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3 years ago

If (2 − 3i) + (x + yi) = 6, what is x + yi? i really need help with these

Mathematics

2 answers:

ra1l [238]3 years ago

8 0

I hope this helps you

2-3i+x+yi =6+0.i

2+x-i(3-y)=6+0.i

2+X=6

X=4

i(3-y)=0.i

Y=3

x+y.i

4+3i

cupoosta [38]3 years ago

3 0

Answer:

x + yi = 4 + 3i .

Step-by-step explanation:

Given : If (2 − 3i) + (x + yi) = 6,

To find : x + yi.

Solution : We have given that (2 − 3i) + (x + yi) = 6

We can rewrite the equation as

(2 − 3i) + (x + yi) = 6 + 0i

On removing parenthesis

2 − 3i + x + yi = 6 + 0i

On removing by 2 both sides

− 3i + x + yi = 6 - 2 + 0i.

− 3i + x + yi = 4 + 0i.

On adding 3i both sides

x + yi = 4 + 3i .

Therefore, x + yi = 4 + 3i .

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Answer:

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3 years ago

Find the first five terms of the sequence. an=-n2+2n

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3 years ago

Read 2 more answers

G(t)=(t+1) ^2-20.25 what are the zeros of the function

Vinil7 [7]

Step-by-step explanation:

g(t) = (t + 1)² − 20.25

0 = (t + 1)² − 20.25

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3 years ago

2. Given a quadrilateral with vertices (−1, 3), (1, 5), (5, 1), and (3,−1):

zlopas [31]

<h2>Explanation:</h2>

In every rectangle, the two diagonals have the same length. If a quadrilateral's diagonals have the same length, that doesn't mean it has to be a rectangle, but if a parallelogram's diagonals have the same length, then it's definitely a rectangle.

So first of all, let's prove this is a parallelogram. The basic definition of a parallelogram is that it is a quadrilateral where both pairs of opposite sides are parallel.

So let's name the vertices as:

$A(-1,3) \\ \\ B(1,5) \\ \\ C(5,1) \\ \\ D(3,-1)$

First pair of opposite sides:

<u>Slope:</u>

$\text{For AB}: \\ \\ m=\frac{5-3}{1-(-1)}=1 \\ \\ \\ \text{For CD}: \\ \\ m=\frac{1-(-1)}{5-3}=1 \\ \\ \\ \text{So AB and CD are parallel}$

Second pair of opposite sides:

<u>Slope:</u>

$\text{For BC}: \\ \\ m=\frac{1-5}{5-1}=-1 \\ \\ \\ \text{For AD}: \\ \\ m=\frac{-1-3}{3-(-1)}=-1 \\ \\ \\ \text{So BC and AD are parallel}$

So in fact this is a parallelogram. The other thing we need to prove is that the diagonals measure the same. Using distance formula:

$d=\sqrt{(y_{2}-y_{1})^2+(x_{2}-x_{1})^2} \\ \\ \\ Diagonal \ BD: \\ \\ d=\sqrt{(5-(-1))^2+(1-3)^2}=2\sqrt{10} \\ \\ \\ Diagonal \ AC: \\ \\ d=\sqrt{(3-1)^2+(-5-1)^2}=2\sqrt{10} \\ \\ \\$

So the diagonals measure the same, therefore this is a rectangle.

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3 years ago