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2 years ago

separate into real and imaginary parts and find the multiplicative inverse of √2+i/√2-i (only 2 is in the √

Mathematics

1 answer:

Archy [21]2 years ago

4 0

The multiplicative inverse for √2-i = 1/(√2+1) and The multiplicative inverse is √2+1 = 1/(√2-1)

<h3>Real and Imaginary Numbers/Complex Numbers</h3>

Given Data

First expression = √2+i
Second expression = √2-i

For√2+i

the real part is is √2 and the imaginary part is i

The multiplicative inverse is √2+1 = 1/(√2-1)

rationalising the denominator we have

= 1/√2-1 * √2-1/√2-1

= √2-1/(√2-1)*(√2-1)

= √2-1/(2-√2-√2+1)

= √2-1/(-2√2+3)

For√2-i

the real part is is √2 and the imaginary part is -i

The multiplicative inverse is √2-1 = 1/(√2+1)

Rationalising the denominator we have

= 1/√2+1 * √2+1/√2+1

= √2+1/(√2+1)*(√2+1)

= √2+1/(2+√2+√2+1)

= √2+1/(2√2+3)

Learn more about complex Numbers here

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Find two vectors in R2 with Euclidian Norm 1<br> whoseEuclidian inner product with (3,1) is zero.

alina1380 [7]

Answer:

$v_1=(\frac{1}{10},-\frac{3}{10})$

$v_2=(-\frac{1}{10},\frac{3}{10})$

Step-by-step explanation:

First we define two generic vectors in our $\mathbb{R}^2$ space:

$v_1 = (x_1,y_1)$
$v_2 = (x_2,y_2)$

By definition we know that Euclidean norm on an 2-dimensional Euclidean space $\mathbb{R}^2$ is:

$\left \| v \right \|= \sqrt{x^2+y^2}$

Also we know that the inner product in $\mathbb{R}^2$ space is defined as:

$v_1 \bullet v_2 = (x_1,y_1) \bullet(x_2,y_2)= x_1x_2+y_1y_2$

So as first condition we have that both two vectors have Euclidian Norm 1, that is:

$\left \| v_1 \right \|= \sqrt{x^2+y^2}=1$

and

$\left \| v_2 \right \|= \sqrt{x^2+y^2}=1$

As second condition we have that:

$v_1 \bullet (3,1) = (x_1,y_1) \bullet(3,1)= 3x_1+y_1=0$

$v_2 \bullet (3,1) = (x_2,y_2) \bullet(3,1)= 3x_2+y_2=0$

Which is the same:

$y_1=-3x_1\\y_2=-3x_2$

Replacing the second condition on the first condition we have:

$\sqrt{x_1^2+y_1^2}=1 \\\left | x_1^2+y_1^2 \right |=1 \\\left | x_1^2+(-3x_1)^2 \right |=1 \\\left | x_1^2+9x_1^2 \right |=1 \\\left | 10x_1^2 \right |=1 \\x_1^2= \frac{1}{10}$

Since $x_1^2= \frac{1}{10}$ we have two posible solutions, $x_1=\frac{1}{10}$ or $x_1=-\frac{1}{10}$ . If we choose $x_1=\frac{1}{10}$ , we can choose next the other solution for $x_2$ .

Remembering,

$y_1=-3x_1\\y_2=-3x_2$

The two vectors we are looking for are:

$v_1=(\frac{1}{10},-\frac{3}{10})\\v_2=(-\frac{1}{10},\frac{3}{10})$

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3 years ago

Classify a triangle with each set of side lengths as acute, right, or obtuse

Serga [27]

Answer: Acute- (2,3,3.5) Right-(2.5,6,6.5) obtuse- (7,8,12), (7,9,14)

Step-by-step explanation:

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3 years ago

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zvonat [6]

Answer:7.375?

Step-by-step explanation:

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3 years ago

*Use the Area formula

Artist 52 [7]

Answer:

5ft

Step-by-step explanation:

Given parameters:

Length of the pool = 25ft

Width of the pool = 15ft

Volume of water the pool can hold = 1875ft³ of water

Unknown:

Depth of the pool = ?

Solution:

The pool is in the shape of a cuboid.

The volume of a cuboid is given as;

Volume of a cuboid = L x width x depth

Insert the parameters and solve for the depth;

1875 = 25 x 15 x depth

depth = $\frac{1875}{25 x 15}$ = 5ft

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3 years ago

A baker is putting cupcakes on trays to cool. She put 2 cupcakes on the first tray, 4 cupcakes on the second tray, 8 cupcakes on

elena-14-01-66 [18.8K]

Answer:

64 cupcakes

Step-by-step explanation:

Tray 1 = 2 cupcakes

Tray 2 = 4 cupcakes

Tray 3 = 8 cupcakes

Tray 4 = 16 cupcakes

Tray 5 = 32 cupcakes

Tray 6 = ?

This is a geometric progression

First term, a = 2

Common ratio, r = 4/2

= 2

6th term = ar^n-1

= 2 × 2^6-1

= 2 × 2^5

= 2 × 32

= 64

The 6th tray will have 64 cupcakes

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3 years ago