Mathematics 2007 Jamb Examination

P contains only odd numbers and Q contains only even numbers. There are no elements common to both sets. Therefore the intersection is the empty set: P ∩ Q = ∅.

First, convert to decimal: 1011₂ = 11, 101₂ = 5.
Then square: 11² = 121, 5² = 25.
Subtract: 121 - 25 = 96.
Convert 96 back to base 2: 1100000₂.
Hence the correct result is 1100000₂.

Perfect squares less than 40: 1, 4, 9, 16, 25, 36. Odd numbers from 1 to 15: 1, 3, 5, 7, 9, 11, 13, 15. The common elements are 1 and 9, so X ∩ Y = {1, 9}.

Radius r = diameter/2 = 14/2 = 7 cm. Angle θ = 90° = $\tfrac{\pi}{2}$ radians. Arc length s = rθ = 7 × $\tfrac{\pi}{2}$ = $\tfrac{7\pi}{2}$ cm.

Find slopes. From 3y = 4x - 1 ⇒ y = (4/3)x - 1/3, so slope m1 = 4/3. From qy = x + 3 ⇒ y = (1/q)x + 3/q, so slope m2 = 1/q. Parallel lines have equal slopes: 1/q = 4/3 ⇒ q = 3/4.

x = 30^o (alternate ∠s b/c PQ//Sr)
Y = 50^o (vertical opposite ∠s)
r = 30^o (alternate ∠s b/c PQ//SR)
But = y + q + t = 180^o (∠s on a straight line)
50 + q + 30 = 180
q + 80 = 180
q = 180 -80
q = 100^o

Answer: C. 7.0 cm. Step-by-step: 1. Convert the mixed number to an improper fraction: \(718\tfrac{2}{3}=718+\tfrac{2}{3}=\tfrac{2154}{3}+\tfrac{2}{3}=\tfrac{2156}{3}.\) 2. Volume of a hemisphere: \(V=\dfrac{2}{3}\pi r^3.\) 3. Set up the equation: \(\dfrac{2}{3}\pi r^3 = \dfrac{2156}{3}.\) 4. Multiply both sides by 3 and divide by 2π: \[r^3 = \dfrac{2156}{2\pi} = \dfrac{1078}{\pi}.\] 5. Compute numerically: \(\dfrac{1078}{\pi}\approx 343.16\). Take cube root: \(r\approx\sqrt[3]{343.16}\approx 7.00\) (note \(7^3=343\), so this is consistent). 6. Therefore the radius ≈ 7.0 cm (option C).

It is a semicircle with ST as diameter (general locus principle).
The locus problem states: all points P that subtend a fixed angle at the segment ST. A classical theorem (the circle angle theorem) says: all points from which a given chord ST is seen under a fixed angle lie on an arc of a circle (an arc of the circle that subtends ST at that angle).
The special and commonly tested case is when the constant angle is 90° (a right angle). In that case the locus is the semicircle with ST as diameter (Thales' theorem).
The available options include the semicircle with ST as diameter, which is the correct expression of this locus for the right-angle case and matches the general statement that the locus is an arc of a circle subtending ST under the constant angle.

Answer: C. -2. Step-by-step: 1. The point (1,−2) lies on both lines. Substitute x=1, y=−2 into the first equation: \[2(-2) - k(1) + 2 = 0.\] Simplify: \(-4 - k + 2 = 0\Rightarrow -2 - k = 0\Rightarrow k = -2.\) 2. Verify in the second equation: substitute x=1, y=−2: \[-2 + 1 - k/2 = 0 \Rightarrow -1 - k/2 = 0 \Rightarrow k/2 = -1 \Rightarrow k = -2.\] 3. Both equations agree, so k = −2 (option C).

Answer: A. $\dfrac{40\sqrt{3}}{3}$ m. Step-by-step: 1. Let the height of the tower be h and the horizontal distance from the man to the tower base be 40 m. 2. Using right-triangle trigonometry: \(\tan(30^\circ)=\dfrac{\text{opposite}}{\text{adjacent}}=\dfrac{h}{40}.\) 3. Since \(\tan 30^\circ=\dfrac{1}{\sqrt{3}}\), we have \(\dfrac{1}{\sqrt{3}}=\dfrac{h}{40}\Rightarrow h=\dfrac{40}{\sqrt{3}}.\) 4. Rationalize the denominator if desired: \(h=\dfrac{40\sqrt{3}}{3}.\) 5. Thus the tower's height is $\dfrac{40\sqrt{3}}{3}$ meters (option A).