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Delvig [45]
3 years ago
10

18. The price of 1 box of popcorn and 1 drink together is The price of popcorn and 1 drink together is What is the cost of 1 dri

nk on 3 q

Mathematics
2 answers:
valkas [14]3 years ago
8 0
A box of popcorn = 8.35-5.10
                            = 3.25

a drink= 5.10-3.25
           =1.85
xz_007 [3.2K]3 years ago
7 0
We know that one box of popcorn and one soda cost $5.10. We also know that two boxes of popcorn and one soda cost $8.35. To find how much a box of popcorn costs we can simply subtract $5.10 from $ 8.35. This will give us $3.25. Now we can subtract $3.25 from $5.10. This will give us $1.85. This means that a soda costs $1.85.
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Simplify the expression by combining like terms.<br> 7.2(x + 3) + 4(x - 3)
siniylev [52]

Answer: 11.2x -9.6

Step-by-step explanation:

7.2(x + 3) + 4(x - 3) -  12       - 7.2* (x + 3) + + 4*(x - 3) then

7.2x + 21.6 + 4x -  12  - Combining like terms 7.2x+ 4x  and numbers + 21.6-  12          

7.2x + 4x +21.6 -12       Then we have

11.2x -9.6

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3 years ago
An 80% increase followed by a 40% decrease
scZoUnD [109]
You start with 100%, add 80% for the increase and you get 180%, take away 40% for the decrease and you get 140%
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3 years ago
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Evgesh-ka [11]

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Step-by-step explanation:

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5 0
2 years ago
Colin has a pad with x pieces of paper on it. For his first class, he wrote on 5 fewer than half of the pieces of paper in the p
PIT_PIT [208]

Answer:

Colin has <em>8 sheets </em>left for his third class.

Step-by-step explanation:

Given that:

Total Number of pieces of papers = x

Number of pieces of papers used for 1st class = 5 fewer than half of the pieces in the pad

Writing the equation:

\text{Number of pieces of papers used for 1st class =} \dfrac{x}{2} -5 ...... (1)

Also, Given that number of pieces of papers used for the 2nd class are 2 more than that of papers used in the 1st class.

\text{Number of pieces of papers used for 2nd class =} \dfrac{x}{2} -5+2 = \dfrac{x}2 -3 ...... (2)

Now, number of pieces of papers left for the third class = Total number of pieces of papers in the pad - Number of pieces of papers used in the first class - Number of pieces of papers used in the first class

\text{number of pieces of papers left for the third class = }x-(\dfrac{x}{2}-5)-(\dfrac{x}{2}-3)\\\Rightarrow x-\dfrac{x}2-\dfrac{x}2+5+3\\\Rightarrow x-x+5+3\\\Rightarrow 8

So, the answer is:

Colin has <em>8</em> <em>sheets </em>left for his third class.

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3 years ago
Solve this problem plzzz
damaskus [11]
Here is how to solve it with long division and synthetic division.

7 0
3 years ago
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