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Phoenix [80]
2 years ago
11

XD PLZ HELP ME I DONT UNDER STAND PLZ IM BEGGING

Mathematics
2 answers:
Virty [35]2 years ago
6 0

Answer:

40 and 202.5 m³

Step-by-step explanation:

(7)

Since the figures are similar then the ratios of corresponding sides are equal, that is

\frac{PQ}{16} = \frac{30}{12} ( cross- multiply )

12 PQ = 480 ( divide both sides by 12 )

PQ = 40 in

-------------------------------------------------------------------

The volume (V) of a pyramid is calculated as

V = \frac{1}{3} Ah ( A is the area of the base and h the height ) , then

V = \frac{1}{3} × 9² × 7.5

   = \frac{1}{3} × 81 × 7.5

   = 27 × 7.5

   = 202.5 m³

Natali [406]2 years ago
5 0

Answer:

the volume for the second page is 607.5

Step-by-step explanation:

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Check whether the relation R on the set S = {1, 2, 3} is an equivalent
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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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