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larisa86 [58]
3 years ago
6

Given 1 = 123° and 8 = 7z + 11. What is the value of z, if lines a and b are parallel?

Mathematics
2 answers:
Murljashka [212]3 years ago
7 0

Answer:

z = 16

Step-by-step explanation:

Since a and b are parallel lines, then

∠ 1 and ∠ 8 are Alternate exterior angles and are congruent, thus

7z + 11 = 123 ( subtract 11 from both sides )

7z = 112 ( divide both sides by 7 )

z = 16

Serga [27]3 years ago
3 0

Answer:

z=16

Step-by-step explanation:

We know for sure that since A and B are parallel that 1, 4, 5, and 8 are equal.

So then the equation is simply 123=7z+11---> 7z=112----> z=16

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chubhunter [2.5K]
The number of Euros received against 600 dollars = 450
Then
For 1 dollar, the number of euros = (450/600) euros
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So the unit rate that describes the exchange rate is 0.75 euros per dollar. So the correct option among all the options given in the question is option "A". I hope the procedure for getting to the correct result is clear enough for you to understand. In future you can use this procedure for solving similar problems.


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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

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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}

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Although the information has been presented in a baffling way in this question, but I think I understood the expressions.

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