taking the cube root of the volume.
In this case, the volume is 125 in3.
So, the length of one side of the cube is:
\( \text{side length} = \sqrt[3]{\text{volume}} = \sqrt[3]{125} = 5 \, \text{inches} \)
Therefore, the length of one side of the cube is 5 inches.
You are given a cube. If the volume of the cube is 125 in3 , you can find the length of one of the sides by
5 answers
Which two perfect squares does √300 lie between?
To find the two perfect squares between which √300 lies, we can calculate the square root of the numbers close to 300.
√289 = 17
√324 = 18
So, √300 lies between 17 and 18 because 17^2 = 289 < 300 < 324 = 18^2.
Therefore, √300 lies between the perfect squares of 17 and 18.
√289 = 17
√324 = 18
So, √300 lies between 17 and 18 because 17^2 = 289 < 300 < 324 = 18^2.
Therefore, √300 lies between the perfect squares of 17 and 18.
x^2 = 9/121 which statement is true when you solve for x?
To solve for x when given that \(x^2 = \frac{9}{121}\), we can take the square root of both sides of the equation:
\[ x = \pm \sqrt{\frac{9}{121}} \]
To simplify the square root of \(\frac{9}{121}\), we can take the square root of the numerator and the denominator separately:
\[ x = \pm \frac{\sqrt{9}}{\sqrt{121}} = \pm \frac{3}{11} \]
So, the solutions for x are \(x = \frac{3}{11}\) and \(x = -\frac{3}{11}\).
The true statement when you solve for x is:
\[ x = \pm \frac{3}{11} \]
\[ x = \pm \sqrt{\frac{9}{121}} \]
To simplify the square root of \(\frac{9}{121}\), we can take the square root of the numerator and the denominator separately:
\[ x = \pm \frac{\sqrt{9}}{\sqrt{121}} = \pm \frac{3}{11} \]
So, the solutions for x are \(x = \frac{3}{11}\) and \(x = -\frac{3}{11}\).
The true statement when you solve for x is:
\[ x = \pm \frac{3}{11} \]