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hodyreva [135]
3 years ago
12

What is the Width, explain how you got your answer (8 points btw)

Mathematics
1 answer:
asambeis [7]3 years ago
3 0

Answer:

Based off of my calculations, I got 126

Step-by-step explanation:

The formula for area would be: A = L * W

So, when finding the width (W), you would need to divide the area (A) by the length (L).

Equation: W = A / L

Now you know the equation, all you need to do is plug in the numbers:

[84 divided by 2/3]

W = 84  /  2/3

After solving this, you would get: 126

So, W = 126.

To check if we got the correct answer, we could just multiply the length times the width:

126 * 2/3

Since this equation gives us 84 as the answer, 126 is the correct width.

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1. Points scored by a basketball player are 12, 15, 8, 12, 15, 10, 3, 14, and 15
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mean : add all of them and divide by how much they are

mode: is what repeats it self like 15 is the most repeated.

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How do you solve this problem: 5(x3)-(-20)
Nimfa-mama [501]
First step: multiply the 5(x3) and it will give you x15 or most commonly written as 15x. you should now have 15x-(-20).

Second step: since you cannot subtract -20 from 15x because it has a variable, the final answer is just 15x-(-20), or 15x+20.
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Check whether the relation R on the set S = {1, 2, 3} is an equivalent
kozerog [31]

Answer:

R isn't an equivalence relation. It is reflexive but neither symmetric nor transitive.

Step-by-step explanation:

Let S denote a set of elements. S \times S would denote the set of all ordered pairs of elements of S\!.

For example, with S = \lbrace 1,\, 2,\, 3 \rbrace, (3,\, 2) and (2,\, 3) are both members of S \times S. However, (3,\, 2) \ne (2,\, 3) because the pairs are ordered.

A relation R on S\! is a subset of S \times S. For any two elementsa,\, b \in S, a \sim b if and only if the ordered pair (a,\, b) is in R\!.

 

A relation R on set S is an equivalence relation if it satisfies the following:

  • Reflexivity: for any a \in S, the relation R needs to ensure that a \sim a (that is: (a,\, a) \in R.)
  • Symmetry: for any a,\, b \in S, a \sim b if and only if b \sim a. In other words, either both (a,\, b) and (b,\, a) are in R, or neither is in R\!.
  • Transitivity: for any a,\, b,\, c \in S, if a \sim b and b \sim c, then a \sim c. In other words, if (a,\, b) and (b,\, c) are both in R, then (a,\, c) also needs to be in R\!.

The relation R (on S = \lbrace 1,\, 2,\, 3 \rbrace) in this question is indeed reflexive. (1,\, 1), (2,\, 2), and (3,\, 3) (one pair for each element of S) are all elements of R\!.

R isn't symmetric. (2,\, 3) \in R but (3,\, 2) \not \in R (the pairs in \! R are all ordered.) In other words, 3 isn't equivalent to 2 under R\! even though 2 \sim 3.

Neither is R transitive. (3,\, 1) \in R and (1,\, 2) \in R. However, (3,\, 2) \not \in R. In other words, under relation R\!, 3 \sim 1 and 1 \sim 2 does not imply 3 \sim 2.

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Cans of corn are on sale at 10 for $4. Find the cost of 15 cans?
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