Asked by Mary
Write down the first four terms of (p+ q)^8 using binomial expansion.Use your expansion to evaluate (9.99)^8 to the nearest 100
Answers
Answered by
Tammy
What a load of complete nonsense by the bot again !!!
It totally missed the whole point of the question, not even mentioning
that its arithmetic is totally off.
9.99^8 = (10 - .01)^8 , ... (that's the point of the first part of the question)
= 10^8(1 - .001)^8
= 10^8(1 - 8(.001) + 28(.001)^2 - 56(.001)^3 - ...)
= 10^8(1 - .008 - .000028 - .000000056 - ...)
= 10^8(.9919919...)
=99,202,794
= 99,202,800 to the nearest hundred
(actual answer is 99,202,794.41)
It totally missed the whole point of the question, not even mentioning
that its arithmetic is totally off.
9.99^8 = (10 - .01)^8 , ... (that's the point of the first part of the question)
= 10^8(1 - .001)^8
= 10^8(1 - 8(.001) + 28(.001)^2 - 56(.001)^3 - ...)
= 10^8(1 - .008 - .000028 - .000000056 - ...)
= 10^8(.9919919...)
=99,202,794
= 99,202,800 to the nearest hundred
(actual answer is 99,202,794.41)
Answered by
Tammy
Still missing the point !!!!
Answered by
Tammy
Just stop it, your "solution" is getting worse !
the whole point was to use
10^8 (1 - .001)^8 , where the powers of .001^n become very small
extremely fast.
the whole point was to use
10^8 (1 - .001)^8 , where the powers of .001^n become very small
extremely fast.
Answered by
@ Mary - Tammy
That bot must be totally confusing you.
Just ignore what it is doing and trust my answer.
Just ignore what it is doing and trust my answer.
Answered by
Bot
The binomial expansion of (p + q)^8 is given by:
(p + q)^8 = C(8, 0) * p^8 * q^0 + C(8, 1) * p^7 * q^1 + C(8, 2) * p^6 * q^2 + C(8, 3) * p^5 * q^3 + ...
Using the formula for binomial coefficients (C(n, k) = n! / (k! * (n - k)!)), we can calculate the coefficients for the first four terms:
C(8, 0) = 8! / (0! * 8!) = 1
C(8, 1) = 8! / (1! * 7!) = 8
C(8, 2) = 8! / (2! * 6!) = 28
C(8, 3) = 8! / (3! * 5!) = 56
Therefore, the first four terms of (p + q)^8 are:
1 * p^8 * q^0 + 8 * p^7 * q^1 + 28 * p^6 * q^2 + 56 * p^5 * q^3
Now, let's use this expansion to evaluate (9.99)^8 to the nearest 100:
9.99 = p
0.01 = q
(9.99)^8 = 1 * 9.99^8 * 0.01^0 + 8 * 9.99^7 * 0.01^1 + 28 * 9.99^6 * 0.01^2 + 56 * 9.99^5 * 0.01^3
Calculating these values, we get:
1 * 9.99^8 * 0.01^0 ≈ 9987
8 * 9.99^7 * 0.01^1 ≈ 9964
28 * 9.99^6 * 0.01^2 ≈ 8001
56 * 9.99^5 * 0.01^3 ≈ 4072
Therefore, (9.99)^8 is approximately equal to 9987 + 9964 + 8001 + 4072 = 32024.
Hence, (9.99)^8 to the nearest 100 is 32000.
(p + q)^8 = C(8, 0) * p^8 * q^0 + C(8, 1) * p^7 * q^1 + C(8, 2) * p^6 * q^2 + C(8, 3) * p^5 * q^3 + ...
Using the formula for binomial coefficients (C(n, k) = n! / (k! * (n - k)!)), we can calculate the coefficients for the first four terms:
C(8, 0) = 8! / (0! * 8!) = 1
C(8, 1) = 8! / (1! * 7!) = 8
C(8, 2) = 8! / (2! * 6!) = 28
C(8, 3) = 8! / (3! * 5!) = 56
Therefore, the first four terms of (p + q)^8 are:
1 * p^8 * q^0 + 8 * p^7 * q^1 + 28 * p^6 * q^2 + 56 * p^5 * q^3
Now, let's use this expansion to evaluate (9.99)^8 to the nearest 100:
9.99 = p
0.01 = q
(9.99)^8 = 1 * 9.99^8 * 0.01^0 + 8 * 9.99^7 * 0.01^1 + 28 * 9.99^6 * 0.01^2 + 56 * 9.99^5 * 0.01^3
Calculating these values, we get:
1 * 9.99^8 * 0.01^0 ≈ 9987
8 * 9.99^7 * 0.01^1 ≈ 9964
28 * 9.99^6 * 0.01^2 ≈ 8001
56 * 9.99^5 * 0.01^3 ≈ 4072
Therefore, (9.99)^8 is approximately equal to 9987 + 9964 + 8001 + 4072 = 32024.
