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KIM [24]
4 years ago
9

A machine can wrap 110 candies per minute. At this production rate, how many candies can the machine wrap in 15 minutes

Mathematics
2 answers:
Gala2k [10]4 years ago
5 0

Answer:

1,650 candies

Step-by-step explanation:

rate: 110 candles per minute

time: 15 minutes

rate x time = work

110 x 15 = 1650

ExtremeBDS [4]4 years ago
3 0

Answer:

Therefore in 15 minutes the machine can wrap 1650 candies

Step-by-step explanation:

<em><u>A machine can wrap 110 candies per minute. At this production rate, how many candies can the machine wrap in 15 minutes</u></em>

We can solve this using proportion.

Let x be the  number of candies the machine can wrap in 15 minutes.

110 candies = 1 minute

     x             = 15 minutes

cross-multiply

x  = 15×110

x =1650 candies

Therefore in 15 minutes the machine can wrap 1650 candies

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Subtract the sum of -5/6 and -1 3/5 from the sum of 2 2/3 and -6 2/5
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- 1.3 is the number we get when we subtract the sum of -5/6 and -1 3/5 from the sum of 2 2/3 and -6 2/5. This can be obtained by finding sum separately and then subtracting them.

<h3>What is the required number:</h3>

Here in the question it is given that,

subtract the sum of -5/6 and -1 3/5 from the sum of 2 2/3 and -6 2/5

By separating them as two parts

  • sum of -5/6 and -1 3/5  
  • sum of 2 2/3 and -6 2/5

⇒ sum of -5/6 and -1 3/5  

- 5/6 + - 1 3/5 = - 5/6 + - 8/5  (∵ a b/c = (ac+b)/c(5+3)/5 = 8/5)

                      = (- 25 - 48)/30  (LCM = 30)

                      = - 73/30

⇒ sum of 2 2/3 and -6 2/5

2 2/3 + -6 2/5 = 8/3 + -32/5

                       = (40 - 96)/15  (LCM = 15)

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subtract the sum of -5/6 and -1 3/5 from the sum of 2 2/3 and -6 2/5

= (sum of 2 2/3 and -6 2/5) - (sum of -5/6 and -1 3/5)

= (- 56/15) - (- 73/30 )  

= - 56/15 + 73/30

= - 112/30 + 73/30   (LCM = 30)

= - 39/30

= - 13/10

= - 1.3

Hence - 1.3 is the number we get when we subtract the sum of -5/6 and -1 3/5 from the sum of 2 2/3 and -6 2/5.

Learn more about fractions here:

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Quadrilateral CDEF is a rhombus. What is DCG?
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Step-by-step explanation:

The diagonals of a rhombus are perpendicular bisectors of each other, and each bisects the angle at a vertex of the rhombus. So, ∠DCG is the complement of ∠CDG, which is congruent to ∠EDG. We believe the marking 28° is intended to apply to ∠EDG, which would mean that ...

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