Respuesta :
Let's write an inequality, such as follows: x < sqrt(50) < y. Square both sides of the equation. We get x^2 < 50 < y^2. Obviously, x is between 7 and 8. Also notice, that for integers a,b, (ab)^2/b^2, equals a^2. So let's try values, like 7.1. Using the previous fact, (7.1)^2, equals (71)^2/100. So, (7.1)^2, equals 50.41. Thus, our number is between 7 and 7.1. We find, with a bit of experimentation, that the square root of 50, is 7.07.
Answer:
the square root of 50 rounded to the nearest hundredth is 7.07.
Step-by-step explanation:
We have to find the square root of 50 that is rounded to the nearest hundredth.
For evaluating the square root, we need the factorize 50 and take common factors out of the square root.
[tex]\begin{aligned}\sqrt{50}&=\sqrt{2 \times 5 \times 5}\\&=5 \sqrt {2}\\&=5 \times 1.414 \;(\rm{approx})\\&=7.07 \;(\rm{approx}) \end{aligned}[/tex]
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