Mathematics Questions and Answers

Volume of a cone = \(\frac{1}{3}\pi r^2 h\)

r = 5 cm

l = 13 cm

Using Pythagoras theorem

⇒ \(13^2 = 5^2 + h^2\)

⇒ \(169 = 25 + h^2\)

⇒ \(169 - 25 = h^2\)

⇒ \(h^2 = 144\)

⇒ \(h = \sqrt144 = 12 cm\)

∴ Volume of the cone = \(\frac{1}{3} \times\pi\times 5^2 x 12 = 100\pi\) cm\(^3\)

First, we find the equation of the boundary line using the two intercepts.The slope is

m = \(\frac{4 - 0}{0 - 5} = {4}{5}\)

The y-intercept is 4

The slope-intercept form of the equation is therefore

y = -\(\frac{4}{5} x + 4\)

\(\implies y + \frac{4}{5} x = 4\)

Multiply both sides by \(\frac{5}{4}\)

\(\implies\frac{5}{4}(y +\frac{4}{5} x) = 4\times\frac{5}{4}\)

\(\implies\frac{5}{4} y + x = 5\)

Multiply both sides by 4

⇒ \(5y + 4x = 20\)

The inequality is therefore either \(5y + 4x ≤ 20\) or \(5y + 4x ≥ 20.\)

Using the test point (0, 0) -The origin

⇒ 5(0) + 4(0) ≤ 20

⇒ 0 ≤ 20 (True)

∴ The inequality is \(5y + 4x ≤ 20\)

Intersecting Chords Theorem states that If two chords intersect in a circle, then the products of the measures of the segments of the chords are equal.

⇒ AE * EB = CE * ED

⇒ 6 * \(x\) = 4 * (\(x\) + 5)

⇒ 6\(x\) = 4\(x\) + 20

⇒ 6\(x\) - 4\(x\) = 20

⇒ 2\(x\) = 20

∴ \(x = \frac{20}{2}\) = 10 units

Area of the composite figure = Area of semi circle + Area of rectangle + Area of triangle

Area of semi circle = \(\frac{1}{2}\pi r^2 = \frac{1}{2}\times\pi\times\frac{d^2}{4} = \frac{1}{2}\times\frac{22}{7}\times\frac{42^2}{4} = 693 m^2\)

Area of rectangle = l x b = 42 x 60  =2520 m\(^2\)

Area of triangle = \(\frac{1}{2}\times b \times h = \frac{1}{2}\times 36 \times 42 = 756 m^2\)

∴ Area of the composite figure = 693 + 2520 + 756 = 3969 m\(^2\)

\(log_2 (6 - x) = 3 - log_2 x\)

⇒ \(log_2 (6 - x) = 3 log_2 2 - log_2 x\) (since \(log_2\) 2 = 1)

⇒ \(log_2 (6 - x) = log_2 2^3 - log_2 x\) \((a log\) c = \(log\) c\(^a)\)

⇒ \(log_2 (6 - x) = log_2 8 - log_2 x\)

⇒\(log_2 (6 - x) = log_2 \frac{8}{x}\) (\(log\) a - \(log\) b = \(log \frac{a}{b})\)

⇒ \(6 - x = \frac{8}{x}\)

⇒ \(x (6 - x) = 8\)

⇒ \(6x - x^2 = 8\)

⇒ \(x^2 - 6x + 8 = 0\)

⇒ \(x^2 - 4x - 2x + 8 = 0\)

⇒ \(x (x - 4) - 2(x - 4) = 0\)

⇒ \((x - 4)(x - 2) = 0\)

⇒ \(x - 4 = 0 or x - 2 = 0\)

∴ x = 4 or 2

Let number of children's ticket at ₦250.00 each = \(x\)

∴ Number of adult tickets at ₦520.00 each = 5\(x\)

Then,

Total amount of money received from children's tickets = 250\(x\)

Total amount of money received from adult tickets = 520(5\(x\))

⇒ 250\(x\) + 520(5\(x\)) = 171,000

⇒ 250\(x\) + 2600\(x\) = 171,000

⇒ 2850\(x\) = 171,000

⇒ \(x = \frac{171,000}{2850} = 60\)

∴ 60 tickets were sold at ₦250.00 and 300 tickets were sold at ₦520.00

The line: \(3y + 6x\) = 48

Divide through by 3

⇒ y + 2\(x\) = 16

⇒ y = -2\(x\) + 16

∴ The gradient of the line = -2

The points: A(-2, k) and B (4, 8)

m =\(\frac{y2 - y1}{x2 - x1} = \frac{8 - k}{4 - (-2)}\)

⇒ m =\(\frac[8 - k}{4 + 2} = {8 - k}{6}\)

Since the line passes through the points

∴ -2 = \(\frac{8 - k}{6}\)

⇒ \(\frac{-2}[1} = \frac{8 - k]{6}\)

⇒ 8 - k = -12

⇒ k = 8 + 12

∴ k = 20

\(PR^2 = PQ^2 + RQ^2 - 2(PQ)(RQ)cos Q\)

\(\implies cos Q = \frac{PQ^2 + RQ^2 - PR^2}{2(PQ)(RQ)}\)

\(\implies cos Q = \frac{8^2 + 5^2 - 7^2}{2\times8\times5}\)

\(\implies cos Q = \frac{64 + 25 - 49}{80}\)

\(\implies cos Q = \frac{40}{80} = 0.5\)

\(\implies Q = cos^{-1} (0.5) = 60^0\)

\(\therefore x = 60^0\)

\(T_2 = \frac{-2}{3};S_\infty \frac {3}{2}\)

\(T_n = ar^n - 1\)

∴ \(T_2 = ar = \frac{-2}{3}\)---eqn.(i)

\(S_\infty = \frac{a}{1 - r} = \frac{3}{2}\)---eqn.(ii)

= 2a = 3(1 - r)

= 2a = 3 - 3r

∴ a = \(\frac{3 - 3r}{2}\)

Substitute \(\frac{3 - 3r}{2}\) for a in eqn.(i)

= \(\frac{3 - 3r}{2} \times r = \frac{-2}{3}\)

= \(\frac{3r - 3r^2}{2} = \frac{-2}{3}\)

= 3(3r - 3r\(^2\)) = -4

= 9r - 9r\(^2\) = -4

= 9r\(^2\) - 9r - 4 = 0

= 9r\(^2\) - 12r + 3r - 4 = 0

= 3r(3r - 4) + 1(3r - 4) = 0

= (3r - 4)(3r + 1) = 0

∴ r = \(\frac{4}{3} or - \frac{1}{3}\)

For a geometric series to go to infinity, the absolute value of its common ratio must be less than 1 i.e. |r| < 1.

∴ r = -\(^1/_3\) (since |-\(^1/_3\)| < 1)

Let the length of the longer side = \(x\) cm

∴ The length of the shorter side = (\(x\) - 6) cm

If we increase each side's length by 2 cm, it becomes

(\(x\) + 2) cm and (\(x\) - 4) cm respectively

Area of a rectangle = L x B

\(A_1 = x(x - 6) = x^2 - 6x\)

\(A_2 = (x + 2)(x - 4) = x^2 - 4x + 2x - 8 = x^2 - 2x - 8\)

\(A_1 + 68 = A_2\) (Given)

⇒ \(x^2 - 6x + 68 = x^2 - 2x - 8\)

⇒ \(x^2 - x^2 - 6x + 2x\) = -8 - 68

⇒ -4\(x\) = -76

⇒ \(x\) = \(\frac{-76}{-4}\) = 19cm

∴ The length of the shorter side = \(x\) - 6 = 19 - 6 = 13 cm