-6.867
You do 13 divided by 15 the plug in the -6 in the front
Answer: 24 and 37
Step-by-step explanation:
Let the numbers be x and y . Also let us pick x as the larger number.
From the first statement :
x + y = 61 ............................ equation 1
From the second statement :
x = y + 13 ............................equation 2
Solving the resulting linear equations by substitution method, substitute x = y + 13 into equation 1, then
x + y = 61 becomes
y + 13 + y = 61
2y + 13 = 61
2y = 48
y = 24
Substitute y = 24 into equation 2 , then
x = y + 3 becomes
x = 24 + 13
x = 37
Therefore the numbers are 24 and 37
Check the picture below.
![\stackrel{\textit{\Large Areas}}{\stackrel{triangle}{\cfrac{1}{2}(6)(6)}~~ + ~~\stackrel{semi-circle}{\cfrac{1}{2}\pi (3)^2}}\implies \boxed{18+4.5\pi} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{pythagorean~theorem}{CA^2 = AB^2 + BC^2\implies} CA=\sqrt{AB^2 + BC^2} \\\\\\ CA=\sqrt{6^2+6^2}\implies CA=\sqrt{6^2(1+1)}\implies CA=6\sqrt{2} \\\\\\ \stackrel{\textit{\Large Perimeters}}{\stackrel{triangle}{(6+6\sqrt{2})}~~ + ~~\stackrel{semi-circle}{\cfrac{1}{2}2\pi (3)}}\implies \boxed{6+6\sqrt{2}+3\pi}](https://tex.z-dn.net/?f=%5Cstackrel%7B%5Ctextit%7B%5CLarge%20Areas%7D%7D%7B%5Cstackrel%7Btriangle%7D%7B%5Ccfrac%7B1%7D%7B2%7D%286%29%286%29%7D~~%20%2B%20~~%5Cstackrel%7Bsemi-circle%7D%7B%5Ccfrac%7B1%7D%7B2%7D%5Cpi%20%283%29%5E2%7D%7D%5Cimplies%20%5Cboxed%7B18%2B4.5%5Cpi%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7Bpythagorean~theorem%7D%7BCA%5E2%20%3D%20AB%5E2%20%2B%20BC%5E2%5Cimplies%7D%20CA%3D%5Csqrt%7BAB%5E2%20%2B%20BC%5E2%7D%20%5C%5C%5C%5C%5C%5C%20CA%3D%5Csqrt%7B6%5E2%2B6%5E2%7D%5Cimplies%20CA%3D%5Csqrt%7B6%5E2%281%2B1%29%7D%5Cimplies%20CA%3D6%5Csqrt%7B2%7D%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7B%5CLarge%20Perimeters%7D%7D%7B%5Cstackrel%7Btriangle%7D%7B%286%2B6%5Csqrt%7B2%7D%29%7D~~%20%2B%20~~%5Cstackrel%7Bsemi-circle%7D%7B%5Ccfrac%7B1%7D%7B2%7D2%5Cpi%20%283%29%7D%7D%5Cimplies%20%5Cboxed%7B6%2B6%5Csqrt%7B2%7D%2B3%5Cpi%7D)
notice that for the perimeter we didn't include the segment BC, because the perimeter of a figure is simply the outer borders.
The answer is 46428 to that
One Triangle = 6 + 6 + 5 = 17.
Two Triangles = 6 + 6 + 5 + 5 = 22.
Three Triangles = 6 + 6 + 5 + 5 + 5 = 27.
