Estimating the Cube Root of 250
To find the interval where the cube root of 250 lies, we need to find two consecutive integers whose cubes bracket the number 250.
We can test the cubes of integers:
- Calculate the cube of 5: $5^3 = 5 \times 5 \times 5 = 125$
- Calculate the cube of 6: $6^3 = 6 \times 6 \times 6 = 216$
- Calculate the cube of 7: $7^3 = 7 \times 7 \times 7 = 343$
Determining the Interval
We observe that 250 is greater than $6^3$ (which is 216) and less than $7^3$ (which is 343).
Mathematically, this can be represented as:
$ 6^3 < 250 < 7^3 $Taking the cube root of all parts of the inequality:
$ \sqrt[3]{6^3} < \sqrt[3]{250} < \sqrt[3]{7^3} $ $ 6 < \sqrt[3]{250} < 7 $Therefore, the cube root of 250 lies between the integers 6 and 7.
Similar Questions
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Important Questions from Cube and Cube Root
If 5 \(\sqrt{3}\) + \(\sqrt{75}\) = 17.32, then the value of 14 \(\sqrt{3}\) + \(\sqrt{108}\) is:
Find the value of (25)3 + (-29)3 + (4)3
The value of \(\frac{\sqrt[3]{-2744} \times \sqrt[3]{-216}}{\sqrt[3]{\frac{64}{729}}}\) is:
A number is cube of 53. When 7 times of 57 is subtracted from the number, then the resultant number which is formed will be divisible by:
The sum of a positive number and its cube is 1740. What is the value of the number?
