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KiRa [710]
3 years ago
7

What happens to the opposite and adjacent sides when theta is moved from one acute angle to the other?

Mathematics
1 answer:
Nesterboy [21]3 years ago
8 0

Answer:

the opposite and adjacent sides should switch places if theta is moved. the hypotenuse will always stay the same.

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HELP PLEASE!!!!! PLEASE EXPLAIN WHEN GIVING THE ANSWER!!!! GIVING BRAINLIEST TO THE BEST EXPLANATION!!!!
atroni [7]

Answer:Simplificando

5 (2j + 3 + j) = 0

Reordene os termos:

5 (3 + 2j + j) = 0

Combine termos semelhantes: 2j + j = 3j

5 (3 + 3j) = 0

(3 * 5 + 3j * 5) = 0

(15 + 15j) = 0

Resolvendo

15 + 15j = 0

a verdadeira questão é por que o vizinho de Sarah quer dentes de tubarão

Step-by-step explanation:

Mova todos os termos contendo j para a esquerda, todos os outros termos para a direita.

Adicione '-15' a cada lado da equação.

15 + -15 + 15j = 0 + -15

Combine termos semelhantes: 15 + -15 = 0

0 + 15j = 0 + -15

15j = 0 + -15

Combine termos semelhantes: 0 + -15 = -15

15j = -15

Divida cada lado por '15'.

j = -1

Simplificando

j = -1

Tubarão ou cação é um tipo de peixe de esqueleto cartilaginoso e um corpo hidrodinâmico (com exceção dos Squatiniformes, Hexanchiformes e Orectolobiformes)

4 0
2 years ago
Read 2 more answers
Give the domain and range.
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Answer:

The correct answer is a.

Step-by-step explanation:

The domain is the plotted points on the x-axis.

The range is the plotted points on the y-axis.

(-3, 3), (0, 0), (2, -2)

Domain {-3, 0, 2}; Range: {3, 0, -2}

8 0
3 years ago
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Amélie runs a bakery. She wants to find out whether her cake sales are affected by the weather conditions. She recorded the dail
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Answer:

the coeffiectans is 100

Step-by-step explanation:

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3 years ago
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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}

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