The standard height from the floor to the bull's-eye at which a standard dartboard is hung is 5 feet 8 inches. A standard dartboard is 18 inches in diameter.
Suppose a standard dartboard is hung at standard height so that the bull's-eye is 11 feet from a wall to its left.
Samuel throws a dart at the dartboard that lands at a point 11.5 feet from the left wall and 5.5 feet above the floor.
Does Samuel's dart land on the dartboard?

From the parameters given to find out if Samuel's dart landed on the dartboard, we can say that;

Samuel's dart does not land on the dartboard.

We are given;

Height from the floor to the bulls eye = 5 ft 8 in = 5.67 ft

Diameter of the standard dartboard = 18 in = 1.5 ft

Thus, radius; r = 1.5/2 = 0.75 ft

Distance of the bulls eye from the wall to the left = 11 ft

Distance of dart thrown by Samuel from the left wall = 11.5 ft

Height of dart thrown by Samuel above the ground = 5.5 ft

We know that general equation of a circle is;

(x - a)² + (y - b)² = r²

Since Height from the floor to the bulls eye is 5.67ft,then a = 5.67 ft

And also since Distance of the bulls eye from the wall to the left = 11 ft, then b = 11 ft

Thus;

(x - 11)² + (y - 5.67)² = 0.75²

(x - 11)² + (y - 5.67)² = 0.5625

Now, since he throws a dart at the dartboard that lands at a point 11.5 feet from the left wall and 5.5 feet above the floor. Thus;

r² = (11.5 - 11)² + (5.5 - 5.67)²

r² = 0.2789

r² of 0.2789 gotten here is lesser than 0.5625 gotten earlier, we can conclude that Samuel's dart does not land on the dart board.

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Yes, Samuel dart does land on the dart board

Step-by-step explanation:

The given parameters are;

The height from the floor to the bulls eye = 5 feet 8 inches

The size of the standard board = 18 inches diameter

The distance of the bulls eye from the wall to the left = 11 feet

The distance the dart Samuel throws land from the left wall = 11.5 feet

The distance above the ground, the dart Samuel throws land = 5.5 feet

Whereby Samuel throws the dart directly at the dartboard, we have;

The location of the bulls eye = (11, 5.67)

The equation of the circle representing the dart board = (x - 11)² + (y - 5.67)² = 0.75² = 0.5625

The unit of the radius is in feet

The position of Samuel's dart is represented by the coordinate, (11.5, 5.5)

Plugging in the coordinates of the position of Samuel's dart into the equation of the circle of the dart board gives;

(11.5 - 11)² + (5.5 - 5.67)² = 0.2789 ≈ 0.528²

Therefore, given that the square of the distance of the position of Samuel's dart is less than the square of the radius of the dart board, we have;

Yes, Samuel dart does land on the dart board.

answer: option a.

step-by-step explanation:

you know that the quadrilateral opqr ( a four sided figure) is inscribed in circle. this means that all its vertices are on the circle.

then, by definition, the opposite angles of the quadrilateral are supplementary, this means that the sum of their measures is 180 degrees.

you can observe in the figure that the angles ∠opq   and the angle ∠orq   are opposite, therefore:

then, can be used to d to calculate the measure of ∠opq

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6 purchases

if you are asking how many of each then:

18 social media ads

6 newspaper ads