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Vsevolod [243]
3 years ago
6

How would you do a,b,c,and d

Mathematics
1 answer:
I am Lyosha [343]3 years ago
4 0

a) You are told the function is quadratic, so you can write cost (c) in terms of speed (s) as

... c = k·s² + m·s + n

Filling in the given values gives three equations in k, m, and n.

28 = k\cdot 10^2+m\cdot 10+n\\21=k\cdot 20^2+m\cdot 20+n\\16=k\cdot 30^2+m\cdot 30+n

Subtracting each equation from the one after gives

-7=300k+10m\\-5=500k+10m

Subtracting the first of these equations from the second gives

2=200k\\\\k=\dfrac{2}{200}=0.01

Using the next previous equation, we can find m.

-5=500\cdot 0.01+10m\\\\m=\dfrac{-10}{10}=-1

Then from the first equation

[tex]28=100\cdot 0.01+10\cdot (-1)+n\\\\n=37[tex]

There are a variety of other ways the equation can be found or the system of equations solved. Any way you do it, you should end with

... c = 0.01s² - s + 37

b) At 150 kph, the cost is predicted to be

... c = 0.01·150² -150 +37 = 112 . . . cents/km

c) The graph shows you need to maintain speed between 40 and 60 kph to keep cost at or below 13 cents/km.

d) The graph has a minimum at 12 cents per km. This model predicts it is not possible to spend only 10 cents per km.

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3 0
3 years ago
Mohamed and Li Jing were asked to find an explicit formula for the sequence -5, -25, -125, -625,....
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Answer:

Li Jing's formula i.e.  \boxed{g_n=-5\cdot \:5^{n-1}}  is right.

Step-by-step explanation:

Considering the sequence

-5,\:-25,\:-125,\:-625,...

A geometric sequence has a constant ratio r and is defined by

g_n=g_0\cdot r^{n-1}

\mathrm{Compute\:the\:ratios\:of\:all\:the\:adjacent\:terms}:\quad \:r=\frac{g_{n+1}}{g_n}

\frac{-25}{-5}=5,\:\quad \frac{-125}{-25}=5,\:\quad \frac{-625}{-125}=5

\mathrm{The\:ratio\:of\:all\:the\:adjacent\:terms\:is\:the\:same\:and\:equal\:to}

r=5

So, the sequence is geometric.

as

\mathrm{The\:first\:element\:of\:the\:sequence\:is}

g_1=-5

r=5

so

g_n=g_1\cdot r^{n-1}

\mathrm{Therefore,\:the\:}n\mathrm{th\:term\:is\:computed\:by}\:

g_n=-5\cdot \:5^{n-1}

Therefore, Li Jing's formula i.e.  \boxed{g_n=-5\cdot \:5^{n-1}}  is right.

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