Check the picture below.
![\textit{area of a regular polygon}\\\\ A=\cfrac{1}{2}asn ~~ \begin{cases} a=apothem\\ n=\stackrel{side's}{number}\\ s=\stackrel{side's}{length}\\[-0.5em] \hrulefill\\ a=7\\ s=8.1\\ n=6 \end{cases}\implies \stackrel{\textit{area of the hexagonal base}}{A=\cfrac{1}{2}(7)(8.1)(6)} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Ctextit%7Barea%20of%20a%20regular%20polygon%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7B1%7D%7B2%7Dasn%20~~%20%5Cbegin%7Bcases%7D%20a%3Dapothem%5C%5C%20n%3D%5Cstackrel%7Bside%27s%7D%7Bnumber%7D%5C%5C%20s%3D%5Cstackrel%7Bside%27s%7D%7Blength%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20a%3D7%5C%5C%20s%3D8.1%5C%5C%20n%3D6%20%5Cend%7Bcases%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20hexagonal%20base%7D%7D%7BA%3D%5Ccfrac%7B1%7D%7B2%7D%287%29%288.1%29%286%29%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\textit{volume of a prism}\\\\ V=Bh ~~ \begin{cases} B=\stackrel{base's}{area}\\ h=height\\[-0.5em] \hrulefill\\ B=\frac{1}{2}(7)(8.1)(6)\\ V=3572.1 \end{cases}\implies 3572.1=\cfrac{1}{2}(7)(8.1)(6)h \\\\\\ 3572.1=170.1h\implies \cfrac{3572.1}{170.1}=h\implies \boxed{21=h}](https://tex.z-dn.net/?f=%5Ctextit%7Bvolume%20of%20a%20prism%7D%5C%5C%5C%5C%20V%3DBh%20~~%20%5Cbegin%7Bcases%7D%20B%3D%5Cstackrel%7Bbase%27s%7D%7Barea%7D%5C%5C%20h%3Dheight%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20B%3D%5Cfrac%7B1%7D%7B2%7D%287%29%288.1%29%286%29%5C%5C%20V%3D3572.1%20%5Cend%7Bcases%7D%5Cimplies%203572.1%3D%5Ccfrac%7B1%7D%7B2%7D%287%29%288.1%29%286%29h%20%5C%5C%5C%5C%5C%5C%203572.1%3D170.1h%5Cimplies%20%5Ccfrac%7B3572.1%7D%7B170.1%7D%3Dh%5Cimplies%20%5Cboxed%7B21%3Dh%7D)
well, the length of the tunnel is "h", now two 8 meters cars, that's 8+8=16 meters plus a 3 meter connector between them, that's 16 + 3 = 19 meters, can those two cars connected like so fit inside the tunnel? sure thing, "h" can fit 19 meters just fine.
Senior tickets (x)
Child tickets (y)
First day: 3x + 5y = 70
second day: 12x + 12y = 216
Solve the system of equations (use elimination)
Multiply first equation by -4 -4(3x + 5y = 70), which makes it
-12x - 20y = -280
(+)12x + 12y = 216 add to second equation
-8y = -64
divide by -8. y = 8
Plugin the y value to either equation ( I will choose first equation)
3x + 5(8) =70
3x+ 40 = 70
3x = 30
x = 10
Senior tickets are $10, child tickets are $8
Answer:
x = 3 and x = -2
Step-by-step explanation:
The given equation is :
We need to solve the given equation.
We can solve it using Completing the Square method.
We take two numbers such that their product is -6 and sum is -1.
So,
so, the values of x are 3 and -2.
