\[ \begin{array}{l} \text { Part - II } \\ \text { Section - D } \end{array} \] Note: answer the following question. 18. From following receipts and payments A/C of Sri Charitable Trust, \[ 1 \times 20=20 \] Additional information A. Subscriptions receivable for the year 2020 R5 2,500 B. Prepaid Rent Rs. 300 C. Outstanding stationary Bills Rs. 150 D. Capitalize donations E. Half of the entrance fees capitalized Interest receivable for the year \( 2020 Rs .200 \)

N.I.=i+iii+iv+ii+vii+ix `=400+150+100+200+500+(-50)` =rs 1300 crore.

Correct Answer - A `(a)` Writing `((200),(r ))=((200),(200-r))`, we have `((100),(0))((200),(50))+((100),(1))((200),(49))+((100),(2))((200),(48))+......+((100),(50))((200),(0))` `=` Coefficient of `x^(50)` in the expansion of `(1+x)^(100)(x+1)^(200)` `=` Coefficient of `x^(50)` in the binomial expansion of `(1+x)^(300)` `=((300),(50))`

Correct option is: (C) All revenue and capital receipts and payments

Given \(\vec A\) = \(\hat i+2\hat j-\hat k\) \(\vec B\) = \(-\hat i-2\hat j+\hat k\) Angle b/w \(\vec A\) and \(\vec B\) cos θ = \(\frac{\vec A.\vec B}{|\vec A|.|\vec B|}\) \(=\frac{(\hat i+2\hat j-\hat k).(-\hat i-2\hat j+\hat k)}{\sqrt{1^2+(2)^2+(1)^2}\sqrt{(-1)^2+(-2)^2+(1)^2}}\) cos θ = \(\frac{-1-4-1}{\sqrt6\times\sqrt6}\) cos θ = -6/6 cos θ = -1 θ = 180° option...

Correct Answer - Subscription credited to Income and Expenditure Account for the year ended on March 31, 2017 is Rs. 69,000. Subscription Outstanding as on 31.3.2017 is Rs. 5,400 and...

Correct Answer - (i) False; (ii) False; (iii) True; (iv) False; (v) True (vi) True (vii) False (viii) True (ix) False (x) True (xi) True (xii) True

Matrix A \(=\begin{bmatrix}1&3\\4&1\end{bmatrix}_{2\times2}\) |3 A'| = 2 \(A'=\begin{bmatrix}1&4\\3&1\end{bmatrix}_{2\times2}\) We know that ⇒ |K A| = (K)3 |A| ⇒ |3 A'| = 32 |A'| \(A'=\begin{bmatrix}1&4\\3&1\end{bmatrix}\) ⇒ |A'| = 1 x 1 - 4 x 3 ⇒ 1 - 12 = -11 ⇒...

I want answer

A = \(\begin{bmatrix}8&-8&-2\\4&-3&-2\\3&-4&1\end{bmatrix}\)  A2 = \(\begin{bmatrix}8&-8&-2\\4&-3&-2\\3&-4&1\end{bmatrix}\)\(\begin{bmatrix}8&-8&-2\\4&-3&-2\\3&-4&1\end{bmatrix}\) = \(\begin{bmatrix}26&-32&-2\\14&-15&-4\\11&-16&3\end{bmatrix}\) A4 = A2. A2 = \(\begin{bmatrix}26&-32&-2\\14&-15&-4\\11&-16&3\end{bmatrix}\)\(\begin{bmatrix}26&-32&-2\\14&-15&-4\\11&-16&3\end{bmatrix}\) = \(\begin{bmatrix}206&-320&-70\\110&-159&20\\95&-160&51\end{bmatrix}\)

Given that \(\vec a, \vec b\) and \(\vec c\)are unit vectors. i.e., |\(\vec a\)| = |\(\vec b\)|  = |\(\vec c\)| = 1 And \(\vec a\) + \(\vec b\) + \(\vec c\) = \(\vec 0\) Then \((\vec a+\vec b+\vec c).(\vec a+\vec b+\vec c)=\vec 0.\vec...