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The general equation of a horizontal ellipse is
(x-h)2/a2 + (y-k)2/b2 = 1, at center (h,k) while a = semi-major axis, b = semi-minor axis. These are related through the distance of the focus from the center,c. a2 = b2 + c2.
If you draw the points on a coordinate plane, the center of the ellipse is at (0,0), so h and k equals 0. Then, the minor axis (2b) spans from 8 to -8 of the y-axis. This is equal to 16 units. Hence,
2b = 16
b = 8
b^2 = 64
The distance between the two foci is 2c. Thus,
2c = 12
c = 6
c^2 = 36
Then, a2 = 64 + 36 = 100. Substituting to the general equation:
x^2/100 + y^2/64 = 1
<span>the statement is ambiguous, it must be written as (p ∧ q) ∨ r
or p ∧ (q ∨ r).</span>
Answer:
14.93
Step-by-step explanation:
For this problem you need to know distance formula, which is
d=√(x2-x1)²+(y2-y1)². You'll want to plug in (0,3) and (-2, 9) and go on to plug in all of them at some point. You'll get 6.32 as the distance between (0,3) and (-2, 9), 3.61 as the distance between (-2, 9) and (-4, 6), and 5 as the distance between (-4, 6) and (0, 3). You add them up and get your answer.
Y=10x+1
x=-1→y=10(-1)+1=-10+1→y=-9
x=0→y=10(0)+1=0+1→y=1
x=1→y=10(1)+1=10+1→y=11
x=5→y=10(5)+1=50+1→y=51
x y
-1 -9
0 1
1 11
5 51
Y = x + 7
<span>y = -2x - 11
x + 7 = -2x - 11
3x + 7 = -11
3x = -18
x = -6
The x-coordinate of the solution is
-6.
I have attached a graph of the two equations since the question really wanted you to graph it and then find the answer.</span>