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STatiana [176]
2 years ago
12

HELP PLS I WILL GIVE BRAINLIEST. NO LINKS!!!

Mathematics
1 answer:
Ksivusya [100]2 years ago
8 0
College and university scholarships are a convenient way to pay for an undergraduate ... How is the money awarded and what can you spend it on? ... Find money to help you fund your education.
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Evaluate the expression,<br> if x = 9, y = 3, and z = 5.<br> z²(x - y)
EastWind [94]
The answer is 150.
(5)^2(9-3)
25(6)
=150
3 0
1 year ago
Please solve this problem.<br> 2x + 34 = 56
maksim [4K]

Answer:

x = 11

Step-by-step explanation:

2x + 34 = 56

2x = 56 - 34

2x = 22

x = 22/2

x = 11

8 0
3 years ago
Read 2 more answers
A+ -12= 45 6th grade
Dmitry_Shevchenko [17]
A = 57
45 + 12 = 57
57 - (-12) = 45
8 0
2 years ago
What is the 7.05 % of 1,051,50
Tasya [4]

Answer: 74130.75

Step-by-step explanation: plz mark brainliest

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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