Answer:
yes
Step-by-step explanation:
The line intersects each parabola in one point, so is tangent to both.
__
For the first parabola, the point of intersection is ...
y^2 = 4(-y-1)
y^2 +4y +4 = 0
(y+2)^2 = 0
y = -2 . . . . . . . . one solution only
x = -(-2)-1 = 1
The point of intersection is (1, -2).
__
For the second parabola, the equation is the same, but with x and y interchanged:
x^2 = 4(-x-1)
(x +2)^2 = 0
x = -2, y = 1 . . . . . one point of intersection only
___
If the line is not parallel to the axis of symmetry, it is tangent if there is only one point of intersection. Here the line x+y+1=0 is tangent to both y^2=4x and x^2=4y.
_____
Another way to consider this is to look at the two parabolas as mirror images of each other across the line y=x. The given line is perpendicular to that line of reflection, so if it is tangent to one parabola, it is tangent to both.
2 , 6 , 17 , 47 , <u>108</u> , ...
t(n) = 2n^3 - 8.5n^2 + 15.5n - 7
An equation between two variables that gives a straight line when plotted on a graph.
2y = x+ x+ 1
2y = 2x + 1
y = (2x+ 1) /2
y = x + 1/2
According to the graph, this equation is linear
Answer:
Below in bold.
Step-by-step explanation:
Arrange the data in ascending order:
1. 1 1 2 3 4 7 9
Median = middle number = 3.
Q1 = 1
Q3 = 7.
2. 26 31 34 38 42 66 74 78
Median = mean of 2 middle numbers = 40.
Q1 = 31.
Q3 = 74.
3. 36 45 49 52 53 64 81 87 93
Median = middle number = 53.
Q1 = 47.
Q3 = 84.