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dangina [55]
2 years ago
14

Mrs.Gunn makes one costume in 3hours and 20 minutes. How long will it take her to make 6 costumes for a school play?

Mathematics
2 answers:
qaws [65]2 years ago
4 0

Answer:

20 hours

Step-by-step explanation:

Tpy6a [65]2 years ago
4 0

We can change the 3 hours and 20 minutes to "minutes" instead of hours:

1 hour is 60 minutes. So 3 hours is 60 * 3 = 180 minutes.

Add the 20 : 180 + 20 = 200 minutes for one costume.

Multiply by 6 for 6 costumes: 200 * 6 = 1200.

So 1200 minutes or 1200/60 = 20 hours to make 6 costumes

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I’m sorry i don’t know
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Photon Lighting Company determines that the supply and demand functions for its most popular lamp are as follows: S(p) = 400 - 4
BlackZzzverrR [31]
S(p) = 400 - 4p + 0.00002p^4
D(p) = 2800 - 0.0012p^3

S(p) = D(p)
400 - 4p + 0.00002p^4 = 2800 - 0.0012p^3
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20 points plz just help me. Explain how you would use fraction bars to find the quotient of StartFraction 10 Over 12 EndFraction
Katarina [22]

Answer:

5/4

Step-by-step explanation:

(10/12) / (4/6)

Reduce the fractions:

(5/6) / (2/3)

Multiply by the reciprocal:

(5/6) × (3/2)

15/12

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5/4

7 0
2 years ago
Read 2 more answers
You are designing a large rectangular fish tank for your tropical fish. The length will be 1 yard 3 inches, the width will be 1.
Keith_Richards [23]

One yard is 36 inches, 1.5 feet is 18 inches.  

So the answer is...

  (36+3)*18*20

= 39*18*28

= 702*28

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4 0
2 years ago
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A bucket that has a mass of 30 kg when filled with sand needs to be lifted to the top of a 30 meter tall building. You have a ro
Rom4ik [11]

Answer:

765 J

Step-by-step explanation:

We are given;

Mass of bucket = 30 kg

Mass of rope = 0.3 kg/m

height of building= 30 meter

Now,

work done lifting the bucket (sand and rope) to the building = work done in lifting the rope + work done in lifting the sand

Or W = W1 + W2

Work done in lifting the rope is given as,

W1 = Force x displacement

W1 = (30,0)∫(0.2x .dx)

Integrating, we have;

W1 = [0.2x²/2] at boundary of 30 and 0

W1 = 0.1(30²)

W1 = 90 J

work done in lifting the sand is given as;

W2 = (30,0)∫(F .dx)

F = mx + c

Where, c = 30 - 15 = 15

m = (30 - 15)/(30 - 0)

m = 15/30 = 0.5

So,

F = 0.5x + 15

Thus,

W2 = (30,0)∫(0.5x + 15 .dx)

Integrating, we have;

W2 = (0.5x²/2) + 15x at boundary of 30 and 0

So,

W2 = (0.5 × 30²)/2) + 15(30)

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W2 = 675 J

Therefore,

work done lifting the bucket (sand and rope) to the top of the building,

W = 90 + 675

W = 765 J

5 0
2 years ago
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