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Nana76 [90]
3 years ago
11

Sprawdz obliczenia i popraw błędy

Mathematics
1 answer:
shepuryov [24]3 years ago
3 0
<span>Yo se la respuesta. será c

Good luck and have a nice day!</span>
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So sorry for asking for so much help but I seriously don't understand what to do for this assignment
astraxan [27]

Answer:

I'll probably come back and do this later.

Step-by-step explanation:

<u>How to Round:</u>

If the number following the number you're going to round up is 5 or above, (5, 6, 7, 8, or 9), then round up. (E.X. 2<u>9</u> rounded up is 30. Since 9 is above 5, you will round up.

On the other hand, if the number following the number you're going to round is 4 or below, (4, 3, 2, or 1), you don't round up, instead you keep it the same, and if they're are decimals, you turn the following numbers to 0s. (E.X. 32.4536 rounded to the nearest hundredth (32.45<u>3</u>6) is 32.45. Since 3 is less than 4, you're not going to round up. The number in the hundredth's place stays the same (5), and the following rest turn to 0s. Leaving you with 32.45

Let me know if this helped. :)

6 0
2 years ago
Read 2 more answers
(PLEASE HELP ME)
Schach [20]
This right answer is D.
because -65 divided by -13 is 5
8 0
2 years ago
Read 2 more answers
Y
qaws [65]

Answer:

(-2, 0.5) first

Step-by-step explanation:

X (-4, 1) → X' (-4*0.5, 1*0.5)

X'(-2, .5)

5 0
2 years ago
Write the equation in slope intercept form of the line passing through the following points (0,5) and (-5, -1).
krek1111 [17]

Answer:

y = 6/5x

Step-by-step explanation:

1. y = \frac{-1-5 }{-5-0}

2. subtract the two number on the top of the fraction bar to get 6

3. do the same for the denominator to get 5

4. your final answer is \frac{6}{5}

7 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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