I think I get 2 and 4, for that one. I think you are supposed to make up an equation, such as xy ≤ x²
Answer:
Step-by-step explanation:
![\sqrt[3]{64}=\sqrt[3]{2*2*2*2*2*2}\\\\ =\sqrt[3]{2^{2}*2^{2}*2^{2}}\\\\=2^{2}=4](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B64%7D%3D%5Csqrt%5B3%5D%7B2%2A2%2A2%2A2%2A2%2A2%7D%5C%5C%5C%5C%20%3D%5Csqrt%5B3%5D%7B2%5E%7B2%7D%2A2%5E%7B2%7D%2A2%5E%7B2%7D%7D%5C%5C%5C%5C%3D2%5E%7B2%7D%3D4)
Answer:
Step-by-step explanation:
Let the three numbers be x, y, z
So from the problem given we can get the equations as
:
x + y+z= 66
x=2y
z=1x/3
Now we will make equation 1 have only one variable using equation 2 and 3.
y=1x/2 since x=2y
Now using equations 1, 3 and 5 we will get
x+1x/2+1x/3=66
11x/6=66
x= 66*6/11
x= 36
y=x/2=36/2=18
z=x/3=36/3=12
Therefore:
x = 36
y=18
z=12
Answer is the graph on the far right.