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.
To find the midpoint, add x1 and x2 then divide by 2, then add y1 and y2 and divide by 2.
Midpoint = (2a +2b /2 , 0 + 2c /2)
Midpoint = a+b, c
Answer:
0
Step-by-step explanation:
6-6 equals 0 and -8 minus 6 equals -14 and 0 divided by equals 0
The answer is
3^2 times 13
Answer:
150
Step-by-step explanation: multiply 10 and 12 and get your rectangle area then multiply .5( 10 and 6) then add them together and get 150
