Page 56 - Hybrid PBD 2021 Form 2 - Matematik

Matematik Tingkatan 2 Bab 3
5. Ungkapkan pemboleh ubah dalam kurungan sebagai perkara rumus. SP 3.1.2 TP3
Express the variable in the brackets as the subject of the formula.

(i) x y = 36 [x] (ii)  = h [m]
2
m
4
Tip Penting
x y = 36 4m = h Kuasa dua
2
1 1 2 Square
m
x y × = 36 × 1 2 = h 2
4
2
y y x x 2
36 4m = h 2
2
x = 1 1
y 4m × = h × Punca kuasa dua
2
Songsangan bagi kuasa 4 4 Square root
 dua bagi nilai itu. m = h Kuasa tiga
x
 = 36 dua ialah punca kuasa 2
2
y
The inverse of square is the
 square root of the value. 4 Songsangan bagi Cube
36
punca kuasa dua ialah
x = x x 3
y kuasa dua bagi nilai itu.
The inverse of a square root
6 is the square of the value. Punca kuasa tiga
x = Cube root
y
(a) p = rq + 2sq 2 [q] k 1
2
(b) x = 5 y  [y] (c) V = s²h [s]
3
rq + 2sq = p
2
2
q (r + 2s) = p 1  y 2 2 2 1 s²h = V
2
k
3
1 1 5 = x
q (r + 2s) × = p × 1 3 3
2
r + 2s r + 2s 25k s²h × = V ×
p = x 2 3 h h
q = y
2
r + 2s 3V
²
s
25k 1 1  =
2
p × = x × h
2
 = y 25k 25k
q
r + 2s 3V
1 x 2 s =
p y = 25k h
q =
r + 2s 25k
y =
x 2
2
2
4 1 (f) p = h – 9 [h]
(d) V = pr 3 [r] (e) e =  [f ]
3 f
h – 9 = p 2
2
4 pr = V 1 h – 9 + 9 = p + 9
3
2
2
3  = e
f
2
2
4 pr × 3 = V × 3 2 h = p + 9
3

1
h
p
+ 9
2
3 4p 4p 1 2 = e 2  = 
2
3V f 2
+ 9
p
r = 1 h = 
3
4p f = e 2
 f = e 1 2
3 3 3V

r
 = 3
4p
3V

r = 3 4p
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Bab 3.indd 29 9/3/21 5:13 PM