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Paha777 [63]
2 years ago
13

You are hiring a limousine. One company charges a $60 reservation fee plus another $15 per hour. A second company charges a $25

reservation fee plus another $20 per hour. Write and solve a linear equation to find the number of hours at which the total cost is the same for both companies.
Mathematics
1 answer:
anzhelika [568]2 years ago
5 0

Answer:

7 hours

Step-by-step explanation:

Step one:

given data

let the total charges be y and the number of hours be x

Company A charges

$60 reservation fee plus another $15 per hour.

so

y=60+15x----------1

Company B charges

$25 reservation fee plus another $20 per hour

y=25+20x------------2

Step two:

equate the 2 expressions above we have

60+15x=25+20x

collect like terms

60-25=20x-15x

35=5x

divide both sides by 5

x= 35/5

x= 7 hours

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How do you do this?<br> Please help, I’ve been staring at this for like 10 min :)
damaskus [11]

Answer & Step-by-step explanation:

The formula is given in the middle. You are supposed to insert the value of x into this formula to find the value of y (like in the first given example):

x=-2\\y=2(-2)+5\\y=-4+5\\y=1

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x=-1\\y=2(-1)+5\\y=-2+5\\y=3

------

x=0\\y=2(0)+5\\y=0+5\\y=5

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x=1\\y=2(1)+5\\y=2+5\\y=7

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:Done

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3 years ago
What are the clear steps for solving for x question (5x + 2)
Alenkinab [10]

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x= -2/5

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5x=-2

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

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6 0
3 years ago
A local hamburger shop sold a combined total of 500 hamburgers and cheeseburgers on Wednesday. There were 50 fewer cheeseburgers
Virty [35]

Hey there! I'm happy to help!

Let's call the hamburgers h and the cheeseburgers c.

h+c=500

c=h-50

Let's plug this value for c into the first equation to solve for h.

h+h-50=500

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Divide both sides by 2.

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Therefore, 275 hamburgers were sold on Wednesday.

Have a wonderful day! :D

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<u>Step-by-step explanation:</u>

We have, the graph of f(x)= 2^{x} , on replacing f(x) by f(x-3) we get:

f(x-3)= 2^{x-3} = \frac{2^{x}}{2^{3}} = \frac{1}{8} 2^{x} = \frac{1}{8} f(x).Below shown are the images for graph of f(x) and f(x-3). Both are functions are exponential , and so having exponential graph but f(x-3) is compressed by a factor of  \frac{1}{8} horizontally . Domain and range of both functions are same i.e. F(x) & f(x-3) domain & range are same , just difference in graph : f(x-3) = \frac{1}{8} f(x).

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