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liberstina [14]
2 years ago
15

Identify the horizontal asymptote of the graph of f(x) = 5^x

Mathematics
1 answer:
Anna [14]2 years ago
4 0
Answer: x=25


*explenation*
5^=5+5+5+5+5= 25
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Find the angle of the sector indicated. Find the correct answer to the problem.
svlad2 [7]

The correct answer is (C)!

Since the area of the circle is 400π, 150π/400π = 135/360.

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3 years ago
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If you worked for 6 days for a total of 67.8 hours. If I worked the same number of hours each day, how many hours did I work in
Hunter-Best [27]
Hi there!

To solve this problem, we need to divide 67.8 by 6 to find the amount of hours each day. 

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Hope this helps!
8 0
3 years ago
♡♡♡please help ♡♡♡<br>thank you ​
ElenaW [278]

Answer:

acute angle

Step-by-step explanation:

<h2><em><u>Finding the angle </u></em></h2>

an obtuse angle is an angle which is more than 90° but less than 180°

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a straight angle is 180°

an acute angle is less than 90°

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7 0
3 years ago
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2. Solve the equation: y- 17 = 6<br> a. y = 11<br> b. y = 23<br> C. y = 12<br> d. y = 33
WARRIOR [948]
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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

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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