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4 years ago

What is the solution of the equation when solved over the complex numbers?

Mathematics

1 answer:

kumpel [21]4 years ago

3 0

Answer:

$x=+i\sqrt{20}\approx+4.472i$

$x=-i\sqrt{20}\approx-4.472i$

Step-by-step explanation:

Starting from $x^2+20=0$ , we do $x^2=-20$ , which means $x=\pm\sqrt{-20}$ , which can be written as $x=\pm\sqrt{(-1)(20)}$ , which gives us $x=\pm\sqrt{(-1)}\sqrt{(20)}$ , which we know is $x=\pm i \sqrt{(20)}$ , so our solutions are:

$x=+i\sqrt{(20)}$

$x=-i\sqrt{(20)}$

Which can be approximated to:

$x\approx+4.472i$

$x\approx-4.472i$

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Which of the following functions are homomorphisms?

Vikentia [17]

Part A:

Given $f:Z \rightarrow Z,$ defined by $f(x)=-x$

$f(x+y)=-(x+y)=-x-y \\ \\ f(x)+f(y)=-x+(-y)=-x-y$

but

$f(xy)=-xy \\ \\ f(x)\cdot f(y)=-x\cdot-y=xy$

Since, f(xy) ≠ f(x)f(y)

Therefore, the function is not a homomorphism.

Part B:

Given $f:Z_2 \rightarrow Z_2,$ defined by $f(x)=-x$

Note that in $Z_2$ , -1 = 1 and f(0) = 0 and f(1) = -1 = 1, so we can also use the formular $f(x)=x$

$f(x+y)=x+y \\ \\ f(x)+f(y)=x+y$

and

$f(xy)=xy \\ \\ f(x)\cdot f(y)=xy$

Therefore, the function is a homomorphism.

Part C:

Given $g:Q\rightarrow Q$ , defined by $g(x)= \frac{1}{x^2+1}$

$g(x+y)= \frac{1}{(x+y)^2+1} = \frac{1}{x^2+2xy+y^2+1} \\ \\ g(x)+g(y)= \frac{1}{x^2+1} + \frac{1}{y^2+1} = \frac{y^2+1+x^2+1}{(x^2+1)(y^2+1)} = \frac{x^2+y^2+2}{x^2y^2+x^2+y^2+1}$

Since, f(x+y) ≠ f(x) + f(y), therefore, the function is not a homomorphism.

Part D:

Given $h:R\rightarrow M(R)$ , defined by $h(a)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)$

$h(a+b)= \left(\begin{array}{cc}-(a+b)&0\\a+b&0\end{array}\right)= \left(\begin{array}{cc}-a-b&0\\a+b&0\end{array}\right) \\ \\ h(a)+h(b)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)+ \left(\begin{array}{cc}-b&0\\b&0\end{array}\right)=\left(\begin{array}{cc}-a-b&0\\a+b&0\end{array}\right)$

but

$h(ab)= \left(\begin{array}{cc}-ab&0\\ab&0\end{array}\right) \\ \\ h(a)\cdot h(b)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)\cdot \left(\begin{array}{cc}-b&0\\b&0\end{array}\right)= \left(\begin{array}{cc}ab&0\\-ab&0\end{array}\right)$

Since, h(ab) ≠ h(a)h(b), therefore, the funtion is not a homomorphism.

Part E:

Given $f:Z_{12}\rightarrow Z_4$ , defined by $\left([x_{12}]\right)=[x_4]$ , where $[u_n]$ denotes the lass of the integer $u$ in $Z_n$ .

Then, for any $[a_{12}],[b_{12}]\in Z_{12}$ , we have

$f\left([a_{12}]+[b_{12}]\right)=f\left([a+b]_{12}\right) \\ \\ =[a+b]_4=[a]_4+[b]_4=f\left([a]_{12}\right)+f\left([b]_{12}\right)$

and

$f\left([a_{12}][b_{12}]\right)=f\left([ab]_{12}\right) \\ \\ =[ab]_4=[a]_4[b]_4=f\left([a]_{12}\right)f\left([b]_{12}\right)$

Therefore, the function is a homomorphism.

7 0

4 years ago

Analyze the diagram below and complete the instructions to follow.<br><br> Find a, b and c

Feliz [49]

Answer/Step-by-step explanation:

✔️Using trigonometric ratio, find b:

Reference angle = 45

opposite = 8

Hypotenuse = b

Thus,

Sin(45) = 8/b

b = 8/sin(45)

b = 8/(1/√2) (sin 45 = 1/√2)

b = 8 × √2/1

b = 8√2

✔️Find a using trigonometric ratio:

Reference angle = 60°

Hypotenuse = b = 8√2

Adjacent = a

Therefore,

Cos(60) = a/8√2

8√2*cos(60) = a

8√2*½ = a

4√2 = a

a = 4√2

✔️Find c using Pythagorean Theorem:

c² = b² - a²

c² = (8√2)² - (4√2)²

c² = 128 - 32

c² = 96

c = √96

c = √(16*6)

c = 4√6

5 0

3 years ago

Since no one knows what type of rock my rock was here's some nice juicy points

Naddik [55]

Answer:

thank you

Step-by-step explanation:

3 0

3 years ago

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50 grams/centimeters= kilograms/meters

Lubov Fominskaja [6]

Answer:

50 grams to kilos is 0.05

50 centimeters to meters is 0.5

4 0

3 years ago

a boat is trying to cross a river that is 2 miles wide, by going straight across. The speed of the boat is 25 mph, but the curre

vodomira [7]

Answer: It will take the boat 0.11 hour to cross the river

From the original starting point, the boat end up 1.85 miles away from that point.

Step-by-step explanation: Please find the attached file for the solution

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3 years ago