Answer:
A'B'C'D' was dilated by 4
Step-by-step explanation:
The distance from B' to C' is 1.
The distance from B to C is 4.
4/1=4
Answer:
Peter Jonathan Winston (March 18, 1958 – disappeared January 26, 1978) was an American chess player from New York City
Step-by-step explanation:
In late 1977, Winston attended a FIDE-rated tournament at Hunter College High School in New York City. Despite being one of the highest-rated players in the tournament, Winston lost all nine of his games. A few months later, on January 26, 1978, following further surprising game losses, Peter Winston vanished and was never heard from again. According to some sources, Winston's disappearance occurred when he left his home without money, identification, or luggage during a severe winter storm. Many chess players who were close to or acquainted with Winston claim that the champion chess player's mental health had deteriorated, along with his game performance, in the last few years of his life, and that the decline in his mental health may have led to his disappearance.
I do not the question. Please explain
<span>The two points that are most distant from (-1,0) are
exactly (1/3, 4sqrt(2)/3) and (1/3, -4sqrt(2)/3)
approximately (0.3333333, 1.885618) and (0.3333333, -1.885618)
Rewriting to express Y as a function of X, we get
4x^2 + y^2 = 4
y^2 = 4 - 4x^2
y = +/- sqrt(4 - 4x^2)
So that indicates that the range of values for X is -1 to 1.
Also the range of values for Y is from -2 to 2.
Additionally, the ellipse is centered upon the origin and is symmetrical to both the X and Y axis.
So let's just look at the positive Y values and upon finding the maximum distance, simply reflect that point across the X axis. So
y = sqrt(4-4x^2)
distance is
sqrt((x + 1)^2 + sqrt(4-4x^2)^2)
=sqrt(x^2 + 2x + 1 + 4 - 4x^2)
=sqrt(-3x^2 + 2x + 5)
And to simplify things, the maximum distance will also have the maximum squared distance, so square the equation, giving
-3x^2 + 2x + 5
Now the maximum will happen where the first derivative is equal to 0, so calculate the first derivative.
d = -3x^2 + 2x + 5
d' = -6x + 2
And set d' to 0 and solve for x, so
0 = -6x + 2
-2 = -6x
1/3 = x
So the furthest point will be where X = 1/3. Calculate those points using (1) above.
y = +/- sqrt(4 - 4x^2)
y = +/- sqrt(4 - 4(1/3)^2)
y = +/- sqrt(4 - 4(1/9))
y = +/- sqrt(4 - 4/9)
y = +/- sqrt(3 5/9)
y = +/- sqrt(32)/sqrt(9)
y = +/- 4sqrt(2)/3
y is approximately +/- 1.885618</span>
Answer:
1 real solution
Step-by-step explanation:
y=2x^2−8x+8
We can use the discriminant to determine the number of real solutions
b^2 -4ac
a =2 b = -8 c=8
(-8)^2 - 4(2)(8)
64 - 64
0
Since the discriminant is 0 there is 1 real solution
>0 there are 2 real solutions
< 0 two complex solutions
