Hello,
to find the total cost (c) of the colored strips we have to multiply the number of strips (n) by the cost of each color strip (99¢), that as an algebraic function is:
[¢]
Now, if we use 3 solid color strips, that means that we use 4 patterned strips, so:
Answer: c=n*99 and Cost per curtain= $7.97
We know that four snacks and three movie tickets cost $40 dollars, and when you take away 2 snacks but still keep the tickets, its $32.
2 snacks= $8
What i did to find the cost of each ticket is I divided 8 by 2 (8/2) and ended getting $4
1 snack= $4
Now to find the cost of each ticket, i divided 2 snacks ($8) and got this equation:
32 - 8= 24
3 tickets= $24
Now to find the amount for each ticket, i divided 24 by 3 (24/3) and got the answer:
1 ticket= $8
Therefore, snacks are $4 and movie tickets are $8
x tickets= 8x
x snacks= 4x
<em>Thank you! Btw can i please get brainly :3</em>
<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>
Alternate exterior angles are the pair of angles that lie on the outer side of the two parallel lines but on either side of the transversal line. Illustration: ... Notice how the pairs of alternating exterior angles lie on opposite sides of the transversal but outside the two parallel lines.
