First you want to subtract 36
so it looks like this ![\sqrt[4] {(4x+164)^3}=64](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%20%7B%284x%2B164%29%5E3%7D%3D64)
Then you want to cancel out the square root 4 by raising that to the 4th power (you must do this to both sides)
which is equal to 
Then you take the cube root to both sides [tex]\sqrt[3]{(4x+164)^3}=\sqrt[3]{16777216}[tex]
Then you end up with the equation 4x+164=256
Then subtract 164 to both sides
4x=92
then divide 92 by 4
Then you get x=23
2504, as 2499 is 5 less, yet rounds to 2.5K. It is the highest possible for rounding to the tens, as 2505 rounds to 2510.
Answer:
2 ≤ x < 7
Step-by-step explanation:
-1 ≤ x – 3 < 4
Add 3 to each side
-1+3 ≤ x – 3+3 < 4+3
2 ≤ x < 7
Answer:
2-5
Step-by-step explanation: