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Gnoma [55]
3 years ago
6

PLEASE HELP!

Mathematics
2 answers:
Blababa [14]3 years ago
8 0
The lateral area is the area of the three triangular sides in white. Do not include the area of the triangular base in grey.

Looking at the net to the right, you see the base and height of each white triangle.

Back side: base = 6 in.; height = 5.2 in.
Right side: base = 3 in.; height = 5.8 in.
Left side: base = 6 in.; height = 5.2 in.

Area of triangle = bh/2

A(back) = (6 in.)(5.2 in.)/2 = 15.6 in.^2
A(right) = (3 in.)(5.8 in.)/2 = 8.7 in.^2
A(left) = (6 in.)(5.2 in.)/2 = 15.6 in.^2

Lateral area = 15.6 in.^2 + 8.7 in.^2 + 15.6 in.^2 =

Lateral area = 39.9 in.^2
slava [35]3 years ago
5 0
If the lateral surface is just the three side triangles: 39.9 in^2

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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)\}

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<span>We cannot deduce about the exact location of P between J and K. But we can conclude: segment JP + segment PK = line JK.
</span><span>JP + PK = JK.
</span><span>Substitute first each.
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Combine like terms.
13z + 20 = 17z - 4
Isolate the variable z.
4z = 24
z = 6
The value of the variable z is then 6 units.</span>
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