Only two prime numbers lie between 50 and 60. They are 53 and 59.
Counting only integer square roots, since sqrt(5) is equal to ±25 and sqrt(6) is equal to ±36, the square roots of the integers between 25 and 36, and between -36 and -25, are between 5 and 6 inclusive.
Let's call this number b 42000 = 24 x 3 x 53 x 7 We have 42000 x b = a2 and a is an integer Then a = sqrt(42000 x b) = sqrt(b) x sqrt(24 x 3 x 53 x 7) sqrt(24 x 3 x 53 x 7) = 22 x sqrt(3) x 5 x sqrt(5) x sqrt(7) For a to be an integer sqrt(b) must be sqrt(3) x sqrt(5) x sqrt(7) at least Thus b = 105
-sqrt(59) < -3, 6 < sqrt(59)
sqrt(6*10) = sqrt(60) = 7.75 (approx)
There are many. One example is -sqrt(175) < -3 < -2 < +sqrt(175)
0 and 1 are consecutive whole numbers that lie between -sqrt(124) and +sqrt(124).
Only two prime numbers lie between 50 and 60. They are 53 and 59.
The square root of 75 lies between 8 and 9.
sqrt(13) = 3.6... so it lies between 3 and 4.
20*pi is significantly larger than 4*sqrt(53).
Counting only integer square roots, since sqrt(5) is equal to ±25 and sqrt(6) is equal to ±36, the square roots of the integers between 25 and 36, and between -36 and -25, are between 5 and 6 inclusive.
Let's call this number b 42000 = 24 x 3 x 53 x 7 We have 42000 x b = a2 and a is an integer Then a = sqrt(42000 x b) = sqrt(b) x sqrt(24 x 3 x 53 x 7) sqrt(24 x 3 x 53 x 7) = 22 x sqrt(3) x 5 x sqrt(5) x sqrt(7) For a to be an integer sqrt(b) must be sqrt(3) x sqrt(5) x sqrt(7) at least Thus b = 105
121 < 128.3 < 144So -12 < negative sqrt(128.3) < -11and11 < sqrt(128.3) < 12121 < 128.3 < 144So -12 < negative sqrt(128.3) < -11and11 < sqrt(128.3) < 12121 < 128.3 < 144So -12 < negative sqrt(128.3) < -11and11 < sqrt(128.3) < 12121 < 128.3 < 144So -12 < negative sqrt(128.3) < -11and11 < sqrt(128.3) < 12
Between which two consecutive whole numbers does \sqrt{83} 83 lie? Fill out the sentence below to justify your answer and use your mouse to drag \sqrt{83} 83 to an approximately correct location on the number line.
53 and 55
251.5 = 253/2 = [ sqrt(25) ]3 = 53 = 125