Question

Class VII DAV MATHEMATICS . Chapter 4 Exponents and powers . Worksheet 4 . please help me friends .Please no SPAMING ❌❌ otherwise I will report​

Answer 1

Answer:

(i)(-3/4)³÷(-3/4)⁵=(-3/4)³-⁵=(-3/4)-²=(-4/3)²=16/9

(ii)(1/(5)²)²×(1/5)=(1/25)²×(1/5)=1/625×1/5=1/3125

(iii)[(-5/6)²]²÷(-5/6)²=(-5/6)⁴÷(-5/6)²=(-5/6)⁴-²=(-5/6)²=25/36

(iv)(2/3)²÷[(2/3)²]⁰=(2/3)²÷(2/3)⁰=(2/3)²÷1=(2/3)²=4/9

(v)[(1/2)⁵÷(1/2)³]-[(1/2)⁶÷(1/2)⁵]=[(1/2)²-(1/2)¹]=(1/4-1/2)=1-2/4=-1/4

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Answer 2

Given data: $(-\frac{3}{4} )^3$÷$(-\frac{3}{4})^5$,  $(\frac{1}{5^2} )^2$×$(\frac{1}{5} )$, $[(-\frac{5}{6} )^2]^2$÷$(-\frac{5}{6} )^2$, $(\frac{2}{3} )^2$÷$[(\frac{2}{3} )^2]^0$ and $\frac{(\frac{1}{2})^5 }{(\frac{1}{2}) ^3}$ - $\frac{(\frac{1}{2})^6 }{(\frac{1}{2}) ^5}$

To find: Simplify the given values

Solution:

(i) $(-\frac{3}{4} )^3$÷$(-\frac{3}{4})^5$

=> $(-\frac{27}{64})$÷$(-\frac{243}{1024})$ => $-\frac{27}{64}$ × $- \frac{1024}{243}$ => $\frac{16}{9}$

(ii) $(\frac{1}{5^2} )^2$×$(\frac{1}{5} )$

=> $(\frac{1}{25} )^2$×$(\frac{1}{5} )$ => $\frac{1}{625}$ × $\frac{1}{5}$ => $\frac{1}{3125}$

(iii) $[(-\frac{5}{6} )^2]^2$÷$(-\frac{5}{6} )^2$

=> $(\frac{25}{36} )^2$÷$(\frac{25}{36} )$ => $\frac{625}{1296}$ × $\frac{36}{25}$ => $\frac{25}{36}$

(iv) $(\frac{2}{3} )^2$÷$[(\frac{2}{3} )^2]^0$

=> $\frac{4}{9}$÷1 (Anything to the power 0 is 1) => $\frac{4}{9}$

(v) $\frac{(\frac{1}{2})^5 }{(\frac{1}{2}) ^3}$ - $\frac{(\frac{1}{2})^6 }{(\frac{1}{2}) ^5}$

=> ${(\frac{1}{2})^5 } X {(\frac{2}{1}) ^3}$ - ${(\frac{1}{2})^6 } X {(\frac{2}{1}) ^5}$ => $\frac{1}{2^2}$ - $\frac{1}{2}$ => $\frac{1-2}{4}$ => $-\frac{1}{4}$