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goldenfox [79]
3 years ago
10

1. Jamie purchases gifts for three friends. She chooses a book for $12.80 and then finds two identical necklaces. Her budget is

$48 before tax. What is the maximum she can spend on each necklace?
2. The school PTA is sponsoring a dance. They decide to price student tickets at $6. It will cost the PTA $250 to put on the dance. How many tickets will the PTA need to sell in order to make a profit?


3. A wedding venue holds no more than 485 guests. If 319 people have said they are attending, how many more people does the venue have room for?


4. A conference is ordering gift bags for each attendee. The cost will be $14.50 per bag plus an $80 set-up fee. If the total exceeds $500 there is a discount. How many gift bags must the conference order to receive the discount? Write and solve an inequality.



5. A taxi cab charges $3 per person plus $1.25 per mile. What is the greatest distance one passenger could travel with a budget of $28? Write and solve an inequality.


6. A 15-passenger van is rented for a family vacation. The van rental is 460 per day, plus a $145 insurance fee. How many days can the van be rented if they want to spend no more than $625 on the van rental. Write and solve an inequality.
Mathematics
1 answer:
Evgen [1.6K]3 years ago
7 0

Step-by-step explanation:

1.

  • (48 - 12.80)/2 = $17.6

2.

  • 250/6 ≈ 42 and more

3.

  • 485 - 319 = 166 more people

4.

  • 14.5x + 80 > 500
  • 14.5x > 420
  • x > 420/14.5
  • x > 29

5.

  • 1.25x + 3 ≤ 28
  • 1.25x ≤ 25
  • x ≤ 25/1.25
  • x ≤ 20 miles

6.  <em>This seems a typo... 460 is likely $60</em>

  • 60x + 145 ≤ 625
  • 60x ≤ 625 - 145
  • 60x ≤ 480
  • x ≤ 480/60
  • x ≤ 8 days

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Step-by-step explanation:

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The distance between the anchor points of the two guy wires holding radio antenna is 181 feet (to the nearest foot)

<h3>How to determine the distance between the two anchor points of the guy wire</h3>

The problem will be solved using SOH CAH TOA

let the distance between the 150 ft long guy wire and the radio antenna be x

let the distance between the 180 ft long guy wire and the radio antenna be y

cos 65° = x / 150

x = cos 65° * 150

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The height of the antenna

sin 65° = height of antenna / 150

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