- 2x2 + 3y2; 3y2 + 3 - x? and x2 + y2 + 5 - 3xy
Answers
Step-by-step explanation:
Answer:
\red { Compound \: interest} \green {= Rs \:7804 }Compoundinterest=Rs7804
Step-by-step explanation:
Principal (P) = Rs \:62500Principal(P)=Rs62500
Time (T) = \frac{3}{2} \: years = 1\frac{1}{2} \: yearsTime(T)=
2
3
years=1
2
1
years
Rate \: of \: InterestAnswer:
\red { Compound \: interest} \green {= Rs \:7804 }Compoundinterest=Rs7804
Step-by-step explanation:
Principal (P) = Rs \:62500Principal(P)=Rs62500
Time (T) = \frac{3}{2} \: years = 1\frac{1}{2} \: yearsTime(T)=
2
3
years=1
2
1
years
Rate \: of \: Interest (r) = 8\% \: per \: annumRateofInterest(r)=8%perannum
rate for \:half \: yearly = \frac{1}{2} \times 8\% =4\%rateforhalfyearly=
2
1
×8%=4%
\begin{gathered}As \: interest \: is \: compounded \:half\\yearly, \:so \: number \: of \:conversion \:periods\\in \: 1\frac{1}{2} \: years \: is \: 3, \:So, \: n = 3\end{gathered}
Asinterestiscompoundedhalf
yearly,sonumberofconversionperiods
in1
2
1
yearsis3,So,n=3
\boxed { \pink { Amount (A) = P\left( 1 + \frac{r}{100} \right )^{n} }}
Amount(A)=P(1+
100
r
)
n
\begin{gathered}= 62500 \left( 1 + \frac{4}{100}\right)^{3}\\= 62500 \left( 1 + \frac{1}{25} \right)^{3}\\= 62500 \left( \frac{25+1}{25}\right)^{3}\\= 62500\times \left(\frac{26}{25}\right)^{3}\\= 62500 \times \frac{25}{26} \times \frac{25}{26} \times \frac{25}{26}\end{gathered}
=62500(1+
100
4
)
3
=62500(1+
25
1
)
3
=62500(
25
25+1
)
3
=62500×(
25
26
)
3
=62500×
26
25
×
26
25
×
26
25
= Rs\: 70304=Rs70304
\begin{gathered}Compound \: interest = A - P \\= 70304 - 62500 \\= Rs\:7804\end{gathered}
Compoundinterest=A−P
=70304−62500
=Rs7804
Therefore.,
\red { Compound \: interest} \green {= Rs \:7804 }Compoundinterest=Rs7804
•••♪ (r) = 8\% \: per \: annumRateofInterest(r)=8%perannum
rate for \:half \: yearly = \frac{1}{2} \times 8\% =4\%rateforhalfyearly=
2
1
×8%=4%
\begin{gathered}As \: interest \: is \: compounded \:half\\yearly, \:so \: number \: of \:conversion \:periods\\in \: 1\frac{1}{2} \: years \: is \: 3, \:So, \: n = 3\end{gathered}
Asinterestiscompoundedhalf
yearly,sonumberofconversionperiods
in1
2
1
yearsis3,So,n=3
\boxed { \pink { Amount (A) = P\left( 1 + \frac{r}{100} \right )^{n} }}
Amount(A)=P(1+
100
r
)
n
\begin{gathered}= 62500 \left( 1 + \frac{4}{100}\right)^{3}\\= 62500 \left( 1 + \frac{1}{25} \right)^{3}\\= 62500 \left( \frac{25+1}{25}\right)^{3}\\= 62500\times \left(\frac{26}{25}\right)^{3}\\= 62500 \times \frac{25}{26} \times \frac{25}{26} \times \frac{25}{26}\end{gathered}
=62500(1+
100
4
)
3
=62500(1+
25
1
)
3
=62500(
25
25+1
)
3
=62500×(
25
26
)
3
=62500×
26
25
×
26
25
×
26
25
= Rs\: 70304=Rs70304
\begin{gathered}Compound \: interest = A - P \\= 70304 - 62500 \\= Rs\:7804\end{gathered}
Compoundinterest=A−P
=70304−62500
=Rs7804
Therefore.,
\red { Compound \: interest} \green {= Rs \:7804 }Compoundinterest=Rs7804
•••♪