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3 years ago

The square root of 105 would lie between which two consecutive integers?

Mathematics

1 answer:

12345 [234]3 years ago

8 0

Answer:

the square root of 105 lies between 10 and 11

Step-by-step explanation:

Which perfect squares "box" 105 in? 10^2 = 100 and 11^2 = 121.

The square roots of 100 and 121 are 10 and 11.

Thus, the square root of 105 lies between 10 and 11.

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$a.\hspace{3} \frac{6}{38} = 0.15789474\\\\b.\hspace{3} \frac{45}{55} = 0.81818181\\\\c.\hspace{3} \frac{4}{28} = 0.14285714\\\\d.\hspace{3} \frac{35}{28} = 1.25\\\\e.\hspace{3} \frac{1030}{2030} = 0.50738916$

Step-by-step explanation:

Rational numbers, expressed as decimal numbers, are obtained from the operation of division between the integer of the numerator and the integer of the denominator. Then:

$a.\hspace{3} \frac{6}{38} = 6\div38 = 0.15789474\\\\ b.\hspace{3} \frac{45}{55} = 45\div55 = 0.81818181\\\\c.\hspace{3} \frac{4}{28} = 4\div28 = 0.14285714\\\\ d.\hspace{3} \frac{35}{28} = 35\div28 = 1.25\\\\ e.\hspace{3} \frac{1030}{2030} = 1030\div2030 = 0.50738916$

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