Mathematics, 29.10.2019 04:31 des7900

Consider the expression 11^n.
a. calculate 11^n, where n = 0, 1, 2, 3, 4.
b. what pattern do you notice in the successive powers?
c. use the binomial theorem to demonstrate why this pattern arises.
d. use a calculator to find the value of 11^5. explain whether this value represents what would be expected based
on the pattern seen in lower powers of 11.

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Answer from: kimlyn58p0wyn0

A. 11^0=1, 11^1=11, 11^2=121, 11^3=1331, 11^4=16461

B. if n is an integer, the digits of the results are the same as the coefficient in the n-row of pascal´s triangle.

C. 11^5=161051 and, no, it does not match the pattern

Step-by-step explanation:

A. Used a calculator 11^0=1, 11^1=11, 11^2=121, 11^3=1331, 11^4=16461

B. Visual, if n is an integer, the digits of the results are the same as the coefficient in the n-row of pascal´s triangle.

C. it´s useful to rewrite 11 as 10+1, now we are able to use the binomial theorem:

$(10+1)^{n}=\sum_{i=0}^{n}{{n\choose{i}}(10^{i})(1^{n-i})$

let´s test this with n=2 and see what happens:

$(10+1)^{2}= {2\choose{0}}(1)+{2\choose{1}}(10) +{2\choose{2}}(100)$

What we observed, is that "10^i" places the combinatoric number in digits based in the value of i (i=0 first digit, i=1 second digit ans so on)

(10+1)^2=1+2(10)+1(100)

11^2=121

Notice the combinatoric numbers are less than 10, so they can be placed into those places as digits. This would work until the combinatoric numbers are greater or equal to 10.

To prove it we need the hypothesis that n≤4 (we assure the combinatoric numbers are less than 10), now we use the binomial theorem.

$(10+1)^{n}=\sum_{i=0}^{n}{{n\choose{i}}(10^{i})(1^{n-i})$

$(10+1)^{n}= {n\choose{0}}(1)+{n\choose{1}}(10)+{n\choose{2}}(100)+{n\choose{3}}(1000)+{n\choose{4}}(10000))$

Considering that the combinatoric numbers are less than 10 (they are single digits) and knowing the term "10^i" places the combinatoric number in digits (i=0 first digit, i=1 second digit ans so on), we can affirm:

$(10+1)^{n}= "{n\choose{4}}"{n\choose{3}}"{n\choose{2}}"{n\choose{1}}"{n\choose{0}}"$

with the combinatoric numbers as digits. And we proved what we were asked for

D. 11^5=161051, and it does not follow the pattern. That´s because ${5\choose{2}}=10$ and we have shown that it only work with the combinatoric numbers being single digits. Because 10 occupies 2 digits places and multiplying it by "10^i" will interfere with the other digits placements.

Answer from: Quest

tickets for the harlem globetrotters

show at michigan state university in

2010 cost $16, $23, or, for vip seats, $40.

total revenue = 16x + 23y + 40z

if 9 times as many $16 tickets were sold

as vip tickets,

x = 9z => x – 9z = 0, if 9 times as many $16 tickets were sold

as vip tickets,

x = 9z => x – 9z = 0, sales of all three kinds of tickets would

total $46,575.

16x + 23y + 40z = 46575, how many of each kind of ticket would

have been sold?

solve for x, y and z

x − 9z = 0

x − y − z = 55

16x + 23y + 40z = 46575

x = 1170

y = 985

z = 130

welcome!

Answer from: Quest

it is a. 4.2 because 7.6-3.4 gets the remaining length of 4.2. i might be wrong but i am pretty sure this is the answer.

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Consider the expression 11^n.
a. calculate 11^n, where n = 0, 1, 2, 3, 4.
b. what patter...

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