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svetoff [14.1K]
3 years ago
6

Simplify- 5+2{3x+7(2-x)}

Mathematics
1 answer:
oksano4ka [1.4K]3 years ago
4 0

Answer:

16x+19

Step-by-step explanation:

i might be wrong and its gonna take so long to explain it

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A circle has an 18 inch radius and a shaded sector with a central angle of 50 degrees. What is the area of the shaded sector
Katyanochek1 [597]
Answer: The area of the sector will be about 141.3 square inches.

First, we need to find the area of the entire circle. The formula is pi(r^2). Inputting 18 for the radius makes the area of the circle about 1017.36 square inches.

Now, we multiply by the area by the size of the sector. It is 50 degree out of the entire 360 degrees. That is about 13.9%.

Multiply 13.9% by 1017.36 to get an area of about 141.3 square inches for the sector.
5 0
3 years ago
Is 1.30 bigger than 1.3?
FromTheMoon [43]
It is the same value
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
Neil goes to the pet shop and
umka2103 [35]

Answer:

b (1/15)

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Write a polynomial equation of least degree with roots 2, -1 and 4.
Rasek [7]
A polynomial with roots a and b is (x - a)(x - b).

(x - 2)(x - (-1))(x - 4) = 0

(x - 2)(x + 1)(x - 4) = 0

has roots 2, -1, and 4.
3 0
3 years ago
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