Math, asked by siddhantshirbhate, 1 month ago

((x - 4) ^ 4 * (x + 5) ^ 3 * (x - 3) ^ 6 * (x + 2) ^ 6)/((x + 4) ^ 3 * (x - 2) ^ 4 * (x + 1) ^ 3) >= 0​

Answers

Answered by jaiswalanjali4999
1

Step-by-step explanation:

Current time that is shown by watch is 10 : 00 AM & Watch is getting slow by 2 minutes for each hour in a day .

Exigency To Find : What will be the time after 12 hrs ?

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀Given that ,

⠀⠀⠀⠀⠀⠀⠀⠀▪︎ ⠀The Current time that is shown by watch is 10 : 00 AM

⠀⠀⠀⠀⠀⠀⠀⠀▪︎ ⠀Watch is getting slow by 2 minutes for each hour .

⠀⠀⠀⠀⠀Now , As we have to find time that is shown by watch after 12 hrs ,

Therefore,

⠀⠀⠀⠀⠀⠀The time that get slower ( or late ) in 12 hrs will be ⠀

\begin{gathered}\qquad \qquad \sf Time \: get \:slower \:= \: 12 \times 2 \: min \:\:\\\end{gathered}

Timegetslower=12×2min

\begin{gathered}\qquad \qquad \bf Time \: get \:slower \:= \: 24 \: min \:\:\\\end{gathered}

Timegetslower=24min

\begin{gathered}\qquad :\implies \pmb{\underline{\purple{\:Time \: get \:slower_{(12 \:hrs\:)} \:= \: 24 \: min\: }} }\:\:\bigstar \\\end{gathered}

:⟹

Timegetslower

(12hrs)

=24min

Timegetslower

(12hrs)

=24min

Therefore ,

\begin{gathered}\dag\:\:\sf{ As,\:We\:know\:that\::}\\\\ \qquad\bigstar\:\:\bf Time \:After \: 12 \: hrs \: : \\\end{gathered}

†As,Weknowthat:

★TimeAfter12hrs:

\begin{gathered}\qquad \dag\:\:\bigg\lgroup \sf{ 10 \:: \:00 \: AM \:+ \: 12 \: hrs \: - \: Time \: get \:slower_{(12\: hrs)} }\bigg\rgroup \\\\\end{gathered}

10:00AM+12hrs−Timegetslower

(12hrs)

\begin{gathered}\qquad:\implies \sf 10 \:: \:00 \: AM \:+ \: 12 \: hrs \: - \: Time \: get \:slower_{(12\: hrs)} \\\end{gathered}

:⟹10:00AM+12hrs−Timegetslower

(12hrs)

\begin{gathered}\qquad:\implies \sf 22 \:: \:00 \: PM \: \: - \: Time \: get \:slower_{(12\: hrs)} \\\end{gathered}

:⟹22:00PM−Timegetslower

(12hrs)

⠀⠀⠀⠀⠀⠀⠀⠀OR ,

\begin{gathered}\qquad:\implies \sf 22 \:: \:00 \: PM \: \: - \: Time \: get \:slower_{(12\: hrs)} \\\end{gathered}

:⟹22:00PM−Timegetslower

(12hrs)

\begin{gathered}\qquad:\implies \sf 10 \:: \:00 \: PM \: \: - \: Time \: get \:slower_{(12\: hrs)} \\\end{gathered}

:⟹10:00PM−Timegetslower

(12hrs)

⠀⠀⠀⠀⠀⠀\begin{gathered}\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\\end{gathered}

⋆NowBySubstitutingtheknownValues:

\begin{gathered}\qquad:\implies \sf 10 \:: \:00 \: PM \: \: - \: Time \: get \:slower_{(12\: hrs)} \\\end{gathered}

:⟹10:00PM−Timegetslower

(12hrs)

\begin{gathered}\qquad:\implies \sf 10 \:: \:00 \: PM \:\: - \: 24 \: min \\\end{gathered}

:⟹10:00PM−24min

\begin{gathered}\qquad:\implies \sf 10 \:: \:00 \: PM \:\: - \: 24 \: min \qquad\bigg\lgroup \sf{ 1 \:hrs \:=\: 60 \: min }\bigg\rgroup \\\end{gathered}

:⟹10:00PM−24min

1hrs=60min

\begin{gathered}\qquad:\implies \sf 10 \:: \:00 \: PM \:\: - \: 24 \: min \qquad\bigg\lgroup \sf{ 60 \:min \:-\: 24 \: min =\: 36 \: min \:}\bigg\rgroup \\\end{gathered}

:⟹10:00PM−24min

60min−24min=36min

\begin{gathered}\qquad:\implies \sf 9\:: \:36 \: PM \:\: \\\end{gathered}

:⟹9:36PM

\begin{gathered}\qquad :\implies \pmb{\underline{\purple{\:Time_{( After\:12\:hrs\:)} = \: 9\:: \:36 \: PM \:\: }} }\:\:\bigstar \\\end{gathered}

:⟹

Time

(After12hrs)

=9:36PM

Time

(After12hrs)

=9:36PM

Therefore,

⠀⠀⠀⠀⠀\begin{gathered}\therefore {\underline{ \mathrm {\:Time\:shown\:after\:12\:hrs \:in\:the\:watch\:will \:be\:\bf{9\:: \:36 \: PM \:\:}}}}\\\end{gathered}

Timeshownafter12hrsinthewatchwillbe9:36PM

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