Question

2) reciprocal of 4 5
_
7
​

Answer 1

$\sf \: in \: the \: \: figure \: \\ \\ \sf \: voltage \: \: V_1 = 35\: V \\ \bf \: registance \: R_1 = 10 \: \Omega \: \\ \bf \: registance \: R_2 = 15 \: \Omega \: \\ \bf \: registance \: R_ 3 = 6 \: \Omega \: \\ \bf \: registance \: R _ 4= 20 \: \Omega \: \\ \bf \: registance \: R_5 = 25 \: \Omega \: \\ \bf \: registance \: R_6= 5 \: \Omega \: \\ \bf \: registance \: R_7= 12\: \Omega \: \\ \bf \: registance \: R_8 = 15 \: \Omega \: \\ \\ \green \odot \: \sf \: find \: the \: value \: of \: \: total \: current \: (I )\:$

Answer 2

Answer:

$answer$

Coefficient of x⁶y³ is 672.

Step-by-step explanation:

General term of expansion (a+b)ⁿ is

\bf \: T_{r+1} = \: \: ^nC_r \: \: \large \frak{ a ^{n−r} b ^r}T

r+1

=

n

C

r

a

n−r

b

r

For (x+2y)⁹,

Putting n =9, a=x, b=2y

\begin{gathered} \bf \: T_{r+1} = \: \: ^{9} C_r (x) ^{9−r} (2y) ^r \\ \\ \bf \: T _{r+1} = \: \: ^{9} C_r (x) ^{9−r} .(y) ^r .(2) ^r\end{gathered}

T

r+1

=

9

C

r

(x)

9−r

(2y)

r

T

r+1

=

9

C

r

(x)

9−r

.(y)

r

.(2)

r

Comparing with x⁶ y³ , we get, r = 3

Therefore,

\begin{gathered} \bf \: T _{r+1} \\ \: \tt ^9C_3 (x)^9−3 .y³ .2³ \\ \tt\: \: 9! (2)³× x⁶ × y³) / (3!.6!) \\ \: \tt 672x⁶ y³\end{gathered}

T

r+1

9

C

3

(x)

9

−3.y³.2³

9!(2)³×x⁶×y³)/(3!.6!)

672x⁶y³