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Reil [10]
2 years ago
8

Please walk me through how to solve these problems. If you can only solve one that’s fine, I just don’t know where to start. (Th

is is geometry)

Mathematics
1 answer:
Wittaler [7]2 years ago
3 0

Answer:

5,-0.5 is the answer to the fist one

Step-by-step explanation:

they want you to give the mid<em>dle </em>point of the line segment when you have graphed the two end points,... find the coordinates of the point the would be in the middle,... Best of luck,... Chow

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The Voyager 2 space probe took 2 years to make its closest approach to Jupiter. How many months did it take Voyager 2 to reach J
Tom [10]
It would be 24 months 
7 0
3 years ago
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Plz help will make brainlyist
siniylev [52]

Answer:

A. 6

B. 143

C. The words "less than", "equal to", and "no more than"

Step-by-step explanation:

For a, we would want to first make an equation to represent the problem. It appears you've already figured that out though, so lets solve it!

Subtract 56 on both sides.

-6m=-36

Now divide -36 by -6 to isolate m.

m=6

It will take her 6 minutes.

For B, we need to make another equation.

2.95b<450-28

First, we can subtract 28 from the total, because she has already spent $28.

Now our equation is 2.95b<422

Lastly, just divide both sides by 2.95 to isolate b.

Our solution is 143. So she can buy 143 batteries.

Lastly, for 13, we can say that the words "less than", "equal to" and "no more than" can indicate an inequality in a real word problem.

Hope this helps!

3 0
3 years ago
Wich state is the farthest state in the southwest region​
stepladder [879]

Answer: Alaska

Step-by-step explanation: Alaska is the farthest state in the southwest region​.

6 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.

3 0
3 years ago
Im not sure about this im confused
Maru [420]

Answer:

Step-by-step explanation:

8 0
3 years ago
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