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Volgvan
3 years ago
11

Find the value of i^9

Mathematics
1 answer:
Tcecarenko [31]3 years ago
6 0

Answer:

The answer is i (i^9)

Step-by-step explanation:

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Okay hi
goblinko [34]

Answer:

30

Step-by-step explanation:

you have 5 marbles and 6 cards just multiply

Hope This Helps!     Have A Nice Day!!

8 0
3 years ago
Read 2 more answers
Rewrite 5a-(a-3b) in standard form
slamgirl [31]
So, -(a-3b) becomes -1(a-3b). Then, this becomes 5a-1a+3b. There is +3b because there is a double negative. So, 5-1=4, so your answer is 4a+3b. Pls brainliest
3 0
3 years ago
Yeah, im clueless please help;-;
svetlana [45]

Answer:

Starting from the left,  top to bottom. first answer is 8, second is 21.3, third is 64, 204.8. Right side is essentially y=x to a power, the answers are from top to bottom, 8,27,64,125

Step-by-step explanation:

Essentially, its just asking you to divide the given y values by the x values.

7 0
3 years ago
Domain of f(x)=(1/4)^x
ANEK [815]

Answer: B

Step-by-step explanation:

The domain of a function is the set of x-values.

6 0
2 years ago
The amount of money in Nadia’s savings account is represented by the recursive formula an=1.025(an-1)+50 if a6=291.51 what are t
Elis [28]
Recursive formula is one way of solving an arithmetic sequence. It contains the initial term of a sequence and the implementing rule that serve as a pattern in finding the next terms. In the problem given, the 6th term is provided, therefore we can solve for the initial term in reverse. To make use of it, instead of multiplying 1.025, we should divide it after deducting 50 (which supposedly is added).
<span>
Therefore, we perform the given formula: A (n) = <span>1.025(an-1) + 50
</span></span>a(5) =1.025 (235.62) + 50 = 291.51

a(4) = 1.025 (181.09) + 50 = 235.62

a(3) = 1.025 (127.89) + 50 = 181.09

a(2) = 1.025 (75.99) + 50 = 127.89

a(1) = 1.025 (25.36) + 50 = 75.99

a(n) = 25.36

The terms before a(6) are indicated above, since a(6) is already given.

So, the correct answer is <span>A. $25.36, $75.99.</span>
3 0
3 years ago
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