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Alexus [3.1K]
2 years ago
11

Kevin works at a restaurant. He stacks glasses as shown. Let x be the number of glasses and f(x) be the height of the stack. If

5.5 inch glasses with 1.2 inch lips are stacked vertically, which table correctly models this situation?

Mathematics
1 answer:
sukhopar [10]2 years ago
8 0

The table that correctly models the linear equation is table A.

<h3>Which table correctly models this situation?</h3>

The first glass added will have a height of 5.5 inches, and then for each glass that we add, we are adding a height of 1.2 inches.

  • This means that the first value on the table must be 5.5 in.
  • The second value must be 5.5 in + 1.2in = 6.7 in.
  • The third value must be: 6.7in + 1.2in = 7.9 in

And so on, so we have the linear relation:

y = 5.5in + x*1.2in

Where x is the number of glasses in the tower.

Then the table that correctly models this situation is table A.

If you want to learn more about linear equations:

brainly.com/question/1884491

#SPJ1

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A whole number first term to render as fifth term a value larger than 10000, should be at least 121

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So if we want to find the first term in the case that the fifth one is greater than 10,000 using this recursive formula, now we have to start backwards, and say that the fifth term is "> 10000" and what the fourth one is.

Notice that if you have this definition for the nth term, we can obtain from it, what the previous term is to find the general rule:

a_n=a_{n-1}*3+6\\a_n-6=a_{n-1}*3\\\frac{a_n-6}{3} = a_{n-1}\\a_{n-1}=\frac{a_n}{3} -2

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a_5>10000\\\frac{a_5}{3} -2>\frac{10000}{3} -2\\a_4>=\frac{10000}{3} -2\\\frac{a_4}{3} -2>\frac{\frac{10000}{3}-2}{3}-2 =\frac{10000}{9}-\frac{8}{3} \\a_3>\frac{10000}{9}-\frac{8}{3} \\\frac{a_3}{3} -2>\frac{\frac{10000}{9}-\frac{8}{3} }{3} -2=\frac{10000}{27} -\frac{8}{9} -2=\frac{10000}{27} -\frac{26}{9}\\a_2=\frac{10000}{27} -\frac{26}{9}\\\frac{a_2}{3} -2>\frac{\frac{10000}{27} -\frac{26}{9}}{3} -2=\frac{10000}{81} -\frac{80}{27} \\a_1>\frac{10000}{81} -\frac{80}{27}\approx 120.49

therefore, the starting first term should be at least about 121 to give a fifth term larger than 10,000

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