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tensa zangetsu [6.8K]
3 years ago
6

3 metal cubes that sit next to each other have sides of 5 yards are next to each other. How much total stuff can fit into the cu

bes?
Mathematics
1 answer:
kap26 [50]3 years ago
5 0

Answer:

375yd^{2}

Step-by-step explanation:

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How do you substitute 4(x) into the inequality 1/2x+5y-10<0 to find the possible value of y?
Veseljchak [2.6K]
You put 4 in the X
1/2(4)+5y-10<0
2+5y-10<0
-8+5y<0
5Y<8
Y<8/5
3 0
3 years ago
Find the smallest relation containing the relation {(1, 2), (1, 4), (3, 3), (4, 1)} that is:
professor190 [17]

Answer:

Remember, if B is a set, R is a relation in B and a is related with b (aRb or (a,b))

1. R is reflexive if for each element a∈B, aRa.

2. R is symmetric if satisfies that if aRb then bRa.

3. R is transitive if satisfies that if aRb and bRc then aRc.

Then, our set B is \{1,2,3,4\}.

a) We need to find a relation R reflexive and transitive that contain the relation R1=\{(1, 2), (1, 4), (3, 3), (4, 1)\}

Then, we need:

1. That 1R1, 2R2, 3R3, 4R4 to the relation be reflexive and,

2. Observe that

  • 1R4 and 4R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 4R1 and 1R2, then 4 must be related with 2.

Therefore \{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(4,1),(4,2)\} is the smallest relation containing the relation R1.

b) We need a new relation symmetric and transitive, then

  • since 1R2, then 2 must be related with 1.
  • since 1R4, 4 must be related with 1.

and the analysis for be transitive is the same that we did in a).

Observe that

  • 1R2 and 2R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 2R1 and 1R4, then 2 must be related with 4.
  • 4R1 and 1R2, then 4 must be related with 2.
  • 2R4 and 4R2, then 2 must be related with itself

Therefore, the smallest relation containing R1 that is symmetric and transitive is

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

c) We need a new relation reflexive, symmetric and transitive containing R1.

For be reflexive

  • 1 must be related with 1,
  • 2 must be related with 2,
  • 3 must be related with 3,
  • 4 must be related with 4

For be symmetric

  • since 1R2, 2 must be related with 1,
  • since 1R4, 4 must be related with 1.

For be transitive

  • Since 4R1 and 1R2, 4 must be related with 2,
  • since 2R1 and 1R4, 2 must be related with 4.

Then, the smallest relation reflexive, symmetric and transitive containing R1 is

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

5 0
2 years ago
Those are the right answer or i don’t think so i really don’t know i. really need help
Whitepunk [10]

Answer:

Triangle is isosceles triangle      X = 9

Step-by-step explanation:

5 0
3 years ago
The length of a rectangle is "a" inches and the width is 3 inches more than its length. find the perimeter and the area
Alex73 [517]

Answer:

Step-by-step explanation:

Givens

Length = a

Width = a + 3

Formulas

P = 2L + 2w

Area = L * w

Solution

Perimeter

P = 2L +  2w

P = 2*a + 2(a + 3)             Remove the Brackets

p = 2a + 2a + 6                Combine

<u><em>P = 4a + 6</em></u>

Area

Area = L * w

Area = a (a + 3)

<u><em>Area = a^2 + 3a</em></u>

7 0
2 years ago
I need to write an algebraic expression that includes 2 constatants, 2 variables, and 2 different operations. It also need it in
belka [17]

Answer:

An example would be

(x+3)-(y+12)

The two constants are 3 and 12 (I'm assuming you meant "constant" as in the number that stands by itself and not "constatant" lol)

The two variables are x and y

The two operations are addition and subtraction

In words you could say "the quantity of x plus three minus the quantity of y plus twelve"

3 0
3 years ago
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