{n}>\mathrm{0} \\ $$

Question Number 215189 Answers: 1 Comments: 0

∫_0 ^∞ ((ln(1+x^(12) ))/(1+x^2 ))dx

$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\left(\mathrm{1}+{x}^{\mathrm{12}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$

Question Number 215188 Answers: 2 Comments: 0

∫_0 ^∞ ((x cos 2πx)/(e^(2π(√x)) −1))dx

$$\int_{\mathrm{0}} ^{\infty} \frac{{x}\:\mathrm{cos}\:\mathrm{2}\pi{x}}{{e}^{\mathrm{2}\pi\sqrt{{x}}} −\mathrm{1}}{dx} \\ $$

Question Number 215185 Answers: 1 Comments: 0

∫_(−∞) ^∞ ((exp(((ax)/(1+x^2 )))exp((a/(1+x^2 ))))/(1+x^2 ))dx,a>0

$$\int_{−\infty} ^{\infty} \frac{\mathrm{exp}\left(\frac{{ax}}{\mathrm{1}+{x}^{\mathrm{2}} }\right)\mathrm{exp}\left(\frac{{a}}{\mathrm{1}+{x}^{\mathrm{2}} }\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx},{a}>\mathrm{0} \\ $$

Question Number 215177 Answers: 2 Comments: 0

x^3 + x^2 + x = − (1/3) Find the value of x ! Help me, please

$$ \\ $$$$\:\:\:{x}^{\mathrm{3}} \:+\:{x}^{\mathrm{2}} \:+\:{x}\:=\:−\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\:\:\mathcal{F}{ind}\:{the}\:{value}\:{of}\:\:{x}\:\:!\: \\ $$$$ \\ $$$$\:\:\:\mathcal{H}{elp}\:{me},\:\:{please} \\ $$$$ \\ $$

Question Number 215168 Answers: 1 Comments: 0

Let two roots of the equation x^2 +2mx+m^2 −2m+3=0 is α, β(For α<β), (2) Find (α−β)^2 Using m. (3) If β−α=2, Solve for m. If you don′t solve it, I will force you to solve.

$$\mathrm{Let}\:\mathrm{two}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:{x}^{\mathrm{2}} +\mathrm{2}{mx}+{m}^{\mathrm{2}} −\mathrm{2}{m}+\mathrm{3}=\mathrm{0}\:\mathrm{is}\:\alpha,\:\beta\left(\mathrm{For}\:\alpha<\beta\right), \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Find}\:\left(\alpha−\beta\right)^{\mathrm{2}} \:\mathrm{Using}\:{m}. \\ $$$$\left(\mathrm{3}\right)\:\mathrm{If}\:\beta−\alpha=\mathrm{2},\:\mathrm{Solve}\:\mathrm{for}\:{m}. \\ $$$$\mathrm{If}\:\mathrm{you}\:\mathrm{don}'\mathrm{t}\:\mathrm{solve}\:\mathrm{it},\:\mathrm{I}\:\mathrm{will}\:\mathrm{force}\:\mathrm{you}\:\mathrm{to}\:\mathrm{solve}. \\ $$

Question Number 215167 Answers: 1 Comments: 0

If the quadratic equation x^2 −2x−m+1=0 has two different positive real roots α, β, Determine the range of m (For example, 0<m<3) or I will force you to determine

$$\mathrm{If}\:\mathrm{the}\:\mathrm{quadratic}\:\mathrm{equation}\:{x}^{\mathrm{2}} −\mathrm{2}{x}−{m}+\mathrm{1}=\mathrm{0}\:\mathrm{has}\:\mathrm{two}\:\mathrm{different}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{roots}\:\alpha,\:\beta, \\ $$$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:{m}\:\left(\mathrm{For}\:\mathrm{example},\:\mathrm{0}<{m}<\mathrm{3}\right)\:\mathrm{or}\:\mathrm{I}\:\mathrm{will}\:\mathrm{force}\:\mathrm{you}\:\mathrm{to}\:\mathrm{determine} \\ $$

Question Number 215166 Answers: 1 Comments: 0

If the quadratic equation 3x^2 +8x+2k=0 has two different negative real roots, Determine the range of k (For example, 0<k<3) or I will force you to determine

$$\mathrm{If}\:\mathrm{the}\:\mathrm{quadratic}\:\mathrm{equation}\:\mathrm{3}{x}^{\mathrm{2}} +\mathrm{8}{x}+\mathrm{2}{k}=\mathrm{0}\:\mathrm{has}\:\mathrm{two}\:\mathrm{different}\:\mathrm{negative}\:\mathrm{real}\:\mathrm{roots}, \\ $$$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:{k}\:\left(\mathrm{For}\:\mathrm{example},\:\mathrm{0}<{k}<\mathrm{3}\right)\:\mathrm{or}\:\mathrm{I}\:\mathrm{will}\:\mathrm{force}\:\mathrm{you}\:\mathrm{to}\:\mathrm{determine} \\ $$

Question Number 215164 Answers: 1 Comments: 0

Let the two roots of the quadratic equation x^2 −12x+k=0 is α, α^2 , Then solve for α and k or I will force you to solve

$$\mathrm{Let}\:\mathrm{the}\:\mathrm{two}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{quadratic}\:\mathrm{equation}\:{x}^{\mathrm{2}} −\mathrm{12}{x}+{k}=\mathrm{0}\:\mathrm{is}\:\alpha,\:\alpha^{\mathrm{2}} , \\ $$$$\mathrm{Then}\:\mathrm{solve}\:\mathrm{for}\:\alpha\:\mathrm{and}\:{k}\:\mathrm{or}\:\mathrm{I}\:\mathrm{will}\:\mathrm{force}\:\mathrm{you}\:\mathrm{to}\:\mathrm{solve} \\ $$

Question Number 215161 Answers: 1 Comments: 4

∫(((x^8 +16x^7 −2x^4 −3)e^x )/((1+x^4 )^(2/3) ))dx

$$ \\ $$$$\:\:\:\:\:\:\int\frac{\left({x}^{\mathrm{8}} +\mathrm{16}{x}^{\mathrm{7}} −\mathrm{2}{x}^{\mathrm{4}} −\mathrm{3}\right){e}^{{x}} }{\left(\mathrm{1}+{x}^{\mathrm{4}} \right)^{\mathrm{2}/\mathrm{3}} }{dx} \\ $$$$ \\ $$

Question Number 215147 Answers: 2 Comments: 0

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