Mathematics 2022 Jamb Examination

According to the laws of indices, any non-zero number raised to the power of 0 is equal to 1 (\(a^0 = 1\)).
For the equation \(\left(\frac{1}{8}\right)^{k+2} = 1\) to be true, the exponent must be zero:\[k + 2 = 0\]Solving for k:\[k = -2\]

First, convert \(101_{two}\) to base ten:\[(1 \times 2^2) + (0 \times 2^1) + (1 \times 2^0) = 4 + 0 + 1 = 5_{ten}\]Now, find the cube of the base ten number:\[5^3 = 5 \times 5 \times 5 = 125_{ten}\]Convert \(125_{ten}\) back to base two by repeated division by 2:
125 ÷ 2 = 62 R 1
62 ÷ 2 = 31 R 0
31 ÷ 2 = 15 R 1
15 ÷ 2 = 7 R 1
7 ÷ 2 = 3 R 1
3 ÷ 2 = 1 R 1
1 ÷ 2 = 0 R 1
Reading the remainders from bottom to top, we get \(1111101_{two}\).

The shaded region is located entirely within set F, but specifically excludes any parts that overlap with set E or set G. In set notation, "only F" is written as the intersection of F with the complements of E and G:\[F \cap E' \cap G' \quad \text{or} \quad E' \cap F \cap G'\]This represents the elements that are in F but not in E and not in G.

When a coin is tossed twice, the sample space (S) is:
S = {HH, HT, TH, TT}. Total outcomes = 4.
The outcomes with "at least one head" are {HH, HT, TH}. Number of successful outcomes = 3.
Probability = \(\frac{3}{4} = 0.75\).

First, find x using the mean formula:\[\frac{2 + 5 + (x + 1) + (x + 2) + 7 + 9}{6} = 6\]\[\frac{26 + 2x}{6} = 6 \implies 26 + 2x = 36\]\[2x = 10 \implies x = 5\]Substitute x into the data: 2, 5, (5+1), (5+2), 7, 9 → 2, 5, 6, 7, 7, 9.
The median of these 6 sorted numbers is the average of the two middle terms (6 and 7):\[\text{Median} = \frac{6 + 7}{2} = 6.5\]

In the third quadrant, both sine and cosine are negative. Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\):\[\left(-\frac{3}{5}\right)^2 + \cos^2 \theta = 1\]\[\frac{9}{25} + \cos^2 \theta = 1 \implies \cos^2 \theta = 1 - \frac{9}{25} = \frac{16}{25}\]\[\cos \theta = \pm\sqrt{\frac{16}{25}} = \pm\frac{4}{5}\]Since θ is in the third quadrant, \(\cos \theta = -\frac{4}{5}\).

To rationalize the denominator, multiply the numerator and denominator by the conjugate \(3 + \sqrt{2}\):\[\frac{1}{3 - \sqrt{2}} \times \frac{3 + \sqrt{2}}{3 + \sqrt{2}} = \frac{3 + \sqrt{2}}{3^2 - (\sqrt{2})^2}\]\[= \frac{3 + \sqrt{2}}{9 - 2} = \frac{3 + \sqrt{2}}{7}\]This can be written as \(\frac{3}{7} + \frac{\sqrt{2}}{7}\).

The distance from the center, the radius, and half the chord form a right-angled triangle. Let \(x\) be half the length of the chord:\[x^2 + 3^2 = 5^2\]\[x^2 + 9 = 25 \implies x^2 = 16 \implies x = 4\text{ cm}\]The full length of the chord is \(2x = 2(4) = 8\text{ cm}\).

To find the angle for a sector in a pie chart, multiply the percentage by 360° (the total angle in a circle):\[\text{Angle} = \frac{28}{100} \times 360^o\]\[= 0.28 \times 360^o = 100.8^o\]

The volume \(V\) of a pyramid is given by the formula:\[V = \frac{1}{3} \times \text{Base Area} \times \text{Height}\]Given Base Area = 24 cm² and Height = 7.5 cm:\[V = \frac{1}{3} \times 24 \times 7.5\]\[V = 8 \times 7.5 = 60.0 \text{ cm}^3\]