01.01.2020

I need to know what the difference of this expression is.

(4 - 5i) - (12 + 11i)

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24.06.2023, solved by verified expert
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-10-16i

Step-by-step explanation:

Given

(4-5i)-(12+11i)

Expand parenthesis

2-5i-12-11i

Put like terms together

(2-12)-(5i+11i)

Combine like terms

-10-16i

Hope this helps!! :)

Please let me know if you have any questions

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Mathematics
Step-by-step answer
P Answered by PhD

-10-16i

Step-by-step explanation:

Given

(4-5i)-(12+11i)

Expand parenthesis

2-5i-12-11i

Put like terms together

(2-12)-(5i+11i)

Combine like terms

-10-16i

Hope this helps!! :)

Please let me know if you have any questions

Mathematics
Step-by-step answer
P Answered by PhD

Let a_n denote the given sequence. a_n has forward differences

{9 - 1, 36 - 9, 100 - 36, ...} = {8, 27, 64, ...} = {2^3, 3^3, 4^3, ...}

If we call the sequence of forward differences b_n, then for n\ge1,

b_n=(n+1)^3

b_n is defined in terms of a_n for all n\ge1 by

b_n=a_{n+1}-a_n

and so a_n is defined recursively by

a_n=\begin{cases}a_1=1\\a_{n+1}=a_n+(n+1)^3&\text{for }n\ge1\end{cases}

We can deduce a pattern for the general n-th term:

a_2=a_1+2^3

a_3=a_2+3^3=a_1+\displaystyle\sum_{i=1}^2(i+1)^3

a_4=a_3+4^3=a_1+\displaystyle\sum_{i=1}^3(i+1)^3

and so on, up to

a_n=a_1+\displaystyle\sum_{i=1}^{n-1}(i+1)^3

We can simplify the right hand side a bit, noticing that a_1=1=1^3 matches (i+1)^3 for i=0:

a_n=\displaystyle\sum_{i=0}^{n-1}(i+1)^3

and to simplify things a bit more, we shift the index of summation:

a_n=\displaystyle\sum_{i=1}^ni^3

You should know that the right side has a nice closed form (look up "Faulhaber's formula" if you don't):

a_n=\dfrac{n^2(n+1)^2}4

Mathematics
Step-by-step answer
P Answered by PhD

Let a_n denote the given sequence. a_n has forward differences

{9 - 1, 36 - 9, 100 - 36, ...} = {8, 27, 64, ...} = {2^3, 3^3, 4^3, ...}

If we call the sequence of forward differences b_n, then for n\ge1,

b_n=(n+1)^3

b_n is defined in terms of a_n for all n\ge1 by

b_n=a_{n+1}-a_n

and so a_n is defined recursively by

a_n=\begin{cases}a_1=1\\a_{n+1}=a_n+(n+1)^3&\text{for }n\ge1\end{cases}

We can deduce a pattern for the general n-th term:

a_2=a_1+2^3

a_3=a_2+3^3=a_1+\displaystyle\sum_{i=1}^2(i+1)^3

a_4=a_3+4^3=a_1+\displaystyle\sum_{i=1}^3(i+1)^3

and so on, up to

a_n=a_1+\displaystyle\sum_{i=1}^{n-1}(i+1)^3

We can simplify the right hand side a bit, noticing that a_1=1=1^3 matches (i+1)^3 for i=0:

a_n=\displaystyle\sum_{i=0}^{n-1}(i+1)^3

and to simplify things a bit more, we shift the index of summation:

a_n=\displaystyle\sum_{i=1}^ni^3

You should know that the right side has a nice closed form (look up "Faulhaber's formula" if you don't):

a_n=\dfrac{n^2(n+1)^2}4

Mathematics
Step-by-step answer
P Answered by Master
Algebra uses abstract symbols to represent concrete elements in the real world.Many great mathematicians were also philosophers.The same variable can represent two different unknowns in two different algebraic expressions.

Step-by-step explanation:

Mathematics
Step-by-step answer
P Answered by PhD

3x-1

Step-by-step explanation:

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