Good morning
Answer:
B: (1, 5)
Step-by-step explanation:
This graph shows the solutions to the inequality: y > (1/3)x - 2
Since, 5 > (1/3)×1 - 2 then the point (1 ,5) is a solution to the inequality.
____________________________
:)
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
Answer: D. 8x² + x + 3
Sum means the answer to an addition problem. To find the sum of polynomials, we will add like terms.
<h2>What are like terms?</h2>
Like terms can be combined using addition or subtraction and have the same variables. Constants are also like terms with each other because they have no variables.
<h2>Solve</h2>
(4x² + 1) + (4x² + x + 2) Starting equation from the question
= 4x² + 1 + 4x² + x + 2 Remove brackets
= 4x² + 4x² + x + 1 + 2 Rearrange to group like terms together
= 8x² + x + 1 + 2 Add like terms with the same 'x²' variables
= 8x² + x + 3 Add like terms that are constants
Learn more about adding polynomials here:
brainly.com/question/1311115
