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Natasha2012 [34]
2 years ago
14

A deli has a special one-day event to celebrate its anniversary. On the day of the event, every eighth customer receives a free

drink. Every twenty-eighth customer receives a free sandwich. If 200 customers show up for the event, how many of the customers will receive both a free drink and a free sandwich?
Mathematics
2 answers:
enyata [817]2 years ago
4 0

Answer:

the number of common customers are 8 in number

garri49 [273]2 years ago
4 0

Answer: The answer is just 3 people will get BOTH the free drink and free sandwich.

Step-by-step explanation:

Every 8th = 1 free drink

Every 28th = 1 free sandwich

200 costumers showed

Add 28 to 28 until you reach the closest to 200 so: 28,56,84,112,140,168,196

Then add 8 to 8 until 200, every number that match the 28th sequence is 1 costumer that receives 1 drink and 1 sandwich

The match numbers are 56,112,168

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For the equation ae^ct=d, solve for the variable t in terms of a,c, and d. Express your answer in terms of the natural logarithm
saveliy_v [14]

We have been given an equation ae^{ct}=d. We are asked to solve the equation for t.

First of all, we will divide both sides of equation by a.

\frac{ae^{ct}}{a}=\frac{d}{a}

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Now we will take natural log on both sides.

\text{ln}(e^{ct})=\text{ln}(\frac{d}{a})

Using natural log property \text{ln}(a^b)=b\cdot \text{ln}(a), we will get:

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We know that \text{ln}(e)=1, so we will get:

ct\cdot 1=\text{ln}(\frac{d}{a})

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Now we will divide both sides by c as:

\frac{ct}{c}=\frac{\text{ln}(\frac{d}{a})}{c}

t=\frac{\text{ln}(\frac{d}{a})}{c}

Therefore, our solution would be t=\frac{\text{ln}(\frac{d}{a})}{c}.

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2 years ago
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Did your explanation include the following?

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The question in part c is not clear, nevertheless, part a and part b would be solved.

Answer:

a. The first twelve terms are:

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144.

b. The first ten terms are:

1.000, 1.000, 1.500, 1.667, 1.600, 1.625, 1.615, 1.619, 1.618, 1.618.

Step-by-step explanation:

a. Given

an + 2 = an + an + 1

where a1 = 1 and a2 = 1.

a3 = a1 + a2

= 2

a4 = a2 + a3

= 3

a5 = a3 + a4

= 5

a6 = a5 + a4

= 8

a7 = a6 + a5

= 13

a8 = a7 + a6

= 21

a9 = a8 + a7

= 34

a10 = a9 + a8

= 55

a11 = a10 + a9

= 89

a12 = a11 + a10

= 144

The first twelve terms are:

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144.

(b)

Given

bn = an+1/an

b1 = a2/a1

= 1/1 = 1.000

b2 = a3/a2

= 2/1 = 1.000

b3 = a4/a3

= 3/2 = 1.500

b4 = a5/a4

= 5/3 = 1.667

b5 = a6/a5

= 8/5 = 1.600

b6 = a7/a6

= 13/8 = 1.625

b7 = a8/a7

= 21/13 = 1.615

b8 = a9/a8

= 34/21 = 1.619

b9 = a10/a9

= 55/34 = 1.618

b10 = a11/a10

= 89/55 = 1.618

The first ten terms are:

1.000, 1.000, 1.500, 1.667, 1.600, 1.625, 1.615, 1.619, 1.618, 1.618.

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The,
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