Mathematics Questions and Answers

No explanation has been provided for this answer.

\( A = P \left(1 + \frac{r}{100}\right)^n \)

Where P = 126, r = 4,n = 2

A=126 \( \left(1 + \frac{4}{100}\right)^2 \text{Using LCM} \)

=126 \( \left(\frac{100+4}{100}\right)^2 = 126 \left(\frac{104}{100}\right)^2 \)


=126 \( \left(1.04^2 \right) \)

= 126 * 1.04 * 1.04

=136.28

A = 136.30 (approx.)

The Amount A, = N136.30k

The slope of the line from (0, 0) passing through (-3, -4) = \(\frac{-4 - 0}{-3 - 0}\)

= \(\frac{4}{3}\)

Equation of a line is given as \(y = mx + b\), where m = slope and b = intercept.

To get the value of b, we use a point on the line, say (0, 0).

\(y = \frac{4}{3} x + b\)

\(0 = \frac{4}{3}(0) + b\)

\(b = 0\)

The equation of the line is \(y = \frac{4}{3} x\)

nth term of a linear sequence (AP) = a+(n − 1)d

first term = 6, last term = 10 sum − 40

i.e. a = 6, l = 10, S = 40

S\(_{n}\)= n/2(2a + (n − 1)d or Sn = ÷2 (a + l)

S\(_{n}\) = n/2(a + l)

40 = n/2(6 + 10)

40 = 8n

8n = 40

8n = 40

n = 40/8

= 5

The number of terms = 5

Using the quadratic formula, we have

\(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\)

From the equation \(3x^{2} - 4x - 5 = 0\), a = 3, b = -4 and c = -5.

\(\therefore x = \frac{-(-4) \pm \sqrt{(-4)^{2} - 4(3)(-5)}}{2(3)}\)

= \(\frac{4 \pm \sqrt{16 + 60}}{6}\)

= \(\frac{4 \pm \sqrt{76}}{6}\)

= \(\frac{4 \pm 8.72}{6}\)

= \(\frac{4 + 8.72}{6} \text{ or } \frac{4 - 8.72}{6}\)

= \(\frac{12.72}{6} \text{ or } \frac{-4.72}{6}\)

\(x = \text{2.12 or -0.79}\)