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## Here's the Solution to this Question

(i) As $x=n+ε, 0≤ε<1$

Given relation is- $\dfrac{55x}{7}=8n$

(a) Let x=14.3, then n=14 as $x=n+ε$

Taking LHS$=\dfrac{55\times 14.3}{7}=112.35$

Taking RHS=$8(14)=112$

(b) Let x=37.7, then n=37

Taking LHS=$\dfrac{55\times 37.7}{7}=296.4$

Taking RHS=$8(37)=296$

(c) Let x=46.9, then n=46

Taking LHS=$\dfrac{55\times 46.9}{7}=368.5$

Taking RHS=$8(46)=368$

(d) Let x=51.6, then n=51

Taking LHS=$\dfrac{55\times 51.6}{7}=405.42$

Taking RHS=$8(51)=408$

The difference between the LHS and RHS is more in part D,

Hence The value of x can not be 51.6

(ii) Given Relation is $\dfrac{33x}{8}=5n$

(a) Let x=5.8 , then n=5

Taking LHS =$\dfrac{33\times 5.8}{8}=23.925$

Taking RHS=$5(5)=25$

(b) Let x=4.7 , then n=4

Taking LHS =\dfrac{33\times 4.7}{8}=19.3875

Taking RHS=5(4)=20

(c) Let x=6.1 , then n=6

Taking LHS $=\dfrac{33\times 6.1}{8}=25.1625$

Taking RHS=$5(6)=30$

(d) Let x=2.5 , then n=2

Taking LHS =$\dfrac{33\times 2.5}{8}=10.3125$

Taking RHS=$5(2)=10$

The difference between the RHS and LHS is significant in part (a) and (C)

Therefore x can not have the values 5.8 and 6.1