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Arada [10]
3 years ago
9

Plzzzzz answer need it plzzz I say

Mathematics
2 answers:
dexar [7]3 years ago
7 0

Answer:

360 square inches

Step-by-step explanation:

12*15 = 360

VMariaS [17]3 years ago
5 0

Answer:

360 square inches^^^^^^

Step-by-step explanation:

the person above is correct

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Debbie has at most $60 to spend on a clothes she wants to buy a pair of jeans for $22 and spend the rest on t-shirts each t-shir
patriot [66]

The inequality to find the maximum number of t-shirts is:

22+8x\leq 60

Step-by-step explanation:

Debbie has at most $60 to spend

Then it will be <=60

Now she has to spend 22 dollars on jeans so 22 will be add to the initial amount for sure.

And

Let x be the number of t-shirts

Then the cost of x t-shirts will be 8x

Putting this in to inequality will be:

22+8x\leq 60

The inequality to find the maximum number of t-shirts is:

22+8x\leq 60

Keywords: Inequality

Learn more about inequalities at:

  • brainly.com/question/763150
  • brainly.com/question/750742

#LearnwithBrainly

7 0
3 years ago
Which of the following expressions is not equivalent to the others?
Zanzabum

Answer:

D) - 24/ -3

Step-by-step explanation:

Because a negative over a negative gives me a positive, its different from the others which are all negative

4 0
3 years ago
Is the following relation a function? (
PolarNik [594]
No, because there are TWO values of y for a certain value of x

say x=5, then y=2 and y=-2

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4 0
3 years ago
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Irene tiene una colección de 50 dvd de películas de 90 minutos de duración cada una .Si el precio de cada uno era de 11€ , ¿cuan
Rainbow [258]

To solve this problem you must apply the proccedure shown below:

1. She has a total of 50 DVDs of 90 minutes each one of them. The cost of each DVD was 11€.

2. Therefore, to calculate the total cost of the collection, you must multiply the cost of each DVD by the total number of them:

total=(11)(50)=550€

3. To calculate the total minutes of the collection, you must multiply 90 minutes by the total number of DVDs:

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4 years ago
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3 years ago
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