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STALIN [3.7K]
3 years ago
14

No, because the bike order does not meet the restrictions of 4c + 6a ≤ 120 and 4c + 4a ≤ 100

Mathematics
2 answers:
Mariana [72]3 years ago
6 0
This is the question:

A bicycle manufacturing company makes a particular type of bike.

Each child bike requires 4 hours to build and 4 hours to test.

Each adult bike requires 6 hours to build and 4 hours to test.

With the number of workers, the company is able to have up to 120 hours of building time and 100 hours of testing time for a week.

If c represents child bikes and a represents adult bikes,

determine which system of inequality best explains whether the company can build 10 child bikes and 12 adult bikes in the week

 

Now you can state the system of inequalities from the statements

1) First inequality based on the hours availble to buiding


Each child bike requires 4 hours, e<span>ach adult bike requires 6 hours to build and </span>the company is able to have up to 120 hours of building =>


4c + 6a ≤ 120


2) Second inequality based of the hours available to testing.


Each child bike requires 4 hours to test, each adult bike 4 hours to test and the company is able to have up 100 hours of testing time for a week =>


4c + 4a ≤ 100

Then the two inequalities are:

4c + 6a ≤ 120
4c + 4a ≤ 100

<span>The answer is Yes, because the bike order meets the restrictions of 4c + 6a ≤ 120 and 4c + 4a ≤ 100

Which you can verify by replacing in both equations 10 for c and 12 for a. Look:

1) 4(10) + 6(12) = 40 + 72 = 112 ≤ 120

2) 4(10) + 4(12) = 40 + 48 = 88 ≤ 100




</span>



exis [7]3 years ago
5 0

Answer:

Yes, because the bike order meets the restrictions of 4c + 6a ≤ 120 and 4c + 4a ≤ 100

Step-by-step explanation:

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Step-by-step explanation:

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40 mph = 40 miles/hour / 60 minutes/hour = 0.6667 miles per minute

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8 0
3 years ago
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6 0
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If you can explain your answer that’d be great!! Thank you!
Rasek [7]

Answer:

A(t)=A_{0}e^{\frac{ln(\frac{1}{2})}{22}t}

Step-by-step explanation:

This half life exponential decay equation goes by the formula:

A(t)=A_{0}e^{kt}

Where

k=\frac{ln(\frac{1}{2})}{Half-Life}

Since half life is given as 22, we plug that into "Half-Life" in the formula for k and then plug in the formula for k into the exponential decay formula:

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Now

A(t)=A_{0}e^{\frac{ln(\frac{1}{2})}{22}t}

third choice is correct.

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3 years ago
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