F(5)= 3.(5)+2=15+2=17
Another example: F(0)=3.(0)+2=0+2=2
You just replace the value in the two sides.
<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>
Given:
ABCD is a quadrilateral
AC = 12, DE = 4 and FB = 5
To find:
The area of the polygon.
Solution:
AC bisects the quadrilateral into two triangles.
Area of triangle:
<u>Area of triangle DAC:</u>
Area of triangle DAC = 24 square units.
<u>Area of triangle BAC:</u>
Area of triangle BAC = 30 square units.
Area of polygon = Area of triangle DAC + Area of triangle BAC
= 24 square units + 30 square units
= 54 square units
The area of the polygon is 54 square units.
Answer:
A. 1/3x + 1/2y <= 4
2x + y <= 12
x >= 0
y >= 0
Step-by-step explanation:
