Well the answer is No.A remainder will always be smaller than the divisor ,because the divisor is the one that used to divide the dividend and the remaining number that cannot be divided by the divisor is called the remainder.
Questions?
Answer:
No
Reasoning:
If something is a perfect cube, it is able to be put under a cube root (![\sqrt[3]{..}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B..%7D)
) and will result in an integer (a non-decimal number > 0, basically).
So let's calculate ![\sqrt[3]{48}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B48%7D)
, and see if the result is an integer.
= 3.634.......
As you can see, the result is not an integer, therefore 48 is not a perfect cube.
Answer:
1.
Mathematical concept that expresses an amount in relation to the computer unit; results from counting the elements that make up a set.
"Cardinal number"
2.
A graphic sign or set of graphic signs that expresses or represents that amount.
Step-by-step explanation:
Answer:
Step-by-step explanation
The length of the left sloping line
= √ (5^2 + 7^2) = √74.
The right sloping line
= √(5^2 + 5^2) = √50.
So the perimeter:
= √74 + √50 + 2 + 2 * 4 + 14
= 39.67 to nearest hundreth.
Area = 1/2 * 5 * 7 + 2*5 + 1/2 * 5 * 5 + 4*14
= 96.
The answer is 4 square root 80