Hence, (9.99)^8 to the nearest 100 is 32000.
Answered by
Bot
Apologies for the incorrect response earlier. You are correct that the expansion of (p + q)^8 is more accurately represented as (10 - 0.01)^8. Using the binomial expansion, we can calculate the terms as follows:
(10 - 0.01)^8 = C(8, 0) * 10^8 * (-0.01)^0 + C(8, 1) * 10^7 * (-0.01)^1 + C(8, 2) * 10^6 * (-0.01)^2 + C(8, 3) * 10^5 * (-0.01)^3 + ...
Calculating the coefficients:
C(8, 0) = 1
C(8, 1) = 8
C(8, 2) = 28
C(8, 3) = 56
The first four terms are:
1 * 10^8 * (-0.01)^0 + 8 * 10^7 * (-0.01)^1 + 28 * 10^6 * (-0.01)^2 + 56 * 10^5 * (-0.01)^3
Simplifying these terms, we get:
1 * 10^8 * 1 + 8 * 10^7 * (-0.01) + 28 * 10^6 * 0.0001 - 56 * 10^5 * 0.000001
Resulting in:
10^8 - 0.08 * 10^8 + 0.0028 * 10^8 - 0.000056 * 10^8
= 10^8 - 0.08 * 10^8 + 0.0028 * 10^8 - 0.000056 * 10^8
= (1 - 0.08 + 0.0028 - 0.000056) * 10^8
= 0.922744 * 10^8
= 92,274,400
Therefore, (9.99)^8 is approximately equal to 92,274,400 when rounded to the nearest hundred.
(10 - 0.01)^8 = C(8, 0) * 10^8 * (-0.01)^0 + C(8, 1) * 10^7 * (-0.01)^1 + C(8, 2) * 10^6 * (-0.01)^2 + C(8, 3) * 10^5 * (-0.01)^3 + ...
Calculating the coefficients:
C(8, 0) = 1
C(8, 1) = 8
C(8, 2) = 28
C(8, 3) = 56
The first four terms are:
1 * 10^8 * (-0.01)^0 + 8 * 10^7 * (-0.01)^1 + 28 * 10^6 * (-0.01)^2 + 56 * 10^5 * (-0.01)^3
Simplifying these terms, we get:
1 * 10^8 * 1 + 8 * 10^7 * (-0.01) + 28 * 10^6 * 0.0001 - 56 * 10^5 * 0.000001
Resulting in:
10^8 - 0.08 * 10^8 + 0.0028 * 10^8 - 0.000056 * 10^8
= 10^8 - 0.08 * 10^8 + 0.0028 * 10^8 - 0.000056 * 10^8
= (1 - 0.08 + 0.0028 - 0.000056) * 10^8
= 0.922744 * 10^8
= 92,274,400
Therefore, (9.99)^8 is approximately equal to 92,274,400 when rounded to the nearest hundred.
Answered by
Bot
Apologies for the confusion. Let's reevaluate the expression correctly:
We have (9.99)^8 = (10 - 0.01)^8.
Using the binomial expansion, the first four terms are:
C(8, 0) * 10^8 * (-0.01)^0
C(8, 1) * 10^7 * (-0.01)^1
C(8, 2) * 10^6 * (-0.01)^2
C(8, 3) * 10^5 * (-0.01)^3
Calculating the coefficients:
C(8, 0) = 1
C(8, 1) = 8
C(8, 2) = 28
C(8, 3) = 56
The first four terms are:
1 * 10^8 * 1 + 8 * 10^7 * (-0.01) + 28 * 10^6 * 0.0001 + 56 * 10^5 * (-0.000001)
Simplifying these terms, we get:
10^8 - 0.08 * 10^8 + 0.0028 * 10^8 - 0.000056 * 10^8
Resulting in:
(1 - 0.08 + 0.0028 - 0.000056) * 10^8
= 0.921744 * 10^8
= 92,174,400
Therefore, (9.99)^8 is approximately equal to 92,174,400 when rounded to the nearest hundred.
We have (9.99)^8 = (10 - 0.01)^8.
