It is an irrational number that lies between 7 and 8
The square root of 53 lies between the consecutive integers 7 and 8. This is because (7^2 = 49) and (8^2 = 64), so (49 < 53 < 64). Therefore, (\sqrt{53}) is greater than 7 but less than 8.
Only two prime numbers lie between 50 and 60. They are 53 and 59.
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
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.
A real number between (\sqrt{13}) and (\sqrt{14}) is (\sqrt{13.5}). This value lies between the two square roots, as (\sqrt{13} \approx 3.60555) and (\sqrt{14} \approx 3.74166). Thus, (\sqrt{13.5} \approx 3.67423) fits within that range.
The square root of 53 lies between the consecutive integers 7 and 8. This is because (7^2 = 49) and (8^2 = 64), so (49 < 53 < 64). Therefore, (\sqrt{53}) is greater than 7 but less than 8.
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).
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
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.
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
A real number between (\sqrt{13}) and (\sqrt{14}) is (\sqrt{13.5}). This value lies between the two square roots, as (\sqrt{13} \approx 3.60555) and (\sqrt{14} \approx 3.74166). Thus, (\sqrt{13.5} \approx 3.67423) fits within that range.
To find the sine of the angle ( \theta ) (feta) in standard position with the point (2, -7) on its terminal side, we first determine the length of the radius (r) using the Pythagorean theorem: ( r = \sqrt{x^2 + y^2} = \sqrt{2^2 + (-7)^2} = \sqrt{4 + 49} = \sqrt{53} ). The sine of the angle is given by the ratio of the y-coordinate to the radius: ( \sin(\theta) = \frac{y}{r} = \frac{-7}{\sqrt{53}} ). Thus, ( \sin(\theta) = -\frac{7}{\sqrt{53}} ).