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Diano4ka-milaya [45]
2 years ago
14

Can anyone solve this?

Mathematics
2 answers:
lesantik [10]2 years ago
7 0

Answer:

44,280

Step-by-step explanation:

I might know. I multiplied the 1st and got 307.5.

then multiplied the 2d. I got 144

multiplied 307.5x144 and I got 44,280

mina [271]2 years ago
3 0
44280
Hope this helps
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GrogVix [38]
96+24^6+9
105+24^6
=163.78775
8 0
4 years ago
Find the value of x (approximately),
tresset_1 [31]

Answer:. If the Area is 177, it should have been given in square centimeters.

If one side is 13 cm, the other side is about 13.6cm.

If one side is 11 cm, the other side should be about 16.1 cm. Even the choice is incorrect!

A/x = w

Step-by-step explanation: to find a side when the Area is given, divide the Area by the length or width given.

5 0
3 years ago
If you do 3.2x6.892 I don’t get the answer
Paul [167]

Answer:

<h2>22.0544</h2>

Step-by-step explanation:

3.2 × 6.892 = 22.0544

I'm always happy to help :)

4 0
4 years ago
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.

3 0
3 years ago
What are the values of x in the equation 4x2 + 4x - 3 = 0?
9966 [12]

Answer:

x=  0.5

or x=  -1.5

Step-by-step explanation:

Let's solve your equation step-by-step.

4x2+4x−3=0

Step 1: Factor left side of equation.

(2x−1)(2x+3)=0

Step 2: Set factors equal to 0.

2x−1=0 or 2x+3=0

x= 1/2 or x= −3/2

5 0
3 years ago
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