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Inga [223]
2 years ago
7

The price of a new version of a computer game is 120% of the price of the original version. The original version cost $48. What

is the cost of the new version?
Mathematics
2 answers:
12345 [234]2 years ago
7 0
120÷100=1.2×48=57.6-48=9$and 6 cents is the correct answer
vesna_86 [32]2 years ago
3 0
It would $72 because you have to subtract the computer game and the original version cost. <span />
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Help asap!!!!!!!!!!!
svlad2 [7]

Answer:

The answer to your question is:

Step-by-step explanation:

12.-

                x² - 6x = 5

                x² - 6x + (3)² = 5 + (3)²

               x² - 6x + 9 = 5 + 9

               (x - 3)² = 14

13.-           x² - 6x = 12

               x² - 6x + (3)² = 12 + (3)²

               x² - 6x + 9 = 12 + 9

              (x - 3)² = 21

              x - 3 = ±√21

             x1 = √21 + 3                     x2 = -√21 + 3

               x1 = 7.58                           x2 = -1.58

14.-           2x² - 24x = - 20            Divide by 2

                x² - 12 x = -10

               x² - 12x + (6)² = - 10 + (6)²

               x² - 12x + 36 = - 10 + 36

              ( x - 6) ² = 26

               x - 6 = ±√26

x1 = √26 + 6                                 x2 = -√26 + 6

x1 = 11.1                                          x2 = 0.9

3 0
3 years ago
What is the intersection of the three sets: A = {1, 3, 5, 7, 9}, B = {1, 2, 3, 5, 7}, and C = {1, 2, 5, 8, 9}?
marysya [2.9K]

ITS A. 1 AND 5 BECAUSE 1 AND 5 ARE IN ALL 3 SETS

7 0
2 years ago
Exercise Plus charges a yearly fee of $75 plus $10 a month. Gym and Swim charges a yearly fee of $50 plus $15 a month. When , if
Natasha2012 [34]

Answer:

5 months will be the same cost at each gym.

Step-by-step explanation:

As

  • Exercise Plus charges a yearly fee of $75 plus $10 a month.
  • Gym and Swim charges a yearly fee of $50 plus $15 a month.

Let x by the number of months

75+10x=50+15x

10x=15x-25

-5x=-25

\frac{-5x}{-5}=\frac{-25}{-5}

x=5

Therefore, 5 months will be the same cost at each gym.

4 0
2 years ago
A glacier advances toward the sea 0.004 mile anually. In 2010 the front of the glacier is 6.9 miles from the mouth of the sea. H
olchik [2.2K]

You know where the glacier is now, and how far it moves in
one year.  The question is asking how close to the sea it will be
after many years.

Step-1 ... you have to find out how many years

Step-2 ... you have to figure out how far it moves in that many years

Step-3 ... you have to figure out where it is after it moves that far

The first time I worked this problem, I left out  the most important
step ... READ the problem carefully and make SURE you know
the real question.  The first time I worked the problem, I thought
I was done after Step-2. 

============================

Step-1:  How many years is it from 2010 to 2030 ?

               (2030  -  2010)  =  20 years .


Step-2:  How far will the glacier move in 20 years ? 

               It moves 0.004 mile in 1 year.

              In 20 years, it moves 0.004 mile 20 times

              0.004 x 20  =  0.08 mile


Step-3: How far will it be from the sea after all those years ?

              In 2010, when we started watching it, it was 6.9 miles
              from the sea.

              The glacier moves toward the sea.
               In 20 years, it will be  0.08 mile closer to the sea.
               How close will it be ?

               6.9 miles  -  0.08 mile  =  6.82 miles  (if it doesn't melt)

7 0
3 years ago
What is the rational number equivalent to 1.28? The 1.28 has a line above the 28
Svetach [21]
<span>the line over the 28 means the 28 repeats forever.
1.282828.... and so on
 let x be the rational number 1.28...
we can use this trick:
100*1.282828....= 128.282828...
(the decimal 28 part repeats) 100x = 128.28...
next: 100x - x = 128.282828... - 1.282828...
the .282828... part will be subtracted away
99x = 127
divide both sides by 99 to get
x= 127/99</span>
8 0
3 years ago
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