Correct Answer: A. (a - 2b)(c - a - 2b)

Correct Answer: B. a semicircle

Correct Answer: D. 5

Explanation

First, arrange all of the data in numerical order: 3,3,4,4,4,5,5,6,6,7,8. Then locate the middle number by using the formula n?12+1, which gives you the location of the median in the ordered data set and where n is the number of terms in the data set. Here, there are 11 terms. So, 11?12+1=102+1=5+1=6. Therefore, our number is the 6th one in the list, which is 5.

Correct Answer: C. \(\frac{1}{2}\)

Correct Answer: A. 14cm

Explanation

Curved surface area = 2\(\pi h\)

704 = 2 x \(\frac{22}{7} \times 8 \times h\)

704 = \(\frac{352}{7}\)h

352h = 704 x 7

h = \(\frac{704 \times 7}{352}\)

= \(\frac{4928}{352}\)

h = 14cm

Correct Answer: A. 5

Correct Answer: B. (1 + a - b)(1 - a + b)

Correct Answer: B. 3

Correct Answer: D.θ = 218\(^o\), 323\(^o\)

Explanation

On the \(y\)-axis, each box is \(\frac{1 - 0}{5} = \frac{1}{5}\) = 0.2unit

On the \(x\)-axis, each box is \(\frac{90 - 0}{6} = \frac{90}{6} = 15^o\)

⇒ \(θ_1 = 180^o + (2.5\times15^o) = 180^o + 37.5^o = 217.5^o ? 218^o \)(2 and half boxes were counted to the right of 180\(^o\))

⇒ \(θ_2 = 270^o + (3.5\times15^o) = 270^o + 52.5^o = 322.5^o ? 323^o \)(3 and half boxes were counted to the right of 270\(^o\))

∴ \(θ = 218^o, 323^o\)

Explanation

(i) \(T_{1} = \begin{pmatrix} -T_{1} \cos 40° \\ T_{1} \sin 40° \end{pmatrix} = \begin{pmatrix} -0.7660T_{1} \\ 0.6428T_{2} \end{pmatrix}\)

\(T_{2} = \begin{pmatrix} T_{2} \cos 30° \\ T_{2} \cos 30° \end{pmatrix} = \begin{pmatrix} 0.8660T_{2} \\ 0.5000T_{2} \end{pmatrix}\)

\(W = \begin{pmatrix} 0 \\ -65 \end{pmatrix}\)

(ii) P is at rest, the vector equation is

\(T_{1} + T_{2} + W = 0\)

(iii) \(\begin{pmatrix} -0.7660T_{1} \\ 0.6428T_{1} \end{pmatrix} + \begin{pmatrix} 0.8660T_{2} \\ 0.5000T_{2} \end{pmatrix} + \begin{pmatrix} 0 \\ -65 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\)

\(0.7660T_{1} = 0.8660T_{2} ... (1)\)

\(0.6428T_{1} + 0.5000T_{2} - 65 = 0 ... (2)\)

From (1), \(T_{1} = \frac{0.8660T_{2}}{0.7660} ... (3)\)

Put (3) in (2) :

\(0.6428(\frac{0.8660T_{2}}{0.7660}) + 0.5000T_{2} = 65 \)

\(0.7627T_{2} + 0.5000T_{2} = 65\)

\(1.2267T_{2} = 65 \implies T_{2} = \frac{65}{1.2267}\)

\(T_{2} = 52.988N \approxeq 53.0N\) (to 1 d.p)

From (3), \(T_{1} = \frac{0.866 \times 52.988}{0.766} = 59.905N\)

\(\approxeq 59.9N\) (to 1 d.p)