One bullring (a perfect circle) has a 55-yard diameter. what is its area?

Mathematics

1 answer:

Anastaziya [24]2 years ago

8 0

2374.625 is the correct answer you need to do pi R squared to get area

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Find the area of a circle circumscribing an equilateral triangle of side 15cm.<br> Take pi = 3.14.

miskamm [114]

Answer:

236 cm²

Step-by-step explanation:

Height of an equilateral triangle (h) = √3 /2 (l)

l = side of the equilateral triangle.

h = √3 /2 (15)

In an equilateral triangle the orthocenter, centroid, circumcenter and incenter are in the same spot

The center of the circle is the centroid and height match with the median. The radius of the circumcircle is equal to two thirds the height.

Formula for the Radius of the circumcircle = 2/3 h

= 2/3 x √3 /2 (15)

= 5 √3 cm (=radius)

Area of the circle = πr^

= 3.14 x( 5 √3 ) ^

=3.14 x(25*3)

=3.14 x 75

=235.5

=236 cm²

8 0

2 years ago

What is the slope between the points (8, -2) and (-7, 10)

lara [203]

Slope formula:
(y2-y1)/(x2-x1)

Substitute in the points into the slope formula.

Slope= [10-(-2)] / [(-7)-8]
Slope= (10+2) / (-15)
Slope= 12/-15
Slope= -12/15
Simplify (3 goes into each)
Slope= -4/5

Hope this helps!! :)

8 0

2 years ago

PLEASE ANSWER AS SOON AS POSSIBLE

Valentin [98]

7 because it is whole number

6 0

2 years ago

Read 2 more answers

On a cold day, hailstones fall with a velocity of 2i − 6k m s−1 . If a cyclist travels through the hail at 10i m s−1 , what is t

Paraphin [41]

Answer:

The velocity of the hail relative to the cyclist is $V_{hc}=-8\hat{i}-6\hat{k}$

The angle at which hailstones falling relative to the cyclist is $\theta = 36.86^\circ$

Step-by-step explanation:

Given : On a cold day, hailstones fall with a velocity of $2\hat{i}-6\hat{k}\text{ m/s}$ . If a cyclist travels through the hail at $10\hat{i} \text{ m/s}$ .