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Question Number 216887 Answers: 0 Comments: 0

Π_(k=1) ^n cos((x/2^k ))=Pn(x) evaluate Pn(x) and P_n (x^2 +1)

$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}{cos}\left(\frac{{x}}{\mathrm{2}^{{k}} }\right)={Pn}\left({x}\right) \\ $$$${evaluate}\:{Pn}\left({x}\right)\:\:{and}\:\:{P}_{{n}} \left({x}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$ \\ $$$$ \\ $$

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An object of mass M, initially at rest at the coordinate origin, explodes into three parts. Fragment A has a mass M/2, and fragments B and C have a mass M/4 each. After the explosion, fragment A moves in the +X direction at 10 m/s and fragment B moves in the +Y direction at 8 m/s. Find the direction and speed of fragment C

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Find the value of ω^7 + ω^8 + ω^(12) where ω is omega function.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:\omega^{\mathrm{7}} \:\:+\:\:\omega^{\mathrm{8}} \:\:+\:\:\omega^{\mathrm{12}} \:\:\mathrm{where} \\ $$$$\omega\:\:\mathrm{is}\:\mathrm{omega}\:\mathrm{function}. \\ $$

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t_(33) ′ab+t_(00) ′ab=? Clearing up defintions: t_(33) ′a=(a−(√(36)))×3 t_(00) ′a=(a−(√(36)))×0 t′a=a−(√(36)) t′a= “transformation” of a

$${t}_{\mathrm{33}} '{ab}+{t}_{\mathrm{00}} '{ab}=? \\ $$$$ \\ $$$$\mathrm{Clearing}\:\mathrm{up}\:\mathrm{defintions}: \\ $$$${t}_{\mathrm{33}} '{a}=\left({a}−\sqrt{\mathrm{36}}\right)×\mathrm{3} \\ $$$${t}_{\mathrm{00}} '{a}=\left({a}−\sqrt{\mathrm{36}}\right)×\mathrm{0} \\ $$$${t}'{a}={a}−\sqrt{\mathrm{36}} \\ $$$${t}'{a}=\:``\mathrm{transformation}''\:\mathrm{of}\:{a} \\ $$

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 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!

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

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