For her parents' anniversary party, Sarah is considering using one of two venues. A hotel in Booneville will cost $904 for a reservation, plus $86 per person. A restaurant in the same city will cost $89 per person, in addition to $901 for the reservation. In order to make the best decision, Sarah figures out how many attendees it would take to have the venues cost the same amount. How many attendees would that be? What would the total cost be?

x = 1

$990

Step-by-step explanation:

First you want to write the linear equations for both scenarios

in the first scenario hotel:

y = 86x + 904 where x = # of people, y = total cost

in the second scenario restaurant:

y= 89x + 901

now you want to set them equal to each other since you want to know how many people (x) will be needed to make total costs equal.

86x + 904 = 89x + 901

subtract 86x on both sides to move all x terms on the right

904 = 3x + 901

subtract 901 on both sides to move numbers to the left

3 = 3x

divide both sides by 3

1 = x or x=1

So it would take 1 attendee.

Now plug in x=1 into either equation to find the total cost that is the same.

y = 86(1) + 904

= 86 + 904

= $990

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geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

step-by-step explanation:

circumference divided by 3.141592654 then timesthe radius twice