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zhuklara [117]
3 years ago
10

Simplify (-8)(-2) + 11 027 0 -1 0-5

Mathematics
1 answer:
Schach [20]3 years ago
4 0
Answer:-5
-8x-2=-16
-16+11=-5
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Step-by-step explanation:

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What is the total area of the velocity sails?
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3 years ago
Is 7, 20, 25 a right triangle?
love history [14]

Answer:

The answer to your question is below

Step-by-step explanation:

Process

To answer if they are right triangles or not use the Pythagorean theorem which states that the square of the hypotenuse equals the sum of the square of the legs.

a)    25² = 20² + 7²

      625 = 400 + 49

     625 ≠ 449              It is not a right triangle

b)   26² = 24² + 10²

      676 = 576 + 100

      676 = 676               it is a right triangle

c)   13² = 8² + (√10)²

     169 = 64 + 10

      169 ≠ 74                 it is not a right triangle

d)   52² = 48² + 20²

      2704 = 2304 + 400

      2704 = 2704               it is a right triangle

3 0
3 years ago
Evaluate to 212 base 3 - 121 base 3 + 222 base 3​
aksik [14]

Answer:

212 bp^3 (base pairs cubed) - 121 bp^3 (base pairs cubed) + 222 bp^3 (base pairs cubed) = 313 bp^3 (base pairs cubed

Step-by-step explanation:

5 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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