Answer:
The last one.
Step-by-step explanation:
The standard form equation of an ellipse with foci on the y-axis is ...
y²/a² +x²/b² = 1
where "a" and "b" are the lengths of the semi-major and semi-minor axes, respectively. If "c" is the distance from the center to the focus, then this relation also holds:
b² +c² = a²
For this ellipse ...
b² + 8² = 17²
289 -64 = b² = 225 . . . . . subtract 8²
b = 15 . . . . . . . . . . . . . . . . . take the square root
The equation of the ellipse is then ...
y²/17² +x²/15² = 1
So until what day do you have to stop or what is your question?
Step-by-step explanation:
I cant show a number line so i'll do my best to explain how its graphed.
15) x - 1 < 15
add 1 to both sides:
x < 16
Because it is less than, rather than less than or equal to, it's graphed with an open circle (not filled in) on 16, and everything less than 16 highlighted
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16) 2(y + 1) - 2 ≥ 12
distribute the 2:
2y + 2 - 2 ≥ 12
combine like terms
2y ≥ 12
isolate y:
y ≥ 6
This one is greater than or equal to, so it's graphed with a closed circle (filled in) and everything above 6 highlighted
<h3>
Answer:</h3>
x² -6x = -13 ⇒ x ∈ {3-2i, 2+2i}
<h3>
Step-by-step explanation:</h3>
To make a=1, divide the equation by the coefficient of x², which is 8.
... x² -6x = -13 . . . . . . your blanks are filled with -6 and -13
Now, to complete the square, add the square of half the x-coefficient:
... (-6/2)² = 9.
... x² -6x +9 = -4 . . . 9 added to both sides
... (x -3)² = -4 . . . . . rewrite as a square
... x -3 = ±2i . . . . . . take the square root
... x = 3 ±2i . . . . . . . add 3
The solutions are the complex numbers x = 3 ±2i.
