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Zina [86]
3 years ago
10

Chloe has 20 unit cubes.How many different rectangular prisms can she build with the cubes?

Mathematics
2 answers:
likoan [24]3 years ago
8 0
If she uses all the cubes, its is 12. 
1.  One that is 20x1x1 cubes (20 is length, 1 is width, and 1 is height)
2. One that is 1x20x1 cubes
3. One that is 1x1x20 cubes
4. One that is 4x5x1 cubes
5. One that is 5x1x4 cubes
6. One that is 1x4x5 cubes
7. One that is 10x1x2 cubes
8. One that is 1x2x10 cubes
9. <span>One that is 2x10x1 cubes
10. </span>One that is 2x2x5 cubes<span>
11. </span>One that is 5x2x2 cubes<span>
12. </span>One that is 2x5x2 <span>cubes</span><span>

</span>
Musya8 [376]3 years ago
4 0

Chloe has 20 unit cubes

No. of different rectangular prisms can she build with the cubes = ?

The number of combinations that are possible are as follows:

20 * 1 * 1

1 * 1 *20

1 * 20 * 1

4 * 5 * 1

5 * 4 *1

1 * 5 *4

1* 4 * 5

1 * 10 * 20

20 * 10 * 1

20 * 1 *10

5 * 2 * 2

2 * 5 * 2

These are the 12 possible ways  

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Which of these relations on the set {0, 1, 2, 3} are equivalence relations? If not, please give reasons why. (In other words, if
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Answer:

(1)Equivalence Relation

(2)Not Transitive, (0,3) is missing

(3)Equivalence Relation

(4)Not symmetric and Not Transitive, (2,1) is not in the set

Step-by-step explanation:

A set is said to be an equivalence relation if it satisfies the following conditions:

  • Reflexivity: If \forall x \in A, x \rightarrow x
  • Symmetry: \forall x,y \in A, $if x \rightarrow y,$ then y \rightarrow x
  • Transitivity: \forall x,y,z \in A, $if x \rightarrow y,$ and y \rightarrow z, $ then x \rightarrow z

(1) {(0,0), (1,1), (2,2), (3,3)}

(3) {(0,0), (1,1), (1,2), (2,1), (2,2), (3,3)}

The relations in 1 and 3 are Reflexive, Symmetric and Transitive. Therefore (1) and (3) are equivalence relation.

(2) {(0,0), (0,2), (2,0), (2,2), (2,3), (3,2), (3,3)}

In (2), (0,2) and (2,3) are in the set but (0,3) is not in the set.

Therefore, It is not transitive.

As a result, the set (2) is not an equivalence relation.

(4) {(0,0), (0,1), (0,2), (1,0), (1,1), (1,2), (2,0), (2,2), (3,3)}

(1,2) is in the set but (2,1) is not in the set, therefore it is not symmetric

Also, (2,0) and (0,1) is in the set, but (2,1) is not, rendering the condition for transitivity invalid.

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Step-by-step explanation:

Total cards that deck contains= 5+4+2+1=12 ( as mentioned in question)

So,

1. Probability of getting blue card is 1/12

2. Probability of getting a red card is 2/12=1/6

hope it helps

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postnew [5]

Answer:

x  ≤  4

Step-by-step explanation:

6x + 3 ≤ 27        subtract 3 both sides

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x ≤ 4                   There's your answer

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