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antiseptic1488 [7]
3 years ago
14

Simplify: (3 + 4i) + (2 - 3i) - (5 - 6i)

Mathematics
2 answers:
Pachacha [2.7K]3 years ago
7 0
The answer is:  " 7 i " .
________________________________________________________
Explanation:
________________________________________________________
<span>(3 + 4i) + (2 - 3i) - (5 - 6i) =

</span>(3 + 4i) + 1(2 - 3i) - 1(5 - 6i) ;   ←  Note that the coefficient, "1" is implied; 
                    since any value, multiplied by "1" , results in that same value.
_____________________________________________________________
Note the "distributive property" of multiplication:
_____________________________________________________________
a(b + c) = ab + ac ;

a(b – c) = ab – ac .
_____________________________________________________________
We have:

(3 + 4i) + 1(2 - 3i) - 1(5 - 6i) ;

Start with:

 " + 1(2 - 3i) "  =  (1*2) - (1*3i)  = 2 - 3i ; 

Then, we examine:

 " - 1(5 - 6i) "  =  (-1*5) + (-1 * -6i) = -5 + (+6i) =  -5 + 6i ; 
______________________________________________________________
And we rewrite the entire expression:
______________________________________________________________
   3 + 4i + 2 - 3i - 5 + 6i  ; 

Combine the "like terms" ; as follows:

   3 + 2 - 5 = 5  - 5 = 0 ; 

   4i - 3i + 6i = 1 i  + 6i = 7 i ; 

     0 + 7i = 7i .
_______________________________________________________
The answer is:  " 7 i " .
_______________________________________________________ 
Sav [38]3 years ago
5 0
(3 + 2 - 5) and (4i - 3i + 6i)
= 0 + 7i
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An area is approximated to be 14 in 2 using a left-endpoint rectangle approximation method. A right- endpoint approximation of t
USPshnik [31]
The trapezoidal approximation will be the average of the left- and right-endpoint approximations.

Let's consider a simple example of estimating the value of a general definite integral,

\displaystyle\int_a^bf(x)\,\mathrm dx

Split up the interval [a,b] into n equal subintervals,

[x_0,x_1]\cup[x_1,x_2]\cup\cdots\cup[x_{n-2},x_{n-1}]\cup[x_{n-1},x_n]

where a=x_0 and b=x_n. Each subinterval has measure (width) \dfrac{a-b}n.

Now denote the left- and right-endpoint approximations by L and R, respectively. The left-endpoint approximation consists of rectangles whose heights are determined by the left-endpoints of each subinterval. These are \{x_0,x_1,\cdots,x_{n-1}\}. Meanwhile, the right-endpoint approximation involves rectangles with heights determined by the right endpoints, \{x_1,x_2,\cdots,x_n\}.

So, you have

L=\dfrac{b-a}n\left(f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1})\right)
R=\dfrac{b-a}n\left(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n)\right)

Now let T denote the trapezoidal approximation. The area of each trapezoidal subdivision is given by the product of each subinterval's width and the average of the heights given by the endpoints of each subinterval. That is,

T=\dfrac{b-a}n\left(\dfrac{f(x_0)+f(x_1)}2+\dfrac{f(x_1)+f(x_2)}2+\cdots+\dfrac{f(x_{n-2})+f(x_{n-1})}2+\dfrac{f(x_{n-1})+f(x_n)}2\right)

Factoring out \dfrac12 and regrouping the terms, you have

T=\dfrac{b-a}{2n}\left((f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1}))+(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n))\right)

which is equivalent to

T=\dfrac12\left(L+R)

and is the average of L and R.

So the trapezoidal approximation for your problem should be \dfrac{14+21}2=\dfrac{35}2=17.5\text{ in}^2
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