<span>Each set of prices can be described by a linear function. In this case, Company A would be y = 5 + .5x, where y is the fare and x is the number of miles driven. Similarly, Company B would b y = 10 + .1x. The definition of intersection tells us that two lines interstect only at that point where their x and y coordinates are identical, which means, of course, that their y coordinates will be identical. So to find the point of intersection we can examine the point where the y coordinates are the same. Which is to say, where 5 + .5x = 10 + .1x. We can simplify this to .5x - .1x = 10 - 5, then .4x = 5, and finally, x = (5/.4) = 12.5. In other words, the fares will be equal when the miles driven is 12.5</span>
Can you please tell me what the first question says? i can read it, its too dark
Answer:
54
Step-by-step explanation:
Tn = an²+bn
T2 = a2²+2b
2 = 4a +2b
Multiply all through by 2
4 = 8a +4b
4b = 4-8a
T4 = a4² + 4b
20 = 16a +4b
4b = 20-16a
4-8a = 20-16a
16a -8a = 20-4
8a = 16
a = 16/8 = 2
4b = 4-8a = 4-8(2) = 4-16 = -12
b = -12/4 = -3
T6 = a6² + 6b
T6 = 2(36) + 6(-3) = 72 -18 = 54
To solve quadratics, the most correct and effective way, in my opinion, is the quadratic formula:
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Plugging it in, we get:
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So the final answer is:
and
Check the picture below.
so the volume will simply be the area of the hexagonal face times the height.
![\textit{area of a regular polygon}\\\\ A=\cfrac{1}{4}ns^2\stackrel{\qquad degrees}{\cot\left( \frac{180}{n} \right)}~~ \begin{cases} n=\stackrel{number~of}{sides}\\ s=\stackrel{length~of}{side}\\[-0.5em] \hrulefill\\ n=6\\ s=12 \end{cases}\implies A=\cfrac{1}{4}(6)(12)^2\cot\left( \frac{180}{6} \right) \\\\\\ A=216\cot(30^o)\implies A=216\sqrt{3} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of the hexagon}}{(216\sqrt{3})}~~\stackrel{height}{(10)}\implies 2160\sqrt{3}~~\approx ~~3741.2~cm^3](https://tex.z-dn.net/?f=%5Ctextit%7Barea%20of%20a%20regular%20polygon%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7B1%7D%7B4%7Dns%5E2%5Cstackrel%7B%5Cqquad%20degrees%7D%7B%5Ccot%5Cleft%28%20%5Cfrac%7B180%7D%7Bn%7D%20%5Cright%29%7D~~%20%5Cbegin%7Bcases%7D%20n%3D%5Cstackrel%7Bnumber~of%7D%7Bsides%7D%5C%5C%20s%3D%5Cstackrel%7Blength~of%7D%7Bside%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20n%3D6%5C%5C%20s%3D12%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Ccfrac%7B1%7D%7B4%7D%286%29%2812%29%5E2%5Ccot%5Cleft%28%20%5Cfrac%7B180%7D%7B6%7D%20%5Cright%29%20%5C%5C%5C%5C%5C%5C%20A%3D216%5Ccot%2830%5Eo%29%5Cimplies%20A%3D216%5Csqrt%7B3%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20hexagon%7D%7D%7B%28216%5Csqrt%7B3%7D%29%7D~~%5Cstackrel%7Bheight%7D%7B%2810%29%7D%5Cimplies%202160%5Csqrt%7B3%7D~~%5Capprox%20~~3741.2~cm%5E3)
