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Stels [109]
2 years ago
15

Find the quotient of -32/15 divided by -4/5

Mathematics
1 answer:
lana [24]2 years ago
8 0

Answer:

Step-by-step explanation:

-32/15/-4/5

-32/15*5/-4

-32*5/15*-4

-160/-60

8/3

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The midpoint of AB is M(2,-2). If the coordinates of A are (-2,1), what ate the coordinates of B?
s344n2d4d5 [400]

Step-by-step explanation:

To go from A to M you need to go 4 to the right and 3 down. If M is the middle you'll need to do the same to get from M to B. So (2+4,-2-3)=(6,-5)

5 0
3 years ago
A 35 foot wire is secured from the top of a flagpole to a stake in the ground. If the stake is 14 feet from the base of the flag
cupoosta [38]

the answer is 21ft because the total is 35feet so 35 - 14 = 21



6 0
3 years ago
Read 2 more answers
Ken went to the store for his mother. He spent $2.37 on groceries. In addition, his mother said he could spend $0.35 on candy. W
Sladkaya [172]

Ken spent $2.37 on groceries and $0.35 on candy, and paid for them with a 5 dollar bill.

To find how much money Ken spent, we add up how much he spent on groceries and candy

Total amount spent = 2.37+0.35= $2.72

Since he paid with a 5 dollar bill, we need to subtract it from how much he spent to find the change

Change = 5 - 2.72 = $2.28

Ken got $2.28 in change.

To fill the boxes, it would be

$

2.

2

8

3 0
3 years ago
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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
a student answers 90% of the questions on a math exam correctly. if he answers 27 questions correctly how many questions are on
kodGreya [7K]

Answer:

30

Step-by-step explanation:

6 0
3 years ago
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