Exercise 2A
Question 1
Evaluate
(a) 90
Sol : 1
(b) $\left(\frac{1}{3}\right)^0$
Sol : 1
(c) $\left(-\frac{64}{21}\right)^0$
Sol : 1
(d) (395.008)0
Sol : 1
(e) $\left(\frac{3xy}{ab}-z\right)^0$
Sol : 1
(f) $\frac{1}{a^0 + b^0 + c^0}$
Sol : $=\frac{1}{3}$
(g) (80 – 20)x 250
Sol :
=(1-1)×1
=0×1=0
(h) (350 + 20 + 30) ÷ 3
Sol : 1
Question 2
Simplifying giving the answer as a fraction:
(a) 6-1
(b) 10-1
Sol : $=\frac{1}{10}$(c) 4-3
Sol : $=\frac{1}{4^3}=\frac{1}{64}$(d) 2-7
Sol :(e) 3-3
Sol :(f) p-1
Sol :(g) (x-2)2
Sol :(h) (10a)-2
Sol :(i) (x+y)-2
Sol :(j) (m5)-3
Sol :Question 3
Find the reciprocal of the following.
(a) (3)-2
Sol :(b) (-2)-5
Sol :(c) $\left(\frac{2}{5}\right)^{-3}$
Sol :
$=\left(\frac{5}{2}\right)^3=\frac{125}{8}$
(d) $\left(\frac{2}{3}\right)^{-4}$
Sol :
$=\left(\frac{3}{2}\right)^4=\frac{81}{16}$
Question 4
Simplify and express the result in power notation with positive exponent.
(a) (-2)-4 x (-2)-6
(b) 37 x 3-9 x 36
Sol : $=\frac{1}{3^7}\times \frac{1}{3^9}\times \frac{1}{3^{6}}$(c) $3^7\times \left(\frac{3}{5}\right)^{-7}$
Sol : $=3^7 \times \left(\frac{5}{3}\right)^7=3^7 \times \frac{5^7}{3^{7}}$
(d) 2-7 x [(2)-3]-4
Sol : $=\frac{1}{2^7}\times 2^{12}$
(f) (4-8 ÷ 4-10) x 4-5
Sol : $=\frac{4^{-8}}{4^{-10}}\times 4^{-5}$ $=\frac{1}{4^2}\times 4^{-5}=4^{-3}=\frac{1}{4^3}$
Question 5
Evaluate:
(a) (80 + 5-1)×$\left(\frac{1}{5}\right)^{-2}$
Sol :
$=\left(1+\frac{1}{5}\right)\times (5^{-1})^{-2}$
$=\frac{5+1}{5}\times 5^2$
$=\frac{6}{5}\times 5^2$
=6×5=30
(b) (300+ 150)×$\left(\frac{1}{2}\right)^{-3}$
Sol :
=(1+1)×(2-1)-3=2×23
=24=16
(c) (7 -1+ 8-1 + 9-1)0÷$\left(-\frac{1}{9}\right)^{-2}$
Sol :
$=\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}\right)^0 \div \left(\frac{9}{-1}\right)^2$
$=1\div \frac{81}{1}=\frac{1}{81}$
(d) $\left[\left(\frac{1}{3}\right)^{-2}+\left(\frac{1}{4}\right)^{-2}+\left(\frac{1}{6}\right)^{-2}\right]$
Sol :
=32+42+62
=9+16+36
=25+36=61
(e) $\left\{\left(-\frac{4}{5}\right)^{-2}\right\}^{-2}$
Sol :
$=\left\{\left(-\frac{5}{4}\right)\right\}^{-2}=\left(-\frac{5}{4}\right)^{-4}$
$=\left(\frac{4}{5}\right)^4=\frac{256}{625}$
(f) $9^{-1}\times \frac{4^3}{3^{-4}}$
Sol :
$=\frac{(3^2)^{-1}\times 4^3}{3^{-4}}=\frac{3^{-2}\times 4^3}{3^{-4}}$
$=\frac{4^3}{3^{-2}}$
=43×32=576
(g) $(5^{-1}\times 4^{-1})\times \left(\frac{3}{2}\right)^{-2}$
Sol :
$=\left(\frac{1}{5}\times \frac{1}{4}\right)\times \left(\frac{2}{3}\right)^2$
$=\frac{1}{20}\times \frac{2^2}{3^2}$
$=\frac{1}{20}\times \frac{4}{9}=\frac{1}{45}$
Question 6
Evaluate:
(a) $\left[\left(\frac{1}{4}\right)^{-1}-\left(\frac{1}{5}\right)^{-1}\right]+1$
Sol :
=(4-5)+1
=-1+1=0
(b) $\left[\left(\frac{11}{12}\right)^{-10} \div \left(\frac{11}{12}\right)^{-12}\right]\times 1\frac{23}{121}$
Sol :
$=\left[\left(\frac{12}{11}\right)^{10}\div \left(\frac{12}{11}\right)^{12}\right]\times \frac{144}{121}$
$=\left(\frac{12}{11}\right)^{-2}\times \left(\frac{12}{11}\right)^{2}$
$=\frac{(11)^2}{(12)^2}\times \frac{(12)^2}{(11)^2}$
=1
Question 7
Find the value of m for which, 72m ÷ 7-4 = 7-6
Sol :
$=\frac{7^{2m}}{7^{-4}}=$7-6
⇒72m=7-6×7-4=7-10
⇒(72)m=(72)-5
⇒m=-5
Waw
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