Answer:
14.14 i hope this helps lol ;)
Answer:
(The image is not provided, so i draw an idea of how i supposed that the problem is, the image is at the bottom)
Ok, we have a rectangle of length x by r.
At the extremes of length r, we add two semicircles.
So the perimeter will be equal to:
Two times x, plus the perimeter of the two semicircles (that can be thought as only one circle).
The radius of the semicircles is r, and the perimeter of a circle is:
C = 2*pi*r
where pi = 3.14
Then the perimeter of the track is:
P = 2*x + 2*pi*r.
b) now we want to solve this for x, this means isolating x in one side of the equation.
P - 2*pi*r = 2*x
P/2 - pi*r = x.
c) now we have:
P = 660ft
r = 50ft
then we can replace the values and find x.
x = 660ft/2 - 3.14*50ft = 173ft
complementary angles add up to 90°, so therefore we know that ∡A + ∡B = 90°, and also they are in a ratio of 3:6.
![\bf \cfrac{A}{B}=\cfrac{3}{6}\implies \cfrac{A}{B}=\cfrac{1}{2}\implies 2A=\boxed{B} \\\\[-0.35em] ~\dotfill\\\\ A+B=90\implies A+\boxed{2A}=90\implies 3A=90\\\\\\ A=\cfrac{90}{3}\implies \blacktriangleright A=30 \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ 2(30)=B\implies \blacktriangleright 60=B \blacktriangleleft](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7BA%7D%7BB%7D%3D%5Ccfrac%7B3%7D%7B6%7D%5Cimplies%20%5Ccfrac%7BA%7D%7BB%7D%3D%5Ccfrac%7B1%7D%7B2%7D%5Cimplies%202A%3D%5Cboxed%7BB%7D%0A%5C%5C%5C%5C%5B-0.35em%5D%0A~%5Cdotfill%5C%5C%5C%5C%0AA%2BB%3D90%5Cimplies%20A%2B%5Cboxed%7B2A%7D%3D90%5Cimplies%203A%3D90%5C%5C%5C%5C%5C%5C%20A%3D%5Ccfrac%7B90%7D%7B3%7D%5Cimplies%20%5Cblacktriangleright%20A%3D30%20%5Cblacktriangleleft%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0A2%2830%29%3DB%5Cimplies%20%5Cblacktriangleright%2060%3DB%20%5Cblacktriangleleft)
Is that considered two equations?
4a + 5a - 15 = 2 + 3a
6a = 17
a = 17/6