Answer:
It is irrational because it can not be represented as a fraction of two integers
Step-by-step explanation:
Given
![H = \sqrt[3]5](https://tex.z-dn.net/?f=H%20%3D%20%5Csqrt%5B3%5D5)
Required
Why is the area irrational?
First, we need to calculate the area
![Area = \frac{1}{2}(3.6 + 12\frac{1}{3}) * \sqrt[3]5](https://tex.z-dn.net/?f=Area%20%3D%20%5Cfrac%7B1%7D%7B2%7D%283.6%20%2B%2012%5Cfrac%7B1%7D%7B3%7D%29%20%2A%20%5Csqrt%5B3%5D5)
<em>It is irrational because it can not be represented as a fraction of two integers</em>
Equation of an ellipse:
(x-h)²/a² + (y-k)²/b² = 1
Since it passes through the origin (0,0) , then h = k = 0 hence the equation:
(x-0)²/a² + (y-0)²/b² = 1
x²/a² + y²/b² = 1
2a = major axis = 2.|5| + |-5| = 10. then a = 5 and a² = 25
2b = minor axis = 2.|3| + |-3| = 6. then b = 3 and b² = 9
Then the final equation is:
x²/25 + y²/9 = 1
Answer:
- (a) no
- (b) yes
- (c) no
- (d) no
Step-by-step explanation:
"Of the order x^2" means the dominant behavior matches that of x^2 as x gets large. For polynomial functions, the dominant behavior is that of the highest-degree term.
For other functions, the dominant behavior will typically be governed in some other way. Here, the rate of growth of the x·log(x) function is determined by log(x), which has decreasing slope as x increases.
Only answer selection B has a highest-degree term of x^2, so only that one exhibits O(x^2) behavior.
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