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sdas [7]
3 years ago
5

Find the distance between the two points in simplest radical form. (-9,3) and (-3,-5)

Mathematics
1 answer:
pentagon [3]3 years ago
5 0

Answer:

wekl..................................

Step-by-step explanation:

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Kind of lost, can anyone please help me find the answer and explain? Id really appreciate it.
stich3 [128]

Answer:

B. 45

Step-by-step explanation:

do 12×12 which is 144 then Davide it by 2 because it's a triangle then do 9×3 which is 27 then subtract that from 72 and you get 45.

5 0
3 years ago
Read 2 more answers
What dose 5h-2+9h equal explain☺
DerKrebs [107]
Start out by combining LIKE TERMS 
5h-2+9h 
in this case 9h and 5h are like terms so you would add both of those together.
9h+5h=14h
-2 has no like terms so we can't do anything with it our final answer would be:
14h-2
8 0
3 years ago
Write and solve an equation to find the value of x and a missing angle in each triangle below. Use the number bank to help you c
jolli1 [7]

Answer:

1. x = 30

REC = 130

Step-by-step explanation:

2x + 70 = 4x + 10

60 = 2x

x = 30

you try the others

Good luck :D

5 0
2 years ago
If integral 1 to 5 f(x)dx =36/15 what is the value of integral from 5 to 1 f(x)dx?<br><br><br>​
fredd [130]

Answer:

  -36/15

Step-by-step explanation:

If the indefinite integral of f(x)dx is F(x), then your integral from 1 to 5 is ...

  F(5) -F(1) = 36/15

The integral in the reverse direction is ...

  F(1) -F(5) = -(F(5) -F(1)) = -36/15

6 0
3 years ago
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
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