Mathematics 2022 Jamb Examination

Standard egg crates contain 30 eggs. First, find the total eggs sold:\[5.5 \times 30 = 165 \text{ eggs sold}\]Next, find the eggs remaining:\[200 - 165 = 35 \text{ eggs left}\]Now, calculate the percentage left:\[\frac{35}{200} \times 100 = 17.5\%\]

Use the identity \(\sec^2 \theta = 1 + \tan^2 \theta\). Substitute this into the equation:\[(1 + \tan^2 \theta) + \tan^2 \theta = 3\]\[1 + 2\tan^2 \theta = 3\]\[2\tan^2 \theta = 2 \implies \tan^2 \theta = 1\]\[\tan \theta = 1 \implies \theta = \tan^{-1}(1) = 45^o\]

Direct substitution gives \(\frac{0}{0}\). We can use L'Hôpital's rule by differentiating the numerator and denominator:\[\lim_{x \to 1} \frac{3x^2 + 6}{1}\]Substitute \(x = 1\):\[3(1)^2 + 6 = 9\]Alternatively, you can factor the numerator: \(x^3 + 6x - 7 = (x-1)(x^2 + x + 7)\). Canceling \((x-1)\) leaves \(x^2 + x + 7\), which equals 9 when \(x=1\).

First, find the gradient of \(2x - y = 5\) by rewriting it as \(y = 2x - 5\). The gradient is 2.
Parallel lines have the same gradient, so our new line is \(y = 2x + c\).
The question specifies an intercept of 5 (usually implying the y-intercept). Thus, \(y = 2x + 5\).
Rearranging to match the options: \(2x - y = -5\).

First, arrange the data in ascending order:
10, 10, 13, 14, 17, 18.
Since there is an even number of values (6), the median is the average of the two middle terms (13 and 14):\[\text{Median} = \frac{13 + 14}{2} = 13.5\]

Convert the integer part: \((1 \times 2^4) + (0 \times 2^3) + (1 \times 2^2) + (1 \times 2^1) + (0 \times 2^0) = 16 + 0 + 4 + 2 + 0 = 22\).
Convert the fractional part: \((0 \times 2^{-1}) + (1 \times 2^{-2}) + (1 \times 2^{-3})\)\[0 + \frac{1}{4} + \frac{1}{8} = 0.25 + 0.125 = 0.375\]Combined, the value is 22.375.

Direct proportion means \(C = kN\).
Find \(k\): \(10,000 = k(20) \implies k = 500\).
Now, find \(N\) when \(C = 1,000,000\):\[1,000,000 = 500N\]\[N = \frac{1,000,000}{500} = 2,000\]

[attachment_2](attachment)
The midpoint formula is \(\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\):\[x_m = \frac{-3 + 2}{2} = -\frac{1}{2}\]\[y_m = \frac{5 + 10}{2} = \frac{15}{2}\]The midpoint is \((-\frac{1}{2}, \frac{15}{2})\).

First, substitute \(n = 3\) into the expressions:\[n^2 + 1 = 3^2 + 1 = 10\]\[n + 5 = 3 + 5 = 8\]Calculate ¹⁰C₈, which is equal to ¹⁰C₂:\[\frac{10 \times 9}{2 \times 1} = 45\]

Velocity (v) is the derivative of displacement (S) with respect to time (t):\[v = \frac{dS}{dt} = 6t + 5\]Initial velocity occurs at \(t = 0\):\[v = 6(0) + 5 = 5 \text{ m/s}\]