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Lorico [155]
2 years ago
5

In question 1, you found the cubes of both positive and negative numbers. Does x3 = 8 have two solutions as x2 = 4 does? Why or

why not?
Mathematics
1 answer:
Temka [501]2 years ago
4 0
No it doesn’t. This is because if you have a negative number and multiply it by itself 3 times, you’ll still have a negative number (-1 times -1 times -1 is still -1). On the other hand, if you take a negative number and multiply it by itself twice (-1 times -1) you’ll have the answer that is positive (1).
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zhuklara [117]
\bf tan^{-1}(3)\iff tan^{-1}\left( \frac{\pm 3}{\pm 1} \right)=\theta
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thus\qquad tan(\theta)=\cfrac{\pm 3}{\pm 1}\cfrac{\leftarrow opposite=y}{\leftarrow adjacent=x}

now, the angle of θ, can only have a "y" value that is positive on, well, y is positive at 1st and 2nd quadrants

and "x" is positive only in 1st and 4th quadrants

now, that angle θ, can only have those two fellows, "y" and "x" to be positive, only in the 1st quadrant, and also both to be negative on the 3rd quadrant.

and that those two fellows, can also be both negative in the 3rd quadrant
3/1 = 3, and -3/-1 = 3

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Need help with these two questions.​
Sindrei [870]

Answer:

Answer is below :)

Step-by-step explanation:

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33.75

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