The amount of the discount was $140.
Using it's concepts, the domain and the range of the graph are given as follows:
- Domain: all real values except x = -1.
<h3>What are the domain and the range of a function?</h3>
- The domain of a function is the set that contains all the values of the input. In a graph, it is given by the values of x, which is the horizontal axis of the graph.
- The range of a function is the set that contains all the values of the output. In a graph, it is given by the values of y, which is the vertical axis of the graph.
In this graph, have that the function is defined for all values of x except x = -1, and assumes all real values, hence the domain and the range of the graph are given as follows:
- Domain: all real values except x = -1.
More can be learned about the domain and the range of a function at brainly.com/question/10891721
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Answer: Last Option
![4x^5\sqrt[3]{3x}](https://tex.z-dn.net/?f=4x%5E5%5Csqrt%5B3%5D%7B3x%7D)
Step-by-step explanation:
To make the product of these expressions you must use the property of multiplication of roots:
![\sqrt[n]{x^m}*\sqrt[n]{x^b} = \sqrt[n]{x^{m+b}}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bx%5Em%7D%2A%5Csqrt%5Bn%5D%7Bx%5Eb%7D%20%3D%20%5Csqrt%5Bn%5D%7Bx%5E%7Bm%2Bb%7D%7D)
we also know that:
![\sqrt[3]{x^3} = x](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E3%7D%20%3D%20x)
So
![\sqrt[3]{16x^7}*\sqrt[3]{12x^9}\\\\\sqrt[3]{16x^3x^3x}*\sqrt[3]{12(x^3)^3}\\\\x^2\sqrt[3]{16x}*x^3\sqrt[3]{12}\\\\x^5\sqrt[3]{16x*12}\\\\x^5\sqrt[3]{2^4x*2^2*3}\\\\x^5\sqrt[3]{2^6x*3}\\\\4x^5\sqrt[3]{3x}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B16x%5E7%7D%2A%5Csqrt%5B3%5D%7B12x%5E9%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7B16x%5E3x%5E3x%7D%2A%5Csqrt%5B3%5D%7B12%28x%5E3%29%5E3%7D%5C%5C%5C%5Cx%5E2%5Csqrt%5B3%5D%7B16x%7D%2Ax%5E3%5Csqrt%5B3%5D%7B12%7D%5C%5C%5C%5Cx%5E5%5Csqrt%5B3%5D%7B16x%2A12%7D%5C%5C%5C%5Cx%5E5%5Csqrt%5B3%5D%7B2%5E4x%2A2%5E2%2A3%7D%5C%5C%5C%5Cx%5E5%5Csqrt%5B3%5D%7B2%5E6x%2A3%7D%5C%5C%5C%5C4x%5E5%5Csqrt%5B3%5D%7B3x%7D)
Answer:
i
Step-by-step explanation:
because you can't technically can't find the square root of -1, so it's an imaginary number
