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Sagot :
You can also figure it out by knowing that 7/8 is 1/8 away from 1 and 11/10 is 1/10 away from 1. since 1/10 is smaller than 1/8, you know that 11/10 is closer to 1 than 7/8.
Let
A-----> [tex] \frac{7}{8} [/tex]
B-----> [tex] \frac{11}{10} [/tex]
C-----> [tex] 1 [/tex]
we know that
Point A-------> Multiply numerator and denominator by [tex] 10 [/tex]
[tex] \frac{7*10}{8*10} =\frac{70}{80} [/tex]
Point B-------> Multiply numerator and denominator by [tex] 8 [/tex]
[tex] \frac{11*8}{10*8} =\frac{88}{80} [/tex]
Point C-------> Multiply numerator and denominator by [tex] 80 [/tex]
[tex] \frac{1*80}{1*80} =\frac{80}{80} [/tex]
Find the distance Point A to Point C
[tex] \frac{80}{80} -\frac{70}{80} =\frac{10}{80} [/tex]
Find the distance Point B to Point C
[tex] \frac{88}{80} -\frac{80}{80} =\frac{8}{80} [/tex]
therefore
[tex] \frac{8}{80} < \frac{10}{80} [/tex]
Point B is closer to Point C
[tex] \frac{11}{10} [/tex] is closer to [tex] 1 [/tex]
the answer is
[tex] \frac{11}{10} [/tex] is closer to [tex] 1 [/tex]
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