The standard form equation for this hyperbola, when vertices are (+-5,0) and one focus is (6,0), is x²/25 - y²/11 = 1.
In the question, we are given a hyperbola with vertices at (+-5,0) and one focus at (6,0). A hyperbola is defined by its distances from a given point to the two different foci, and its standard form equation along the x-axis can be written as
(x-h)²/a² - (y-k)²/b² = 1
, where (h, k) is the center of the hyperbola, a represents the distance from the center to each vertex, and b represents the distance from the center to each co-vertex. In this case,
h = 0
, since the center of the hyperbola is at the origin. The value of
a = 5
is the distance from the center to each vertex. Finally, the square of the distance c from the center to each focus is defined as
c² = a² + b²
, so we can find
b = sqrt(c² - a²)
. Here, c = 6, so b = sqrt(6² - 5²) = sqrt(11). Thus, the standard form equation of this hyperbola is
x²/25 - y²/11 = 1
.
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Answer:
the total cost of paint and tiles needed to decorate the community hall is £105 + £900 = £1005.
Step-by-step explanation:
Area of the walls:
- The community hall is shaped like a cuboid, so the walls are the four rectangular sides.
- The length and height of each wall are given as 20m and 4m, respectively.
- The total area of the walls is the sum of the areas of all four walls: 2 * (length * height + width * height).
- Substituting the given values, we have: 2 * (20m * 4m + 15m * 4m).
2. Area of the ceiling:
- The ceiling is also a rectangle, with the same length and width as the floor.
- The area of the ceiling is given by length * width: 20m * 15m.
3. Area of the floor:
- The area of the floor is the same as the area of the ceiling: 20m * 15m.
Now, let's calculate the areas:
1. Area of the walls:
- 2 * (20m * 4m + 15m * 4m) = 2 * (80m^2 + 60m^2) = 2 * 140m^2 = 280m^2.
2. Area of the ceiling:
- 20m * 15m = 300m^2.
3. Area of the floor:
- 20m * 15m = 300m^2.
Next, we can calculate the number of tins of paint and tiles needed and their costs:
- Number of tins of paint needed = Area of walls / Coverage per tin = 280m^2 / 40m^2/tin = 7 tins.
- Cost of paint = Number of tins * Cost per tin = 7 tins * £15/tin = £105.
- Number of tiles needed = Area of floor / Area per tile = 300m^2 / 1m^2/tile = 300 tiles.
- Cost of tiles = Number of tiles * Cost per tile = 300 tiles * £3/tile = £900.
Therefore, the total cost of paint and tiles needed to decorate the community hall is £105 + £900 = £1005.
yes, i am a student
The statement is valid because the measure of the vertices located in the center of the pentagon is the quotient of 360 and 5, and the sum of two base angles in the given isosceles triangle is 108.
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Explanation:
Check out the diagram below. I have added x and y such that
x = base angle
y = vertex angle (located adjacent to center of polygon)
The pentagon is sliced up like a cake into 5 equal portions. Each vertex angle is y = 360/5 = 72 degrees.
The two base angles x must add with the vertex angle y to get 180
(angle1)+(angle2)+(angle3) = 180
x+x+y = 180
2x+y = 180
2x+72 = 180
2x = 180-72
2x = 108
This is true for any one of the five triangles. Notice that for angle LMN, we can divide it into LMQ and QMN where Q is the center of the polygon. Both of these angles are x. Since we've shown 2x = 108, we can see that LMN must also be 108 as well.
Choice A is close, but we wouldn't use the exterior angle theorem. Choice B is the better answer.
Answer: 7(5v-w+6x)
Step-by-step explanation:
GCD/HCF = 7
8x = 5x + 27 Add 12 to both sides
3x = 27 Subtract 5x from both sides
x = 9 Divide both sides by 3