Answer:
24.
Step-by-step explanation:
We can use the factorial function to find the number of unique arrangements in a word.
There are 4 unique characters in 'OPAL'. As such, 4! = 24.
In the case that there were repeating characters, such as 'OPALA', then we would divide by the factorial of the number of repeating letters. That looks something like this:
Hope this helped.
Answer:
57
Step-by-step explanation:
fudge it tgyshhvt8 to t up 7
-3 + -3 = 0
-5 + 1 = -4
vector = < 0,-4 >
Answer:
The brainly terms of service states that you can not use Brainly for tests or exams. Please do not do this.
Step-by-step explanation:
Answer:
option c is correct.
Step-by-step explanation:
![7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{16x}\right)-3\left(\sqrt[3]{8x}\right)](https://tex.z-dn.net/?f=7%5Cleft%28%5Csqrt%5B3%5D%7B2x%7D%5Cright%29-3%5Cleft%28%5Csqrt%5B3%5D%7B16x%7D%5Cright%29-3%5Cleft%28%5Csqrt%5B3%5D%7B8x%7D%5Cright%29)
WE need to simplify this equation.
Solve the parenthesis of each term.
![=7\left\sqrt[3]{2x}\right-3\left\sqrt[3]{16x}\right-3\left\sqrt[3]{8x}\right](https://tex.z-dn.net/?f=%3D7%5Cleft%5Csqrt%5B3%5D%7B2x%7D%5Cright-3%5Cleft%5Csqrt%5B3%5D%7B16x%7D%5Cright-3%5Cleft%5Csqrt%5B3%5D%7B8x%7D%5Cright)
Now, We will find factors of the terms inside the square root
factors of 2: 2
factors of 16 : 2x2x2x2
factors of 8: 2x2x2
Putting these values in our equation:![=7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{2X2X2X2 x}\right)-3\left(\sqrt[3]{2X2X2 x}\right)\\=7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{2X2X2} \sqrt[3] {2 x}\right)-3\left(\sqrt[3]{2X2X2} \sqrt[3]{x}\right)\\=7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{2^3} \sqrt[3] {2 x}\right)-3\left(\sqrt[3]{2^3} \sqrt[3]{x}\right)\\=7\left(\sqrt[3]{2x}\right)-3*2\left(\sqrt[3] {2 x}\right)-3*2\left(\sqrt[3]{x}\right)\\=7\left(\sqrt[3]{2}\sqrt[3]{x}\right)-6\left(\sqrt[3] {2}\sqrt[3]{x})-6\left(\sqrt[3]{x}\right)](https://tex.z-dn.net/?f=%3D7%5Cleft%28%5Csqrt%5B3%5D%7B2x%7D%5Cright%29-3%5Cleft%28%5Csqrt%5B3%5D%7B2X2X2X2%20x%7D%5Cright%29-3%5Cleft%28%5Csqrt%5B3%5D%7B2X2X2%20x%7D%5Cright%29%5C%5C%3D7%5Cleft%28%5Csqrt%5B3%5D%7B2x%7D%5Cright%29-3%5Cleft%28%5Csqrt%5B3%5D%7B2X2X2%7D%20%5Csqrt%5B3%5D%20%7B2%20x%7D%5Cright%29-3%5Cleft%28%5Csqrt%5B3%5D%7B2X2X2%7D%20%5Csqrt%5B3%5D%7Bx%7D%5Cright%29%5C%5C%3D7%5Cleft%28%5Csqrt%5B3%5D%7B2x%7D%5Cright%29-3%5Cleft%28%5Csqrt%5B3%5D%7B2%5E3%7D%20%5Csqrt%5B3%5D%20%7B2%20x%7D%5Cright%29-3%5Cleft%28%5Csqrt%5B3%5D%7B2%5E3%7D%20%5Csqrt%5B3%5D%7Bx%7D%5Cright%29%5C%5C%3D7%5Cleft%28%5Csqrt%5B3%5D%7B2x%7D%5Cright%29-3%2A2%5Cleft%28%5Csqrt%5B3%5D%20%7B2%20x%7D%5Cright%29-3%2A2%5Cleft%28%5Csqrt%5B3%5D%7Bx%7D%5Cright%29%5C%5C%3D7%5Cleft%28%5Csqrt%5B3%5D%7B2%7D%5Csqrt%5B3%5D%7Bx%7D%5Cright%29-6%5Cleft%28%5Csqrt%5B3%5D%20%7B2%7D%5Csqrt%5B3%5D%7Bx%7D%29-6%5Cleft%28%5Csqrt%5B3%5D%7Bx%7D%5Cright%29)
Adding like terms we get:
![=7\left(\sqrt[3]{2}\sqrt[3]{x}\right)-6\left(\sqrt[3] {2}\sqrt[3]{x})-6\left(\sqrt[3]{x}\right\\=(\sqrt[3] {2}\sqrt[3]{x})-6\left(\sqrt[3]{x}\right)\\](https://tex.z-dn.net/?f=%3D7%5Cleft%28%5Csqrt%5B3%5D%7B2%7D%5Csqrt%5B3%5D%7Bx%7D%5Cright%29-6%5Cleft%28%5Csqrt%5B3%5D%20%7B2%7D%5Csqrt%5B3%5D%7Bx%7D%29-6%5Cleft%28%5Csqrt%5B3%5D%7Bx%7D%5Cright%5C%5C%3D%28%5Csqrt%5B3%5D%20%7B2%7D%5Csqrt%5B3%5D%7Bx%7D%29-6%5Cleft%28%5Csqrt%5B3%5D%7Bx%7D%5Cright%29%5C%5C)
![(\sqrt[3] {2}\sqrt[3]{x})-6\left(\sqrt[3]{x}\right)\\can\,\,be \,\, written\,\, as\,\,\\(\sqrt[3] {2x})-6\left(\sqrt[3]{x}\right)](https://tex.z-dn.net/?f=%28%5Csqrt%5B3%5D%20%7B2%7D%5Csqrt%5B3%5D%7Bx%7D%29-6%5Cleft%28%5Csqrt%5B3%5D%7Bx%7D%5Cright%29%5C%5Ccan%5C%2C%5C%2Cbe%20%5C%2C%5C%2C%20written%5C%2C%5C%2C%20as%5C%2C%5C%2C%5C%5C%28%5Csqrt%5B3%5D%20%7B2x%7D%29-6%5Cleft%28%5Csqrt%5B3%5D%7Bx%7D%5Cright%29)
So, option c is correct
