Answer:
13. c,
14. a.
Step-by-step explanation:
13. The length of a side PQ with coordinates of
P as (x,y) and Q as (w,z)
is: 
- so, the length of the side AB =
now, the easiest one will be when the vertices are on the coordinate axes.
- so, option c will be the most appropriate one.
14. If u see the figure clearly, the lines l and FH are parallel.
the parallel postulate, i.e, the alternate interior angles will always be congruent.
- here, the alternate interior angles are angle1 and angle4.
- therefore to prove this step, he used parallel postulate as a reason.
- so, the correct option is "a"
Answer:
B) 24
Step-by-step explanation:
A right triangle is 90°
90-42=48
48/2=24
<span>x^2/8 - x/4 + 1/8 = 0
A parabola is defined as the set of all points such that each point has the same distance from the focus and the directrix. Also the parabola's equation will be a quadratic equation of the form ax^2 + bx + c. So if we can determine 3 points on the parabola, we can use those points to calculate the desired equation.
First, let's draw the shortest possible line from the focus to the directrix. The midpoint of that line will be a point on the desired parabola. Since the slope of the directrix is 0, the line will have the equation of x=1. This line segment will be from (1,2) to (1,-2) and the midpoint will be ((1+1)/2, (2 + -2)/2) = (2/2, 0/2) = (1,0).
Now for the 2nd point, let's draw a line that's parallel to the directrix and passing through the focus. The equation of that line will be y=2. Any point on that line will have a distance of 4 from the directrix. So let's give it an x-coordinate value of (1+4) = 5. So another point for the parabola is (5,2). And finally, if we subtract 4 instead of adding 4 to the x coordinate, we can get a third point of 1-4 = -3. So that 3rd point is (-3,2).
So we now have 3 points on the parabola. They are (1,0), (5,2), and (-3,2). Let's create some equations of the form ax^2 + bx + c = y and then substitute the known values into those equations. SO
ax^2 + bx + c = y
(1) a*1^2 + b*1 + c = 0
(2) a*5^2 + b*5 + c = 2
(3) a*(-3)^2 + b*(-3) + c = 2
Let's do the multiplication for those expressions. So
(4) a + b + c = 0
(5) 25a + 5b + c = 2
(6) 9a - 3b + c = 2
Equations (5) and (6) above look interesting. Let's subtract (6) from (5). So
25a + 5b + c = 2
- 9a - 3b + c = 2
= 16a + 8b = 0
Now let's express a in terms of b.
16a + 8b = 0
16a = -8b
a = -8b/16
(7) a = -b/2
Now let's substitute the value (-b/2) for a in expression (4) above. So
a + b + c = 0
-b/2 + b + c = 0
And solve for c
-b/2 + b + c = 0
b/2 + c = 0
(8) c = -b/2
So we know that a = -b/2 and c = -b/2. Let's substitute those values for a and c in equation (5) above and solve for b.
25a + 5b + c = 2
25(-b/2) + 5b - b/2 = 2
-25b/2 + 5b - b/2 = 2
2(-25b/2 + 5b - b/2) = 2*2
-25b + 10b - b = 4
-16b = 4
b = -4/16
b = -1/4
So we now know that b = -1/4. Using equations (7) and (8) above, let's calculate a and c.
a = -b/2 = -(-1/4)/2 = 1/4 * 1/2 = 1/8
c = -b/2 = -(-1/4)/2 = 1/4 * 1/2 = 1/8
So both a and c are 1/8. So the equation for the parabola is
x^2/8 - x/4 + 1/8 = 0
Let's test to make sure it works. First, let's use an x of 1.
x^2/8 - x/4 + 1/8 = y
1^2/8 - 1/4 + 1/8 = y
1/8 - 1/4 + 1/8 = y
1/8 - 2/8 + 1/8 = y
0 = y
And we get 0 as expected. Let's try x = 2
x^2/8 - x/4 + 1/8 = y
2^2/8 - 2/4 + 1/8 = y
4/8 - 1/2 + 1/8 = y
4/8 - 1/2 + 1/8 = y
1/2 - 1/2 + 1/8 = y
1/8 = y.
Let's test if (2,1/8) is the same distance from both the focus and the directrix. The distance from the directrix is 1/8 - (-2) = 1/8 + 2 = 1/8 + 16/8 = 17/8
The distance from the focus is
d = sqrt((2-1)^2 + (1/8-2)^2)
d = sqrt(1^2 + -15/8^2)
d = sqrt(1 + 225/64)
d = sqrt(289/64)
d = 17/8
And the distances match again. So we do have the correct equation of:
x^2/8 - x/4 + 1/8 = 0</span>
Answer:
2
1/3
Step-by-step explanation:
A(xA, yA) = A (3,1)
B(xB,yB)=B(5,5)
the gradient=(yB-yA)/ (xB-xA)=(5-1)/(5-3)=4/2=2
C(4,7),D(10,9)
the gradient=(yD-yC)/(xD-xC)=(9-7)/(10-4)=2/6=1/3
ITS B OR 92 DEGREES enjoy!!!
