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tangare [24]
3 years ago
13

Consider the arithmetic sequence where the 12th term is 41 and the 4th term is 1. a. Find the formula of the nthterm of the sequ

ence. b. Find the sum of the first 20 terms.
Mathematics
1 answer:
svlad2 [7]3 years ago
3 0
The common difference is 6. The first term is x−d=37−6=31.
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What is 0.000025(800)^2 - 0.04(800) +40..?
Scilla [17]

Answer:

24

Step-by-step explanation:

6 0
2 years ago
What is 2700 divided 90​
crimeas [40]

Answer:

30

Step-by-step explanation:

2700/90 = 30

27/9 = 3

100/10 = 10

3*10 = 30

8 0
3 years ago
The product of two consecutive odd integers is 782. What are the integers?
Alex Ar [27]

Call the smaller of the two odds = n

Call the next number in the sequence = n + 2

n*(n +2) = 782           Remove the brackets.

n^2 + 2n = 782          Subract 782 from both sides.

n^2 + 2n - 782 = 0    We are going to have to factor this.

Discussion

This problem can't be done the way it is written. The product of an odd integer with another odd integer is and odd integer. There are no exceptions to this. So you need to give a number that has two factors very near it's square root for this question to work.

For example, you could use 783, (which factors) instead of 782 .

Solve

n^2 + 2n - 783 = 0

(n + 29)(x - 27) = 0

<u>Solution One</u>

n - 27 = 0

n = 27

The two odd consecutive integers are 27 and 29.

<u>Solution Two</u>

n + 29 = 0

n = - 29

The two solution integers are -29 and - 27 Notice that - 29 is smaller than - 27.

7 0
4 years ago
What are the possible numbers of positive, negative, and complex zeros of f(x) = −3x4 −
Anna007 [38]

Answer:

b.

Step-by-step explanation:

We have to look at sign changes in f(x) to determine the possible positive real roots.

f(x)=-3x^4-5x^3-x^2-8x+4

There is only one sign change here, between the -8x and the +4.  So that means there is only 1 possible real positive root.

Now we have to look at sign changes in f(-x) to determine the possible negative real roots.

f(-x)=-3x^4+5x^3-x^2+8x+4

There are 3 sign changes here.  That means there are either 3 negative roots or 3-2 = 1 negative root.  So we have:

1 positive

3 or 1 negative

We need to pair them up now with all the possible combinations.

If we have 1 positive and 1 negative, we have to have 2 imaginary

If we have 1 positive and 3 negative, we have to have 0 imaginary

Keep in mind that the total number or roots--positive, negative, imaginary--have to add up to equal the degree of the polynomial.  This is a 4th degree polynomial, so we will have 4 roots.

7 0
3 years ago
How many outcome sequences are possible when a die is rolled 4 times where we say for instance that the outcome is 3 4 3 1 if th
STatiana [176]

Answer:

the possible outcome sequences when a die is rolled 4 times is 1296

Step-by-step explanation:

 Given the data in the question;

a die is rolled 4 times

and outcomes are { 3, 4, 3, 1 }

we know that; possible number of outcomes on a die is n = 6{ 1,2,3,4,5,6 }

Now when we roll a die lets say, r times

then the total number of possible outcomes will be;

N = n^r

given that; r = 4

Hence if we roll a die 4 times;

Total number of possible outcome N = 6⁴

N = 1296

Therefore, the possible outcome sequences when a die is rolled 4 times is 1296

3 0
3 years ago
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