Let point (x, y) be any point on the graph, than the distance between (x, y) and the focus (3, 6) is sqrt((x - 3)^2 + (y - 6)^2) and the distance between (x, y) and the directrix, y = 4 is |y - 4|
Thus sqrt((x - 3)^2 + (y - 6)^2) = |y - 4|
(x - 3)^2 + (y - 6)^2 = (y - 4)^2
x^2 - 6x + 9 + y^2 - 12y + 36 = y^2 - 8y + 16
x^2 - 6x + 29 = -8y + 12y = 4y
(x - 3)^2 + 20 = 4y
y = 1/4(x - 3)^2 + 5
Required answer is f(x) = one fourth (x - 3)^2 + 5
Answer:
15/32 cm²
Step-by-step explanation:
Area of rectangle = l × b
=> 3/4 × 5/8
=> 15/32
Therefore,
Area of rectangle is 15/32 cm²
Answer: The slant height of the cone is 65.6 m
Step-by-step explanation:
Given: The diameter of a cone = 10 m
Surface area of cone = 190.6 m²
To find: Slant height
Diameter of cone = 10 m
Therefore Radius of cone = 
As we know that surface area of a cone is given by
Where S.A. is surface area , r is the radius of cone and l is the slant height of the cone.
Let Slant height = l
So we have
Hence the slant height of the cone is 65.6 m
Answer: 
Step-by-step explanation:
By definition, the volume of a rectangular prism can be calculated with the following formula:
Where "l" is the length, "w" is the width and "h" is the height of the rectangular prism.
In this case, you can identify that the length, the width and the height of this rectangular prism given in the exercise, are:
Then, knowing its dimensions, you can substitute them into the formula:
Finally, evaluating, you get that the volume of that rectangular prism is:
All you need to solve these is a knowledge of the 3 main trigonometrical ratios.
I a right angled triangle where A is a non right angle:-
sin A = opposite / hypotenuse (hypotenuse is the sloping side)
cos A = adjacent / hyptenuse ( the adjacent side is the one the angle 'sits' on)
tan A = opposite / adjacent ( the opposite side is the one opposite the angle)
So to find the length of segment x in the first problem use
sin 61 = x / 50
then
x = 50*sin61 = 43.73
and to find the length of y:-
cos 61 = y / 50
y = 60 cos 61 = 24.24
The other questions can be solved in a similar way.
