Question

The standard height from the floor to the bull’s-eye at which a standard dartboard is hung at 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 10 feet from the wall to its left. Sasha throws a dart at the dartboard that land at point 10.25 Feet from the left wall and 5 feet above the floor. Does Sasha’s dart land on the dartboard? Drag the choices into the boxes to correctly complete the statements.

Answer 1

Answer:

Hello! I'm sorry I couldn't get to your question sooner. I just completed this quiz!

The equation of the circle that represents the dartboard is (x-10)^2 + (y-17/3)^2 = 9/16, where the origin is the lower-left corner of the room and the unit of the radius is feet.  

The position of Sasha's dart is represented by the coordinates (10.25,5). Sash's dart does land on the dartboard.

This quiz was completed on k12, lesson 3.03.

Answer 2

The question is an illustration of equation of circles.

• The equation of the dartboard circle is: $\mathbf{(x - 10)^2 + (y - \frac{17}3)^2 = \frac 9{16}}$
• Sasha's dart lands on the dartboard because

From the question, we understand that:

$\mathbf{h = 5\ ft\ 8\ in }$ ---- the height at which the dartboard was hung

$\mathbf{d = 18\i n }$ ---- the diameter of the dartboard

$\mathbf{B = 10ft}$  --- the bull's eye

$\mathbf{D = (10.25ft, 5ft)}$ --- Sasha's dart

Equation of the circle

First, we convert all units to feet

This is done by dividing inches units by 12

$\mathbf{h = 5\ ft\ 8\ in }$

$\mathbf{h = 5\ ft\ + \frac{8}{12}\ ft }$

$\mathbf{h = 5\ ft\ + \frac{2}{3}\ ft }$

Take LCM

$\mathbf{h = \frac{15 + 2}{3}\ ft }$

$\mathbf{h = \frac{17}{3}\ ft }$

$\mathbf{d = 18\i n }$

$\mathbf{d = \frac{18}{12}ft}$

$\mathbf{d = \frac{3}{2}ft}$

Divide by 2 to calculate radius

$\mathbf{r = \frac{3}{2*2}ft}$

$\mathbf{r = \frac{3}{4}ft}$

The equation of the circle is represented as:

$\mathbf{(x - a)^2 + (y - b)^2 = r^2}$

In this case:

$\mathbf{a = B = 10ft}$ -- the distance between the bull's eye and the wall

$\mathbf{b = h = \frac{17}{3}\ ft }$ ---- the height at which the dartboard was hung

So, we have:

$\mathbf{(x - a)^2 + (y - b)^2 = r^2}$

$\mathbf{(x - 10)^2 + (y - \frac{17}3)^2 = (\frac 34)^2}$

Evaluate the exponents

$\mathbf{(x - 10)^2 + (y - \frac{17}3)^2 = \frac 9{16}}$

Hence, the equation of the circle is: $\mathbf{(x - 10)^2 + (y - \frac{17}3)^2 = \frac 9{16}}$

Does Sasha’s dart land on the dartboard?

Yes her dart lands on the dartboard because

$\mathbf{D = (10.25ft, 5ft)}$ is within the circumference of the dartboard

Read more about equation of circles at:

https://brainly.com/question/23988015