Vertices (3,0),(-3,0) co-vertices (0,-5),(0,5)
transverse axis (line passing vertices) is on(or parallel to) x-axis then formula is
(x-h)^2/a^2 - (y-k)^2/b^2 = 1
..notice.. x^2 is on positive / y^2 is on negative
center (h,k) is midway between vertices = (0,0)
we have h = k = 0 and now formula is
x^2/a^2 - y^2/b^2 = 1
a is the distance from a vertex to center = 3
b is the distance from a co-vertex to center = 5
the formula is
x^2/3^2 - y^2/5^2 = 1 ... answer is the 1st
Answer:
y -11 = -2(x+4)
or
y = -2x+3
Step-by-step explanation:
We can use the point slope form of a line
y - y1 = m(x-x1) where m is the slope and (x1 ,y1) is a point on the line
y - 11 = -2(x- -4)
y -11 = -2(x+4)
We can rewrite it in slope intercept form
y-11 = -2x-8
Add 11 to each side
y = -2x+3
Answer:
The quadratic mean of 2 real positive numbers is greater than or equal to the arithmetic mean.
Step-by-step explanation:
x and y Quadratic Mean Arithmetic mean
3 and 3 3 3
2 and 3 2.55 2.5
3 and 6 4.74 4.5
2 and 5 3.8 3.5
2 and 17 12.1 9.5
18 and 28 23.5 23
10 and 48 34.7 29
The quadratic mean is always greater than the arithmetic mean except when x and y are the same.
When the difference between the pairs is small the difference in the means is also small. As that difference increases the difference in the means also increases.
So we conjecture that the quadratic mean is always greater than or equal to the arithmetic mean.
Proof.
Suppose it is true then:
√(x^2 + y^2) / 2) ≥ (x + y)/2 Squaring both sides:
(x ^2 + y^2) / 2 ≥ (x + y)^2 / 4 Multiply through by 4:
2x^2 +2y^2 ≥ (x + y)^2
2x^2 +2y^2 >= x^2 + 2xy + y^2
x^2 + y^2 >= 2xy.
x^2 - 2xy + y^2 ≥ 0
(x - y)^2 ≥ 0
This is true because the square of any real number is positive so the original inequality must also be true.
Y=7/6x+ (any number) will be a line parallel to the one above.. It has to contain the same slope 7/6
Y=7/6x+_
