Correct Answer: C. y - x - 6=0

Explanation

From y= 2x- 5x 10 ; dy/dx= 4x-5

But at (1, 7); m\(_1\) = (dy/dx)\(_{1,7}\)

= 4(1) - 5 = -1

Using m\(_1\)m\(_2\),= -1; m\(_2\) = 1.

Gradient m\(_2\), of normal at (1,7) is 1.

Using y - y\(_1\) = m(x - x\(_1\))

y - 7 = 1(x - 1) ;

y - 7 = x - 1

y - x - 6 = 0

Correct Answer: B. log₁₀40

Correct Answer: (d) 1/3

Correct Answer: B. 13

Explanation

(a) \(\frac{\frac{1}{3}c^{2} - \frac{2}{3}cd}{\frac{1}{2}d^{2} - \frac{1}{4}cd}\)

= \(\frac{\frac{1}{3}c(c - 2d)}{\frac{1}{4}d(2d - c)}\)

= \(\frac{\frac{1}{3}c (c - 2d)}{-\frac{1}{4}d (c - 2d)}\)

= \(-\frac{4c}{3d}\)

(b)(i) In \(\Delta XYP\),

\(\frac{7}{XP} = \sin 44°\)

\(XP = \frac{7}{\sin 44}\)

= \(10.08 cm\)

< XPM + 44° + 46° = 180°

< XPM = 90°.

In \(\Delta XPM\),

\(XM^{2} = XP^{2} + PM^{2}\)

\(XM^{2} = (10.08)^{2} + 9^{2}\)

= \(101.6064 + 81 = 182.6064\)

\(XM = \sqrt{182.6064}\)

= \(15.513 cm \approxeq 13.5 cm\)

In \(\Delta XYP\),

\(\frac{7}{YP} = \tan 44\)

\(YP = \frac{7}{\tan 44}\)

\(YP = 7.249 cm \approxeq 7.2 cm\)

In \(\Delta MPF\),

\(\frac{PF}{9} = \cos 46\)

\(PF = 9 \cos 46 = 6.2519 cm\)

\(YF = YP + PF = 7.249 + 6.2519\)

= \(13.5009 cm \approxeq 13.5 cm\)

(ii)

\(\frac{10.1}{9} = \tan \theta\)

\(\tan \theta = 1.122\)

\(\theta = \tan^{-1} (1.122)\)

= 48.296°

Explanation

(a)(i) Position vector of M = 3i + 7j i.e \(\overrightarrow{OM} = 3i + 7j\).

Position vector of N = 4i + 5j i.e \(\overrightarrow{ON} = 4i + 5j\)

\(\overrightarrow{MN} = \overrightarrow{ON} - \overrightarrow{OM}\)

\((4i + 5j) - (3i + 7j) = i - 2j\)

Velocity of M = 5i + 6j

Distance covered in time t = \((5i + 6j)t = 5ti + 6tj\)

When t = 3 seconds, = \(15i + 18j\)

Velocity of N = 2i + 3j

Distance covered in time t = \((2i + 3j)t = 2ti + 3tj\)

When t = 3 seconds, = \(6i + 9j\)

\(\overrightarrow{MN} = (6i + 9j) - (15i + 18j)\)

= \(-9i - 9j\)

\(|MN| = \sqrt{(-9)^{2} + (-9)^{2}} = \sqrt{162}\)

= \(9\sqrt{2} m\)

(b) \(F_{1} = 5i + pj ; F_{2} = qi + j ; F_{3} = -2pi + 3j ; F_{4} = -4i + qj\)

If the particle remains in equilibrium, then

\(F_{1} + F_{2} + F_{3} + F_{4} = 0\)

\(5i + pj + qi + j - 2pi + 3j - 4i + qj = 0\)

Equating by components,

\(i : 5 + q - 2p - 4 = 0 \implies q - 2p = -1 ...(1)\)

\(j : p + 1 + 3 + q = 0 \implies p + q = -4 ... (2)\)

\((2) - (1) : 3p = -3 \implies p = -1\)

\(-1 + q = -4 \implies q = -3\)

Correct Answer: D. 13/9

Explanation

(³/₄ +¹/₃) x 4¹/₃ \(\div\) 3¹₄

\(\begin{pmatrix} 9 + 4 \\ 12 \end{pmatrix}\) x \(\frac{13}{3}\) \(\frac{4}{13}\) = 1⁴₉

Correct Answer: C. 8cm

Explanation

volume of cube = L x L x L

512cm³ = L³

L³ = 512cm³

L = 3\(\sqrt{512}\)

L (512)^{\(\frac{1}{3}\)}

= (2⁹)^{\(\frac{1}{3}\)}

2³ = 8cm

Correct Answer: A. 25.75^o

Explanation

25° 45' = 25° + \(\frac{45}{60}\)°

= 25° + 0.75°

= 25.75°

Correct Answer: B. 11100\(_{three}\)

Explanation

Convert from base 5 to base 10

432\(_{five}\) = (4 x 5\(^2\)) + (3 x 5\(^1\)) + (2 x 5\(^0\))

= (4 x 25) + (3 x 5) + (2 x 1)

= 100 + 15 + 2

= 117\(_{ten}\)

Then convert from base 10 to base 3

Selecting the remainders from bottom to top:

117\(_{ten}\) = 11100\(_{three}\)

Hence; 432\(_{five}\) = 11100\(_{three}\)