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pentagon [3]
3 years ago
13

For what value of c does the following system have no solution? 1/2x+1/5y=2 5x+2y=c

Mathematics
2 answers:
USPshnik [31]3 years ago
4 0

Answer:

c(=/)20

Step-by-step explanation:

c doesn't equal 20. Put the equal sign with dash through it.

Minchanka [31]3 years ago
3 0

Answer:

Step-by-step explanation:

\frac{1}{2}x+\frac{1}{5}y = 2 \mid 5x + 2y = c

First let's multiply the 1st equation by 10.

5x + 2y = 20 \mid 5x + 2y = c

So, we can see that the equations have the same coefficients and that implies  they are equal.

So the equation has no solutions for. c \in R \setminus{20}

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Tamara says she used the GCF to factor the expression 21x + 56xy as 7(3x + 8xy). Is she correct? Explain.
cestrela7 [59]

Answer:

Tamara  incorrectly factored the whole expression.

Step-by-step explanation:

Note that

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Mark in bold all common factors, then GCF(21x,56xy)=7·x=7x.

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3 years ago
During a scuba dive, Lainey descended to a point 20 feet below the ocean surface.She countined her descent at a rate of 20 feet
lorasvet [3.4K]

Answer:

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Let x be the number of minutes she can continue, since she's already 20 feet below the ocean surface and is diving at the rate of 20 ft/minute, her depth with respect to x would be

20 + 20x

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4 0
3 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.

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