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UkoKoshka [18]
2 years ago
10

Help please how do I find x and y

Mathematics
1 answer:
Dafna1 [17]2 years ago
7 0

Answer:

Option 3

Step-by-step explanation:

tan33°=x/7

x=4.5

4.5^2+7^2=y^2

20.25+49=y^2

y=√69.25

y=8.3

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1000*0+321*15678*9(-4563)
Vikki [24]

Answer:

-206675344746

Step-by-step explanation:

1000*0+321*15678*9(-4563) = -206675344746

7 0
2 years ago
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The seventh grade students are having an end-of-year party in the school cafeteria. There are 754 students attending the party.
valina [46]

Answer:

Determine the number of tables by dividing 754 by 6. If there is a remainder, the answer will need to be rounded tenths or hundredths . There is a remainder, so 125.6 tables are needed.

Step-by-step explanation:

5 0
2 years ago
Mixed numbers from 0-2 with intervals of 1/3
prisoha [69]
0, 1/3, 2/3, 1, 1 1/3, 1 2/3, 2
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2 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
HELP PLEASE<br> Find the value of y. Round to<br> the nearest tenth.<br> у<br> 42°<br> 85°<br> 31
MaRussiya [10]
31
Explanation
Nothing
6 0
2 years ago
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