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Vikentia [17]
2 years ago
7

An archer misses 60% of his shots how many shots must he make before he gets 100 hits?​

Mathematics
1 answer:
Serga [27]2 years ago
4 0

Answer:

250

Step-by-step explanation:

60% is 3/5. Meaning, for every 5 shots taken, the archer only makes 3.

So, they would need to make 250 shots before they get 100.

Hope this helps!

Plz name brainliest if possible!

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Jordan builds a miniature catapult for the science fair. He documents how far the catapult can throw a small rock for his report
GrogVix [38]

The rate to convert yards to meters is

1 \text{ yard}=0.9144\text{ meters}

So, if you multiply both sides by 8, you get

8 \text{ yards} = 8\times 0.9144 = 7.3152 \text{ meters}

To round the answer to the nearest tenth, we need to choose between 7.3 and 7.4. Since the hundreths digit is 1, which is less than 5, 7.3 is nearest to 7.3152 than 7.4, so the correct answer is 7.3

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2 years ago
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To reduce wear and tear on the family vehicle, you decide to rent a car. Happy Harry’s Rentals has a $500 fee for any rental up
yan [13]

Let's use 8 days as the maximum time we are going to be renting the car.

Putting that into the equation means, 500$ for Harry's Rentals and Smilin' Sam's at $600.

Therefore, Happy Harry's Rentals are better for the 7th and 8th days while Smiling Sam's are the better from day's 1 to 6.

Work:

500$ is a fixed value so it doesn't change (constant)

200 + 50x

x = days

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200 + 50(8)

200 + 400 = 600

8 0
2 years ago
APEX HELP ASAP!!!!
oee [108]
<span>1. The two boats picked for the trip are the steamboat and the tall ship. Let us assume that we will take the steamboat going to the island, and then we will take the tall ship for the return trip. We will then relate the distances travelled by both ships to each other.

2. We know that the steamboat takes five hours to complete the trip. The tall ship takes more time, at ten hours to complete the trip. We do not have the exact speeds of the steamboat or of the tall ship, but we do know that the tall ship is 10 knots slower than the steamboat. We likewise do not know the exact distance travelled by either ship, but we do know that both travel the same distance. We want to find out how fast each boat travels. We expect the answers to be in knots, with a difference of 10.

3. We know that distance is equivalent to the product of speed of a boat multiplied by the time of travel. For the trip going to the island, we will use the steamboat. Let its speed be x knots (equivalent to x nautical miles per hour), and let the distance going to the island be d nautical miles. Given that the time takes is 5 hours, this means that d = 5x.

4. If we let x be the speed of the boat you are taking to the island (the steamboat), then we know that the speed of the other boat (the tall ship) is 10 knots less than the steamboat's. So the speed of the tall ship (for the return trip) is (x - 10) knots.

5. Similar to part 3: we will multiply speed by time to determine the distance from the island. From part 4, we have determined that the speed of the tall ship to be used in returning is (x - 10) knots. Meanwhile, the given in the problem says that the tall ship will take 10 hours to make the trip. Therefore the distance will be equal to d = 10(x - 10) = 10x - 100 nautical miles.

6. We can assume that the distance travelled going to the island is the same distance travelled coming back. Therefore, we can equate the formula for distance from part 3 for the steamboat, to the distance from part 5 for the tall ship.
5x = 10x - 100

7. Solving for x: 5x = 10x - 100
-5x = -100
x = 20
Since x is the speed of the steamboat, x = 20 means that the steamboat's speed is 20 knots.

8. We determined in part 4 that the speed of the second boat (in our case, the tall ship) is (x - 10) knots. Since we have calculated in part 7 that the steamboat travels at x = 20 knots, then the speed of the tall ship is (x - 10) = 20 - 10 = 10 knots.</span>
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UkoKoshka [18]

Answer:

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Factor out the 4.

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Your answer is A. Megan is correct. When two lines have different slopes, they must intersect, producing one solution.

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