Lesson 8, Topic 1
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# Formulae & Substitution

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One of the major uses of Algebra is to express mathematical relationships in form of formulae.

A formulae is an equation that describes the relationship between two or more quantities.

Letters are used to represent unknown values in a formulae.

For example the frequency f in cycles per unit time i.e.

F = $$\frac {\omega}{2\pi}$$

This formulae shows the relationship between frequency (f) and circular velocity ($$\scriptsize \omega$$)

F is called the subject of the formulae or equation.

i. If, F = ma, find F when  m = 10 and  a = 6

F = ma

F = 10 x 6

F = 60N

ii. If y = m + c,  find y when m = 5, = -2,   c  = -3.

y = mx   +  c

y = 5 x -2 + -3

y = -10 – 3

y  = -13

iii. If, V = mgh  find v when  m = 10, g = 9.8  and  h = 5

V  = mgh

V = 10 x 9.8 x 5   =  490

iv. If y =  16x2  – 9z2,  find y when x = 7, and  z  =  5

y =  16 (7)2 – 9 (5)2

y = 16 x 49 – 9 x 25

y = 784 – 225

y = 559

v. The sum of terms in a geometric progression is given as

s = $$\frac{a( 1 \; – \; r^n)}{1 \; – \; r}$$

Calculate s when  n = 4, a = 3 and r = 3

s = $$\frac{a( 1 \; – \; 3^4)}{1 \; – \; 3}$$

s = $$\frac{3( 1 \; – \; 81)}{1 \; – \; 3}$$

s = $$\frac{ 3(-80)}{-2}$$

s = $$\frac{-240}{-2}$$

s = 120

vi. Here’s a formulae for finding the area of a triangle;

A = $$\scriptsize \sqrt{s(s \; – \; a)(s \; – \; b)(s \; – \; c)}$$

Where s = $$\frac{a + b + c}{2}$$  and a, b and c are the three sides

Find the area of a triangle when a = 5cm, b = 8cm, c = 11cm

s = $$\frac{a + b + c}{2} = \frac{5 + 8 + 11}{2} \\ = \frac{24}{2} \\ \scriptsize = 12$$

A = $$\scriptsize \sqrt{s(s \; – \; a)(s \; – \; b)(s \; – \; c)}$$

A = $$\scriptsize \sqrt{12(12 \; – \; 5)(12 \; – \; 8)(12 \; – \; 11)}$$

A = $$\scriptsize \sqrt{12(7)(4)(1)}$$

A = $$\scriptsize \sqrt{12 \; \times \; 28} \\ \scriptsize = \sqrt{336} \\ \scriptsize = 18.33$$

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