Answer:
i think it is 1% sry if wrong
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
Answer:
1) D. 
2) E. The coefficient of
is -16.
Step-by-step explanation:
1) A polynomial is an algebraic expression of the form
, where
is the grade of the polynomial. In this case, the only equation that satisfies this definition is:

Correct answer: D
2) Let
, we proceed to expand the expression by algebraic means:


We see that coefficients for
and
are -16 and 0, respectively. Hence, the incorrect answer is E.
Answer:
35%
Step-by-step explanation:
The cd players price decresed by 35%
Answer:
x = -1
x = 5
Step-by-step explanation:
Use pythagorean theorem: a² + b² = c²
x² + (2x + 2)² = (2x + 3)²
Since these are quantities, you'll have to make them into quadratic equations.
(2x + 2)(2x + 2) = 4x² + 4x + 4x + 4
(2x + 3)(2x + 3) = 4x² + 6x + 6x + 9
x² + 4x² + 4x + 4x + 4 = 4x² + 6x + 6x + 9
Combine like terms
5x² + 8x + 4 = 4x² + 12x + 9
Move one side to set the equation equal to 0
x² - 4x - 5 = 0
Solve
x² - 5x + x - 5 = 0
x(x - 5) + 1(x - 5) = 0
(x + 1)(x - 5) = 0
x = -1, 5
<em>We</em><em> </em><em>can</em><em> </em><em>check</em><em> </em><em>that</em><em> </em><em>these</em><em> </em><em>are</em><em> </em><em>correct</em><em> </em><em>by</em><em> </em><em>plugging</em><em> </em><em>them</em><em> </em><em>in</em><em> </em><em>for</em><em> </em><em>x</em><em> </em><em>and</em><em> </em><em>seeing</em><em> </em><em>if</em><em> </em><em>they</em><em> </em><em>are</em><em> </em><em>equal</em>
<em>For</em><em> </em><em>example</em>
<em>(</em><em>-1</em><em>)</em><em>²</em><em> </em><em>+</em><em> </em><em>(</em><em>2</em><em>(</em><em>-1</em><em>)</em><em> </em><em>+</em><em> </em><em>2</em><em>)</em><em>²</em><em> </em><em>=</em><em> </em><em>(</em><em>2</em><em>(</em><em>-1</em><em>)</em><em> </em><em>+</em><em> </em><em>3</em><em>)</em><em>²</em>
<em>1</em><em> </em><em>=</em><em> </em><em>1</em>