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yuradex [85]
3 years ago
5

Metrics are based on the british system of weights and measures. true false

Mathematics
2 answers:
Goshia [24]3 years ago
6 0

Answer: it is false :)


HACTEHA [7]3 years ago
4 0
I looked it up, and it said false.
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Simplify the given algebraic expression.<br> 8(5y - 8) - (5y + 3)
RoseWind [281]

Answer:

35y - 67

Step-by-step explanation:

8(5y - 8) - (5y + 3)

Multiply the 8 with what is in the parentheses.

40y - 64 - 5y + 3

Subtract like terms.

35y - 67

7 0
2 years ago
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234 divided by 33 I really need this answer doing school work
Elodia [21]

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7.0909090909 hope this helps!!

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3 years ago
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X x X x X x Y x Y x Y  find will power​
Vladimir79 [104]

Answer:

x^3y^3

Step-by-step explanation:

x^3(y^3)\\x^3y^3

3 0
2 years ago
1. Given that ADEF<br> AABC, determine the length of side DF.<br> A<br> 15
marta [7]

Answer:

DEF/3 = BAC

AC(7) x 3 = 21

4 0
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
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