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Kay [80]
3 years ago
12

If the endpoints of `bar(AB)` have the coordinates A(9, 8) and B(-1, -2), what is the midpoint of `bar(AB)`?

Mathematics
1 answer:
alina1380 [7]3 years ago
5 0

To solve this problem you must apply the proccedure shown below:

1. You have the following points given in the problem above:

A(9,8)\\ B(-1,-2)

2. To find the midpoint, you must apply the formula that is used for calculate it. This is:

m=(\frac{x1+x2}{2}, \frac{y1+y2}{2} )

Where:

x1=9\\ x2=-1\\ y1=8\\ y2=-2

3. Now, you must substitute these values into the formula for calculate the midpoint, as following:

m=(\frac{9+(-1)}{2} ,\frac{8+(-2)}{2} )\\ m=(4,3)

Therefore, the answer is: m=(4,3)

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over six games, a hockey team scored 18 goals. On average, how many goals did the team score each game
Ede4ka [16]

Answer:

3

Step-by-step explanation:

Average = total sum of all numbers/number of items in set

Average = 18/6

Average = 3

4 0
3 years ago
Read 2 more answers
Determine formula of the nth term 2, 6, 12 20 30,42​
nalin [4]

Check the forward differences of the sequence.

If \{a_n\} = \{2,6,12,20,30,42,\ldots\}, then let \{b_n\} be the sequence of first-order differences of \{a_n\}. That is, for n ≥ 1,

b_n = a_{n+1} - a_n

so that \{b_n\} = \{4, 6, 8, 10, 12, \ldots\}.

Let \{c_n\} be the sequence of differences of \{b_n\},

c_n = b_{n+1} - b_n

and we see that this is a constant sequence, \{c_n\} = \{2, 2, 2, 2, \ldots\}. In other words, \{b_n\} is an arithmetic sequence with common difference between terms of 2. That is,

2 = b_{n+1} - b_n \implies b_{n+1} = b_n + 2

and we can solve for b_n in terms of b_1=4:

b_{n+1} = b_n + 2

b_{n+1} = (b_{n-1}+2) + 2 = b_{n-1} + 2\times2

b_{n+1} = (b_{n-2}+2) + 2\times2 = b_{n-2} + 3\times2

and so on down to

b_{n+1} = b_1 + 2n \implies b_{n+1} = 2n + 4 \implies b_n = 2(n-1)+4 = 2(n + 1)

We solve for a_n in the same way.

2(n+1) = a_{n+1} - a_n \implies a_{n+1} = a_n + 2(n + 1)

Then

a_{n+1} = (a_{n-1} + 2n) + 2(n+1) \\ ~~~~~~~= a_{n-1} + 2 ((n+1) + n)

a_{n+1} = (a_{n-2} + 2(n-1)) + 2((n+1)+n) \\ ~~~~~~~ = a_{n-2} + 2 ((n+1) + n + (n-1))

a_{n+1} = (a_{n-3} + 2(n-2)) + 2((n+1)+n+(n-1)) \\ ~~~~~~~= a_{n-3} + 2 ((n+1) + n + (n-1) + (n-2))

and so on down to

a_{n+1} = a_1 + 2 \displaystyle \sum_{k=2}^{n+1} k = 2 + 2 \times \frac{n(n+3)}2

\implies a_{n+1} = n^2 + 3n + 2 \implies \boxed{a_n = n^2 + n}

6 0
2 years ago
Subtraction (must show work) 1. d - 27 = 45 2. h - 114 = 28
Arturiano [62]

Answer:

Step-by-step explanation:

1.

d - 27 = 45

add 27

d = 72

2.

h - 114 = 28

add 114

h = 142

3.

-4 + x = 15

add 4

x = 19

4.

-39 + g = 72

add 39

g = 111

7 0
2 years ago
Lyra and Donna are testing the two-way radios they built for their high school science project. Lyra goes to the top of a buildi
scZoUnD [109]
Using <span>Pythagorean theorem: c^2 = a^2 + b^2, and if a = 22 and b = 50
c^2=484+2500
c=sqrt 2984
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8 0
3 years ago
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The real number square root of 16 belongs to which sets of numbers?
klasskru [66]

Answer: Square root of 16 is +4 or -4. Since -4 is not a natural number, the square root can be described as an integer.

Step-by-step explanation:

The square root of 16 is a rational number. The square root of 16 is 4, an integer

3 0
3 years ago
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