So think of |x+6| as a single object. Move the -6|x+6| to the left side and move the 8 to the right side to get 11|x+6| = 0. Then divide both sides by 11 to get |x+6| = 0 and in this instance the || can be dropped because +0 and -0 are the same. so x + 6 = 0 or x=-6
Answer:
167.27 mg.
Step-by-step explanation:
We have been given that the half-life of Radium-226 is 1590 years and a sample contains 400 mg.
We will use half life formula to solve our given problem.
, where N(t)= Final amount after t years,
= Original amount, t/2= half life in years.
Now let us substitute our given values in half-life formula.


Therefore, the remaining amount of Radium-226 after 2000 years will be 167.27 mg.
Answer:
Option (2)
Step-by-step explanation:
Given expression is ![\sqrt[3]{64a^6b^7c^9}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B64a%5E6b%5E7c%5E9%7D)
By simplifying this expression,
![\sqrt[3]{64a^6b^7c^9}=\sqrt[3]{(4)^3(a^2)^3(b^2)^3(b)(c^3)^3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B64a%5E6b%5E7c%5E9%7D%3D%5Csqrt%5B3%5D%7B%284%29%5E3%28a%5E2%29%5E3%28b%5E2%29%5E3%28b%29%28c%5E3%29%5E3%7D)
![=(4a^2b^2c^3)\sqrt[3]{b}](https://tex.z-dn.net/?f=%3D%284a%5E2b%5E2c%5E3%29%5Csqrt%5B3%5D%7Bb%7D)
Option (1)
= ![2ab^2c^2\sqrt[3]{a^2c}](https://tex.z-dn.net/?f=2ab%5E2c%5E2%5Csqrt%5B3%5D%7Ba%5E2c%7D)
Option (2)
[Fully simplified form]
Option (3)
[Fully simplified form]
Option (4)
![8a^2b^2c^3(\sqrt[3]{b})](https://tex.z-dn.net/?f=8a%5E2b%5E2c%5E3%28%5Csqrt%5B3%5D%7Bb%7D%29)
Expression given in Option (2) is equivalent to the given expression.
Option (2) will be the answer.
I think it should be 5w+7. I'm not 100% sure though
Answer:
When you multiply through by the LCD and solve the resulting quadratic equation, you get solutions x=2 and x=1. However when we try to check the solution x=2, it causes the first and last denominators to become 0, which is undefined. However x=1 checks.
Step-by-step explanation: