As Chris mentioned:
I calculated the distances between the points (side lengths) as
5 7.81 and 8.25
thus p = 10.53
sqrt(10.53(10.53-5)(10.53-7.81)(10.53-8.25)) = 19.0 units^2
(just as Chris found!)
You're welcome.....Keep up the good work!
The picture is EXPANDED to a larger size (expanded by a factor of 12)
Yes, 18 degrees....you have it correct!
Wrong diagram, TM31....that is the lampost problem.... where is the OTHER diagram?
Just work this as a simple ratio problem:
5 ft 3 inches (5.25 ft) is to 6 feet as 'x' is to 18 feet
5.25/6 = x / 18 solve for 'x'
5.25/6 * 18 = x
15.75 ft = x
Is that supposed to be 3/20 and 2/5 ?
Good point x^2 I was working off of yangg's example and not the question. my mistake....BUT since the questioner asked about the quadratic formula, I guess we can assume = 0
The weight given in this question is irrelavent....not needed....
The surface area of a sphere (bowling ball) 4 pi r^2
You are given DIAMETER...use 1/2 of this for the RADIUS then 4 pi r^2 in^2
Can you take it from here?
Remember the equation for a circle with center (h,k) and radius r is
(x-h)^2 + (y-k)^2 = r^2
so given the info in the question, this equals:
(x-(-4))^2 + (y-3)^2 = 2.5^2
(x+4)^2 + (y-3)^2 = 6.25