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posledela
3 years ago
5

The numbers of seats in the first 12 rows of a high-school auditorium form an arithmetic sequence. The first row has 9 seats. Th

e second row has 11 seats. a) Write a recursive formula to represent the sequence. b) Write an explicit formula to represent the sequence. c) How many seats are in the 12th row?
Mathematics
1 answer:
Daniel [21]3 years ago
7 0
A₁ = 9
a₂ = 9+2  → a₂ = a₁ + 2. Since it's an AP, d, the common difference is 2

1) Then the recursive formula is :
a(n) = a(n-1) + 2

2) Explicit formula:
It's an AP, with first term a₁, d= common difference and n = number of term, which is the number of rows in this problem.
the value of the nth term is given by the formula is:
value of nth term  = a₁ + (n-1).d

3) Number of seats in the 12th row:

9 + (12-1).2 = 31 seats


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  • Pizza Duo = $5.60

Those are the costs for a whole small pizza.

==========================================================

Explanation:

Let

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  • y = cost of whole small pizza from Pizza Duo

x and y are some dollar amount, so they cannot be negative numbers. It also doesn't make sense to have them be 0 either. So we'll make them positive.

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x = 5.25

Pizza Uno charges $5.25 for a whole small pizza.

----------------------------

Pizza Duo charges $4.20 for 3/4 of a small pizza, meaning the equation we need to solve is

(3/4)y = 4.20

We'll use the same idea as the last equation to get...

(3/4)y = 4.20

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Pizza Duo charges $5.60 for a whole small pizza.

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3 years ago
Determine whether
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Answer:

(a) and (b) are not equivalent

(c) is equivalent

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Given

\frac{25^m}{5}

Required

Determine an equivalent or nonequivalent expression

(a)\ 25^{m-1

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25^{m-1

Apply law of indices

25^{m-1} = \frac{25^m}{25}

<em>This is not equivalent to </em>\frac{25^m}{5}<em></em>

(b)\ 25^{2m - 1}

We have:

25^{2m - 1}

Apply law of indices

25^{2m - 1} = \frac{25^{2m}}{25}

<em>This is not equivalent to </em>\frac{25^m}{5}<em></em>

<em></em>

<em />(c)\ 5^{2m-1}<em />

We have:

<em />5^{2m-1}<em />

Apply law of indices

<em />5^{2m-1} = \frac{5^{2m}}{5^1}<em />

<em />5^{2m-1} = \frac{5^{2m}}{5}<em />

<em>Evaluate the numerator</em>

<em />5^{2m-1} = \frac{25m}{5}<em />

<em />

<em>This is an equivalent expression</em>

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2 years ago
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