36y + 7 stars are great but I don’t have a bad experience
Answer:
x = 35
y = 17.5
Step-by-step explanation:
By Central angle and its corresponding arc theorem:
x° = 360° - (175° + 150°)
x° = 360° - 325°
x° = 35°
x = 35
By inscribed angle theorem:
![y \degree = \frac{1}{2} x \\ \\ y\degree = \frac{1}{2} \times 35 \degree \\ \\ y\degree = 17.5 \degree \\ \\ y \: = 17.5](https://tex.z-dn.net/?f=y%20%5Cdegree%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20x%20%5C%5C%20%20%5C%5C%20y%5Cdegree%20%3D%20%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%5Ctimes%2035%20%5Cdegree%20%5C%5C%20%20%5C%5C%20y%5Cdegree%20%3D%20%2017.5%20%5Cdegree%20%5C%5C%20%20%5C%5C%20y%20%5C%3A%20%20%3D%2017.5)
Answer:
The expression that represents the total time Louis spends commuting to and from the gym each day is 8(x-2y)
Step-by-step explanation:
Given that the time Louis spends commuting to the gym can be represented by two expressions:
Time to gym: ![2x-6y\hfill (1)](https://tex.z-dn.net/?f=2x-6y%5Chfill%20%281%29)
Time from gym: ![6x-10y\hfill (2)](https://tex.z-dn.net/?f=6x-10y%5Chfill%20%282%29)
To find the expression that represents the total time Louis spends commuting to and from the gym each day :
The total time Louis spends commuting to and from the gym each day is equal to the sum of the Time to gym and Time from gym
Total time Louis spends commuting to and from the gym each day=2x-6y+6x-10y
=8x-16y
=8(x-2y)
Therefore Total time Louis spends commuting to and from the gym each day is 8(x-2y)
The expression that represents the total time Louis spends commuting to and from the gym each day is 8(x-2y)
check the picture below on the top side.
we know that x = 4 = b, therefore, using the 30-60-90 rule, h = 4√3, and DC = 4+8+4 = 16.
![\bf \textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} a,b=\stackrel{bases}{parallel~sides}\\ h=height\\[-0.5em] \hrulefill\\ a=8\\ b=\stackrel{DC}{16}\\ h=4\sqrt{3} \end{cases}\implies A=\cfrac{4\sqrt{3}(8+16)}{2} \\\\\\ A=2\sqrt{3}(24)\implies \boxed{A=48\sqrt{3}}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Barea%20of%20a%20trapezoid%7D%5C%5C%5C%5C%0AA%3D%5Ccfrac%7Bh%28a%2Bb%29%7D%7B2%7D~~%0A%5Cbegin%7Bcases%7D%0Aa%2Cb%3D%5Cstackrel%7Bbases%7D%7Bparallel~sides%7D%5C%5C%0Ah%3Dheight%5C%5C%5B-0.5em%5D%0A%5Chrulefill%5C%5C%0Aa%3D8%5C%5C%0Ab%3D%5Cstackrel%7BDC%7D%7B16%7D%5C%5C%0Ah%3D4%5Csqrt%7B3%7D%0A%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Ccfrac%7B4%5Csqrt%7B3%7D%288%2B16%29%7D%7B2%7D%0A%5C%5C%5C%5C%5C%5C%0AA%3D2%5Csqrt%7B3%7D%2824%29%5Cimplies%20%5Cboxed%7BA%3D48%5Csqrt%7B3%7D%7D)
now, check the picture below on the bottom side.
since we know x = 9, then b = 9, therefore DC = 9+6+9 = 24, and h = b = 9.
![\bf \textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} a,b=\stackrel{bases}{parallel~sides}\\ h=height\\[-0.5em] \hrulefill\\ a=6\\ b=\stackrel{DC}{24}\\ h=9 \end{cases}\implies A=\cfrac{9(6+24)}{2} \\\\\\ A=\cfrac{9(30)}{2}\implies \boxed{A=135}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Barea%20of%20a%20trapezoid%7D%5C%5C%5C%5C%0AA%3D%5Ccfrac%7Bh%28a%2Bb%29%7D%7B2%7D~~%0A%5Cbegin%7Bcases%7D%0Aa%2Cb%3D%5Cstackrel%7Bbases%7D%7Bparallel~sides%7D%5C%5C%0Ah%3Dheight%5C%5C%5B-0.5em%5D%0A%5Chrulefill%5C%5C%0Aa%3D6%5C%5C%0Ab%3D%5Cstackrel%7BDC%7D%7B24%7D%5C%5C%0Ah%3D9%0A%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Ccfrac%7B9%286%2B24%29%7D%7B2%7D%0A%5C%5C%5C%5C%5C%5C%0AA%3D%5Ccfrac%7B9%2830%29%7D%7B2%7D%5Cimplies%20%5Cboxed%7BA%3D135%7D)