French, asked by Prakeetha, 9 months ago


1 Cochez les bonnes réponses :
Sophie étudie en:
a.dixième
b.sixième
c.septième

Sophie habite dans un
a.appartement
b.hôtel
c.bureau

La ville de Sophie est :
a. Lyon
b.paris
c.rouen

d. Sophie joue avec ses :
a. cousins
b.amis
c.parents

Elle joue dans un
a. parc
b.jardin
c.complexe sportif​

Answers

Answered by gungunbajpai061105
0

Answer:

désolé je ne connais pas la réponse mec

Answered by Anonymous
15

QUESTION:-

 \large\bold Q. ғɪɴᴅ ᴛʜᴇ ᴀᴍᴏᴜɴᴛ ᴀɴᴅ ᴛʜᴇ ᴄᴏᴍᴘᴏᴜɴᴅ ɪɴᴛᴇʀᴇsᴛ ᴏɴ ʀᴜᴘᴇᴇs 64000 ғᴏʀ 1 1/2 ʏᴇᴀʀs ᴀᴛ 15% ᴘᴇʀ ᴀɴɴᴜᴍ, ᴄᴏᴍᴘᴜɴᴅᴇᴅ ʜᴀʟғ-ʏᴇʀʟʏ

ANSWER✔

\Large\underline\bold{solution,}

\Large\bold{given,}

 \sf\dashrightarrow  time(T)= 1 \times \dfrac{1}{2} years = \dfrac{3}{2}

\sf\underline\bold{\therefore compound\:for\:half\:yearly,}

 \sf\implies  rate= \dfrac{15}{2} \% ............. \frac{1}{2} \:a\:year

 \sf\implies  time= \dfrac{3}{2} \times 2 =3\:years  ............. \frac{1}{2} yearly

\large{\boxed{\sf{amount=principle \times \bigg[ 1+ \dfrac{rate}{100} \bigg] \times T}}}

\sf\underline\bold{now,}

 \sf\implies  amount=64000 \times \bigg[ \bigg( 1+ \dfrac{ \frac{15}{2}}{100} \bigg)^3 \times 100 \bigg]

 \sf\implies  amount=64000 \times \bigg[ \bigg( \dfrac{215}{200} \bigg)^3 \times 200 \bigg]

\sf\implies  amount= \cancel {64000} \times \bigg[ \bigg(  \dfrac{\cancel {215}}{ \cancel {200}} \bigg)^3 \times \cancel {200} \bigg]

ᴀᴍᴏᴜɴᴛ ( ᴀ )= ʀs.79507

\large{\boxed{\sf{amount=rs.79507}}}

___________________

✯ADDITIONAL INFORMATION,

✯DERIVATION,

DERIVING, compound interest formula,

LET,

 \sf\therefore principal\:amount\:=p

 \sf\therefore time= n\:years

 \sf\therefore rate =R

 \sf\large\therefore simple \:interest\:for\:firsr\:year,

 \sf\therefore SI_1= \dfrac{P \times R \times T }{100}

 \sf\therefore amount\:after\:1_st \:year=p+SI_1

 \sf\implies P+ \dfrac{P times R \times T }{100}

 \sf\therefore P \times \bigg( 1+ \dfrac{R}{100} \bigg)=P_2

 \sf\therefore simple\:intrest\:for\:2_nd \:= \dfrac{P_2 \times R \times T }{100}

 \sf\therefore Amount\:after\:second\:year,

 \sf\implies P_2 +SI_2

 \sf\implies P_2+ \dfrac{P_2 \times R \times T }{100}

 \sf\implies  P_2 \bigg( 1+ \dfrac{R}{100} \bigg)

 \sf\implies P \bigg( 1+ \dfrac{R}{100} \bigg) \times \bigg( 1+ \dfrac{R}{100}  \bigg)

 \sf\therefore P \bigg( 1+ \dfrac{R}{100} \bigg)^2

 \sf\therefore simlarly\:if\:we\:continued,\:for\:"n"\:years.

THEN,

 \sf\therefore A=P \times \bigg( 1+ \dfrac{R}{100} \bigg)^n

 \sf\large\therefore CI=A-P

 \sf\implies p \times \bigg[ \bigg( 1+ \dfrac{R}{100} \bigg)^n \bigg]

\sf{\boxed{\sf{p \times \bigg[ \bigg( 1+ \dfrac{R}{100} \bigg)^n \bigg]}}}

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