Answer:
Zero
Step-by-step explanation:
Slope is:
Since there is no change of the y value, there is no slope.
Answer:
The ratio is the same
Step-by-step explanation:
Bianca- 20:14
= 10:7
Logan- 10:7
The ratio is the same
Answer:
1.) Triangle ABC is congruent to Triangle CDA because of the SAS theorem
2.) Triangle JHG is congruent to Triangle LKH because of the SSS theorem
Step-by-step explanation:
Alright. Let's start with the 1st figure. How do we prove that triangles ABC and CDA (they are named properly) are congruent? First, we can see that segments BC and AD have congruent markings, so that can help us. We also see a parallel marking for those segments as well, meaning that the diagonal AC is also a transversal for those parallel segments. That means we can say that angle CAD is congruent to angle ACB because of the alternate interior angles theorem. Then, the 2 triangles also share the side AC (reflexive property).
So, we have 2 congruent sides and 1 congruent angle for each triangle. And in the way they are listed, this makes the triangles congruent by the SAS theorem since the angle is adjacent to the 2 sides that are congruent.
The second figure is way easier. As you can clearly see by the congruent markings on the diagram, all the sides on one triangle are congruent to the other. So, since there are 3 sides congruent, we can say the triangles JHG and LKH are congruent by the SSS theorem.
(x^m) times (x^n)=x^(m+n)
(1/8)^5 times (1/8)^3=(1/8)^(5+3)=(1/8)^8=1/(8^8)
<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>
