2. Using the given Figure, name the following angles :
(a) alternate interior angles
(b) corresponding angles
(d) angle alternate to angle 3
(e) angle corresponding to angle 4 (f) exterior angles
(g) pairs of interior angles on the same side of the transversal
Answers
Answer:
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Step-by-step explanation:
Let Principal be 'P'
and Rate of interest be 'R'
Time (n) = 2 years
Simple Interest (S.I) = Rs. 400
Compound Interest (C.I) = Rs. 410
\begin{gathered}\begin{gathered}S.I = \frac{P \times R \times T}{100} \\ \\ = \frac{P \times R \times T}{100} = 400 \\ \\ = \frac{PR \times 2}{100} = 400 \\ \\ = \frac{PR}{50} = 400 \\ \\ = PR = 400 \times 50 \\ = PR = 20000\end{gathered}\end{gathered}
S.I=
100
P×R×T
=
100
P×R×T
=400
=
100
PR×2
=400
=
50
PR
=400
=PR=400×50
=PR=20000
S.I=100P×R×T=100P×R×T=400=100PR×2=400=50PR=400=PR=400×50=PR=20000
\begin{gathered}\begin{gathered}C.I = P( \: \: ( {1 + \frac{R}{100} })^{n} - 1) \\ \\ = P( \: \: ( {1 + \frac{R}{100} })^{n} - 1) = 410 \\ \\ = P( \: \: ( {1 + \frac{R}{100} })^{2} - 1) = 410 \\ \\ = P( \: \: {(1)}^{2} + 2 \times 1 \times \frac{R}{100} + { (\frac{R}{100}) }^{2} - 1 ) = 410 \\ \\ = P( \: \: 1 - 1 \times \frac{R}{50} + { (\frac{R}{100}) }^{2} \: \: \: ) = 410 \\ \\ = P( \: \frac{R}{100} + \frac{ {R}^{2} }{10000} \: \: ) = 410 \\ \\ (take \: \: common \: R \: ) \\ \\ = PR ( \frac{1}{50} + \frac{R}{10000} ) = 410 \\ (put \: the \: value \: of \: PR \: = 20000) \\ \\ = 20000( \frac{1}{50} + \frac{R}{10000} ) = 410 \\ \\ = 20000( \frac{200 + R }{10000} ) = 410 \\ \\ = \frac{R + 200}{10000} = \frac{410}{20000 } \\ \\ = \frac{R + 200}{10000} = \frac{41}{2000 } \\ \\ = R + 200 = \frac{41}{2000} \times 10000 \\ \\ = R + 200 = 41 \times 5 \\ = R + 200 = 205 \\ = R = 205 - 200 \\ = R = 5\end{gathered}\end{gathered}
C.I=P((1+
100
R
)
n
−1)
=P((1+
100
R
)
n
−1)=410
=P((1+
100
R
)
2
−1)=410
=P((1)
2
+2×1×
100
R
+(
100
R
)
2
−1)=410
=P(1−1×
50
R
+(
100
R
)
2
)=410
=P(
100
R
+
10000
R
2
)=410
(takecommonR)
=PR(
50
1
+
10000
R
)=410
(putthevalueofPR=20000)
=20000(
50
1
+
10000
R
)=410
=20000(
10000
200+R
)=410
=
10000
R+200
=
20000
410
=
10000
R+200
=
2000
41
=R+200=
2000
41
×10000
=R+200=41×5
=R+200=205
=R=205−200
=R=5
C.I=P((1+100R)n−1)=P((1+100R)n−1)=410=P((1+100R)2−1)=410=P((1)2+2×1×100R+(100R)2−1)=410=P(1−1×50R+(100R)2)=410=P(100R+10000R2)=410(takecommonR)=PR(501+10000R)=410(putthevalueofPR=20000)=20000(501+10000R)=410=20000(10000200+R)=410=10000R+200=20000410=10000R+200=200041=R+200=200041×10000=R+200=41×5=R+200=205=R=205−200=R=5
Rate of interest = 5%
Now Principal,
as PR
\begin{gathered}\begin{gathered}Now Principal, \\ as P \times R = 20000 \\ = > P \times 5 = 20000 \\ = > P = \frac{20000}{5} \\ \\ = > P = 4000\end{gathered}\end{gathered}
NowPrincipal,
asP×R=20000
=>P×5=20000
=>P=
5
20000
=>P=4000
NowPrincipal,asP×R=20000=>P×5=20000=>P=520000=>P=4000
Principal = Rs. 4000
Q. (7)
A man invested Rs. 1000 for 3 years at 11% Simple Interest .....................(CASE 1)
Principal (P) = Rs. 1000
Rate of interest (R) = 11% per annum
Time (n) = 3 years
\begin{gathered}\begin{gathered}S.I = \frac{P \times R \times T}{100} \\ \\ = \frac{1000 \times 11 \times 3}{100} \\ \\ = 10 \times 11 \times 3 \\ = 330\end{gathered}\end{gathered}
S.I=
100
P×R×T
=
100
1000×11×3
=10×11×3
=330
S.I=100P×R×T=1001000×11×3=10×11×3=330
S.I = Rs. 330
He also invested Rs. 1000 at 10% compound interest per annum compounded annually for 3 years .........................( CASE 2 )
Principal (P) = Rs. 1000
Rate of interest (R) = 10% per annum
Time (n) = 3 years
\begin{gathered}\begin{gathered}C.I = P( \: \: {(1 + \frac{R}{100}) }^{n} - 1) \\ \\ = 1000( \: \: {(1 + \frac{10}{100}) }^{3} - 1) \\ \\ = 1000( \: \: {(1 + \frac{1}{10}) }^{3} - 1) \\ \\ = 1000( \: \: { (\frac{10 + 1}{10} )}^{3} - 1 ) \\ \\ = 1000( \: { (\frac{11}{10}) }^{3} - 1 ) \\ \\ = 1000 \times (\frac{1331}{1000} - 1) \\ \\ = 1000 \times ( \frac{1331 - 1000}{1000} ) \\ \\ = 1000 \times \frac{331}{1000} \\ \\ = 331\end{gathered}\end{gathered}
C.I=P((1+
100
R
)
n
−1)
=1000((1+
100
10
)
3
−1)
=1000((1+
10
1
)
3
−1)
=1000((
10
10+1
)
3
−1)
=1000((
10
11
)
3
−1)
=1000×(
1000
1331
−1)
=1000×(
1000
1331−1000
)
=1000×
1000
331
=331
Answer:
the figure is not provided