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Amanda [17]
3 years ago
9

Paul had 45 flyers to post around town. Last week, he posted 2/3 of them. This week, he posted 1/5 of the remaining fliers. How

many fliers has he still not posted?
Mathematics
2 answers:
Ahat [919]3 years ago
8 0
45 times 2/3= 30
30x 1/5=6

36 posted
45-36=9

Hope this helps :)
fenix001 [56]3 years ago
4 0
The answer before me was right at first but missed a certain detail.

45/3=15

15x2=30
therefore last week he posted 30 fliers. 45-30 = 15

up to this point 15 are left. 15/5=3
he posted 1/5th of the REMAINING posters which are 15 not 30. 30 is the amount of posters posted the first week

3x1=3 so this week he only posted 3
15-3 = 12
12 are still not posted
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andre [41]

Answer:

<h2>Zero</h2>

Step-by-step explanation:

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3 years ago
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Mr. Smith traveled to a city 300 miles from his home to attend a meeting. Due to car trouble, his average speed returning was 6
Kryger [21]

Answer: The speed at which he traveled to the city is 69.8 mph

Step-by-step explanation:

Let x represent the speed at which he traveled to the city.

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Time = distance/speed

Time taken to travel to the city is

300/x

Due to car trouble, his average speed on returning was 6 mph less than his speed going. This means that the speed at which he returned is (x - 6) mph. Time taken to return from the city is

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If the total time for the round trip was 9 hours, it means that

300/x + 300/(x - 6) = 9

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9x² - 54x - 300x - 300x - 1800 = 0

9x² - 654x -1800 = 0

Applying the general quadratic equation,

x = [- b ± √(b² - 4ac)]/2a

From the equation given,

a = 9

b = - 654

c = 1800

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x = [- - 654 ± √(- 654² - 4 × 9 × 1800)]/2 × 9

x = [654 ± √(427716 - 64800)]/18

x = [654 ± √362916.4]/18

x = (654 + 602.4)/2 or x = (654 - 602.4)/18

x = 1256.4/18 or x = 51.6/18

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By checking,

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Time spent returning =

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3 0
3 years ago
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m_a_m_a [10]
If you cannot read it, let me know.

3 0
3 years ago
4(x−3)≥56=<br> Please helppppp
pickupchik [31]

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Let's solve your inequality step-by-step.

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Answer:

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5 0
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