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swat32
3 years ago
6

Which pair of angles are vertical angles

Mathematics
1 answer:
8_murik_8 [283]3 years ago
3 0

Answer:

angles J and N

Step-by-step explanation:

vertical angles are congruent angles that are vertically across from each other

angle j and angle n are both right angles and they are both vertically across from each other therefore A is your answer

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Which of the given points would be in a table generated by the equation below?
bixtya [17]
D is the answer. Putting this because it has to be 20 characters
7 0
3 years ago
What is the quotient of the following in simplest form? 5/9 divided by 2/3
Assoli18 [71]

Answer:

5/6

Step-by-step explanation:

You write the question as such: 5/9 divided by 2/3.

After that, you keep the first fraction, flip the sign and the fraction.

You will get: 5/9 multiplied by 3/2.

When you get this, you then cross multiply and simplify as such.

5/3 multiplied by 1/2.

5/6 would be your solution.

4 0
2 years ago
Let’s see who knows this short quiz
Zanzabum

Answer:

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5 0
3 years ago
Simplify.
kenny6666 [7]
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4 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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