since T is the midpoint of SU, then ST = TU.
![\bf \stackrel{10x-14}{\boxed{S}\rule[0.35em]{10em}{0.25pt}} T\stackrel{5x+16}{\rule[0.35em]{10em}{0.25pt}\boxed{U}} \\\\\\ \stackrel{ST}{10x-14}=\stackrel{TU}{5x+16}\implies 5x-14=16\implies 5x=30\implies x=\cfrac{30}{5}\implies x=6 \\\\[-0.35em] ~\dotfill\\\\ \stackrel{ST}{10(6)-14\implies 46}~\hfill \stackrel{TU}{TU=ST=46}~\hfill \stackrel{SU}{ST+TU=92}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B10x-14%7D%7B%5Cboxed%7BS%7D%5Crule%5B0.35em%5D%7B10em%7D%7B0.25pt%7D%7D%20T%5Cstackrel%7B5x%2B16%7D%7B%5Crule%5B0.35em%5D%7B10em%7D%7B0.25pt%7D%5Cboxed%7BU%7D%7D%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7BST%7D%7B10x-14%7D%3D%5Cstackrel%7BTU%7D%7B5x%2B16%7D%5Cimplies%205x-14%3D16%5Cimplies%205x%3D30%5Cimplies%20x%3D%5Ccfrac%7B30%7D%7B5%7D%5Cimplies%20x%3D6%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7BST%7D%7B10%286%29-14%5Cimplies%2046%7D~%5Chfill%20%5Cstackrel%7BTU%7D%7BTU%3DST%3D46%7D~%5Chfill%20%5Cstackrel%7BSU%7D%7BST%2BTU%3D92%7D)
Answer:
The sidewalk is 2 feet wide
Step-by-step explanation:
The area of the sidewalk is given as 80 square feet
The length and width of pool
including side walks will be;
10 + 2x and 6 + 2x
To get the area of the side walk
That will be area of total minus area of the pool
That will be ;
(10 + 2x)(6 + 2x) - 10(6) = 80
60 + 12x + 20x + 4x^2 - 60 = 80
4x^2 + 32x -80 = 0
x^2 + 8x -20
x^2 + 10x - 2x - 20 = 0
x(x + 10) -2(x + 10) = 0
(x-2)(x + 10) = 0
x = 2 or -10
since width cannot be negative;
The sidewalk is 2 feet wide
Answer:The answer is 8
Step-by-step explanation:
Z = 4 , w = 6
z = x , w = 8
Therefore, z = (4*8) / 6
= 32 / 6
= 16/3 = 5.333
Answer:
-14
Step-by-step explanation:
a(3)=a(2)-17=a(1)-17-17=20-17-17=-14
