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Otrada [13]
2 years ago
11

A cross country racer records a time of exactly 4 minutes to complete a race that is a distance of 1,000 meters. What is the rac

er's speed, in meters/second, during the race?
Mathematics
1 answer:
lions [1.4K]2 years ago
6 0

Answer:

.24 meters/second

Step-by-step explanation:

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A hot air balloon is dropping from a height above the ground at a steady rate of 8 feet per second. It starts from a height of 7
pickupchik [31]
I believe the equation would be 765 -8t = 421. Then you would solve for t and get 43 seconds. I hope this helps!! :)
3 0
2 years ago
Solve the following equation for y<br> 1/2x+y=24<br> what is y?
Basile [38]

Answer:

y=-1/2x+24

Step-by-step explanation:

you solve for y by isolating the y on one side of the equation and the rest of the equation on the opposite side of the equal sign.

1/2x-1/2x+y=24-1/2x

y=24-1/2x

rewritten: y=-1/2x+24

5 0
3 years ago
Find the slope of the line that passes through the following points: (5, 3) and (-2, -4)
Aleksandr [31]

Answer:

1

Step-by-step explanation:

To find the slope, you must subtract the second y axis from the first y axis then the second x axis from the first x axis.

Problem: (5, 3) (-2, -4)

Solution: Step 1.   - 4 - 3 = 7  (subtract the second y axis from the first y axis)

               Step 2.  - 2 - 5 = 7  (subtract the second x axis from the first x axis)

Answer: 7/7

Final Answer: 1

Mark me as brainliest!!

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
I need help with this ​
zaharov [31]
Yes it’s a rectangle because the diagonals bisect each other, and they are equal in length.
7 0
2 years ago
Read 2 more answers
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