JAMB - Mathematics (1983 - No. 8)
If 0.0000152 x 0.00042 = A x 10\(^{B}\), where 1 \(\leq\) A < 10, find A and B
A = 9, B = 6.38
A = 6.38, B = -9
A = -6.38, B = -9
A = -9, B = -6.38
Explanation
0.0000152 x 0.00042 = A x 10\(^{B}\)
1 \(\leq\) A < 10, it means values of A include 1 - 9
0.0000152 = 1.52 x 10\(^{-5}\)
0.00042 = 4.2 x 10\(^{-4}\)
1.52 x 4.2 = 6.384
10\(^{-5}\times 10^{-4}\)
= 10\(^{-5-4}\)
= 10\(^{-9}\)
= 6.38 x 10\(^{-9}\)
A = 6.38, B = -9
1 \(\leq\) A < 10, it means values of A include 1 - 9
0.0000152 = 1.52 x 10\(^{-5}\)
0.00042 = 4.2 x 10\(^{-4}\)
1.52 x 4.2 = 6.384
10\(^{-5}\times 10^{-4}\)
= 10\(^{-5-4}\)
= 10\(^{-9}\)
= 6.38 x 10\(^{-9}\)
A = 6.38, B = -9
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