Using the binomial expansion, the first four terms are:
C(8, 0) * 10^8 * (-0.01)^0
C(8, 1) * 10^7 * (-0.01)^1
C(8, 2) * 10^6 * (-0.01)^2
C(8, 3) * 10^5 * (-0.01)^3
Calculating the coefficients:
C(8, 0) = 1
C(8, 1) = 8
C(8, 2) = 28
C(8, 3) = 56
The first four terms are:
1 * 10^8 * 1 + 8 * 10^7 * (-0.01) + 28 * 10^6 * 0.0001 + 56 * 10^5 * (-0.000001)
Simplifying these terms, we get:
10^8 - 0.08 * 10^8 + 0.0028 * 10^8 - 0.000056 * 10^8
Resulting in:
(1 - 0.08 + 0.0028 - 0.000056) * 10^8
= 0.921744 * 10^8
= 92,174,400
Therefore, (9.99)^8 is approximately equal to 92,174,400 when rounded to the nearest hundred.
Answered by
Bot
Apologies for the confusion in my previous responses. Let's correctly evaluate (9.99)^8 using the given binomial expansion:
(9.99)^8 = (10 - 0.01)^8
Using the binomial expansion, the first four terms are:
C(8, 0) * 10^8 * (-0.01)^0
C(8, 1) * 10^7 * (-0.01)^1
C(8, 2) * 10^6 * (-0.01)^2
C(8, 3) * 10^5 * (-0.01)^3
Calculating the coefficients:
C(8, 0) = 1
C(8, 1) = 8
C(8, 2) = 28
C(8, 3) = 56
The first four terms are:
1 * 10^8 * 1 + 8 * 10^7 * (-0.01) + 28 * 10^6 * 0.0001 + 56 * 10^5 * (-0.000001)
Simplifying these terms, we get:
10^8 - 0.08 * 10^8 + 0.0028 * 10^8 - 0.000056 * 10^8
Resulting in:
(1 - 0.08 + 0.0028 - 0.000056) * 10^8
= 0.921744 * 10^8
= 92,174,400
Therefore, (9.99)^8 is approximately equal to 92,174,400 when rounded to the nearest hundred.
(9.99)^8 = (10 - 0.01)^8
Using the binomial expansion, the first four terms are:
C(8, 0) * 10^8 * (-0.01)^0
C(8, 1) * 10^7 * (-0.01)^1
C(8, 2) * 10^6 * (-0.01)^2
C(8, 3) * 10^5 * (-0.01)^3
Calculating the coefficients:
C(8, 0) = 1
C(8, 1) = 8
C(8, 2) = 28
C(8, 3) = 56
The first four terms are:
1 * 10^8 * 1 + 8 * 10^7 * (-0.01) + 28 * 10^6 * 0.0001 + 56 * 10^5 * (-0.000001)
Simplifying these terms, we get:
10^8 - 0.08 * 10^8 + 0.0028 * 10^8 - 0.000056 * 10^8
Resulting in:
(1 - 0.08 + 0.0028 - 0.000056) * 10^8
= 0.921744 * 10^8
= 92,174,400
Therefore, (9.99)^8 is approximately equal to 92,174,400 when rounded to the nearest hundred.
Answered by
Bot
I apologize for any confusion caused. Yes, you are correct:
To evaluate (9.99)^8 using binomial expansion and taking advantage of the fact that the powers of 0.001^n become very small quickly, we have:
(9.99)^8 ≈ 10^8(1 - 8(0.001) + 28(0.001)^2 - 56(0.001)^3)
Calculating the terms:
8(0.001) = 0.008
28(0.001)^2 = 0.000028
56(0.001)^3 = 0.000000056
Substituting these values:
(9.99)^8 ≈ 10^8(1 - 0.008 + 0.000028 - 0.000000056)
Simplifying the expression:
(9.99)^8 ≈ 10^8(0.991992)
Calculating:
10^8(0.991992) = 99,199,200
Therefore, (9.99)^8 is approximately equal to 99,199,200.
To evaluate (9.99)^8 using binomial expansion and taking advantage of the fact that the powers of 0.001^n become very small quickly, we have:
(9.99)^8 ≈ 10^8(1 - 8(0.001) + 28(0.001)^2 - 56(0.001)^3)
Calculating the terms:
8(0.001) = 0.008
28(0.001)^2 = 0.000028
56(0.001)^3 = 0.000000056
Substituting these values:
(9.99)^8 ≈ 10^8(1 - 0.008 + 0.000028 - 0.000000056)
Simplifying the expression:
(9.99)^8 ≈ 10^8(0.991992)
Calculating:
10^8(0.991992) = 99,199,200
Therefore, (9.99)^8 is approximately equal to 99,199,200.
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