The general equation of an ellipse in which case it is not centered at the origin and it is tilted is; (x-h)²/a² + (y-k)²/b² = 1.
<h3>What is the general equation of a tilted ellipses not centered at the origin?</h3>
It follows from the task content that the plane shape in discuss is an ellipse which is described by the characteristics that it is tilted and not centered at the origin.
It follows from convention that the general equation of such an ellipse is;
(x-h)²/a² + (y-k)²/b² = 1.
In which case, such an ellipse has center given as point; (h, k).
Read more on ellipse;
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Answer:
The first question is (n*8) - 4
The second question is no.
Step-by-step explanation:
The first question, you add 8. The first is 1. 1 * 8 = 8 - 4 = 4.
Therefore, the equation is (n*8) - 4
The second question, you can plug in numbers. You can solve for a term. -100 + 17 = -83. -83 is not a multiple of -4. -100 is not a term.
Pv=4,725×((1−(1+0.10÷2)^(−2
×15))÷(0.10÷2))=72,634.83
Answer:
x=11
Step-by-step explanation:
48-4=44
44/4=11
x=11
11+1=12
12•4= 48
Part A: f(t) = t² + 6t - 20
u = t² + 6t - 20
+ 20 + 20
u + 20 = t² + 6t
u + 20 + 9 = t² + 6t + 9
u + 29 = t² + 3t + 3t + 9
u + 29 = t(t) + t(3) + 3(t) + 3(3)
u + 29 = t(t + 3) + 3(t + 3)
u + 29 = (t + 3)(t + 3)
u + 29 = (t + 3)²
- 29 - 29
u = (t + 3)² - 29
Part B: The vertex is (-3, -29). The graph shows that it is a minimum because it shows that there is a positive sign before the x²-term, making the parabola open up and has a minimum vertex of (-3, -29).
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Part A: g(t) = 48.8t + 28 h(t) = -16t² + 90t + 50
| t | g(t) | | t | h(t) |
|-4|-167.2| | -4 | -566 |
|-3|-118.4| | -3 | -364 |
|-2| -69.6 | | -2 | -194 |
|-1| -20.8 | | -1 | -56 |
|0 | -28 | | 0 | 50 |
|1 | 76.8 | | 1 | 124 |
|2 | 125.6| | 2 | 166 |
|3 | 174.4| | 3 | 176 |
|4 | 223.2| | 4 | 154 |
The two seconds that the solution of g(t) and h(t) is located is between -1 and 4 seconds because it shows that they have two solutions, making it between -1 and 4 seconds.
Part B: The solution from Part A means that you have to find two solutions in order to know where the solutions of the two functions are located at.
