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Karo-lina-s [1.5K]
3 years ago
8

From start to finish, a machine can fill a bottle of water, seal it, label it, and pack it in 8.6 seconds. About how many bottle

s of water are packed in 480 seconds?
Mathematics
2 answers:
levacccp [35]3 years ago
7 0
Or take about 55 bottles
sdas [7]3 years ago
7 0

Answer:

The answer is 50

Step-by-step explanation: because if you divide 480 by 8.6  it would be 55 and 55 is closer to 50

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$528 was raised by a team of 24 students. If it's divided evenly among them, how much money will each student earn?
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PLEASE HELP ASAP TENTH GRADE GEOMETRY
wel

Answer: SU = 4(1) + 1 = 5

Step-by-step explanation:

Since T is on segment SU we know the whole is eqaul to the sum of it’s parts.

ST + TU = SU substitute

3x - 1 + 3x = 4x + 1 simplify

6x - 1 = 4x + 1 solve for x

2x = 2

x = 1

ST = 3(1) -1 = 2

TU = 3(1) = 3

SU = 4(1) + 1 = 5

5 0
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The explicit formula for an arithmetic sequence is an = 20 + (n − 1)(2). What is the 200th term?
Andrews [41]

Answer:

a_{200}=418

Step-by-step explanation:

The explicit formula for an arithmetic sequence is given by :

a_n=20+(n-1)2 ...(1)

We need to find the 200th term of the sequence.

Put n = 200 in equation (1)

a_{200}=20+(200-1)2\\\\=20+199\times 2\\\\=418

So, the 200th term of the sequence is 418.

7 0
3 years ago
A supplier sells 2 1/4 pounds of mulch for every 1 1/3 pounds of gravel. The supplier sells 172 pounds of mulch and gravel combi
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Read 2 more answers
Average Temperatures Suppose the temperature (degrees F) in a river at a point x meters downstream from a factory that is discha
mr Goodwill [35]

Answer:

Step-by-step explanation:

Average Temperatures Suppose the temperature (degrees F) in a river at a point x meters downstream from a factory that is discharging hot water into the river is given by

T(x) = 160-0.05x^2

a. [0, 10]

For x = 0

T(0) = 160 - 0.05 × 0^2

T(0) = 160

For x = 10

T(10) = 160 - 0.05 × 10^2

T(10) = 160 - 5 = 155

The average temperature

= (160 + 155)/2 = 157.5

b. [10, 40]

For x = 10

T(10) = 160 - 0.05 × 10^2

T(10) = 160 - 5 = 155

For x = 40

T(10) = 160 - 0.05 × 40^2

T(10) = 160 - 80 = 80

The average temperature

= (80 + 155)/2 = 117.5

c. [0, 40]

For x = 0

T(0) = 160 - 0.05 × 0^2

T(0) = 160

For x = 40

T(10) = 160 - 0.05 × 40^2

T(10) = 160 - 80 = 80

The average temperature

= (160 + 80)/2 = 120

6 0
3 years ago
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