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3 years ago

Find the fifth roots of 32(cos 280° + i sin 280°).

1 answer:

7 0

I) Let z = 32 (cos 280 degrees + i sin 280 degrees)
= 32 (cos (k* 360 + 280) + isin (k*360 + 280)), where k = integer
II) z^ (1/5) = (32 (cos (k*360 + 280) + i sin (k*360 + 280))) ^ (1/5)
then,
z^ (1/5) = 2(cos ((k*360 + 280)/5) + i sin((k*360 + 280)/5))
We apply De Moivre's theorem:
z^ (1/5) = 2(cos(72k + 56) + i sin (72k +56))
We can get the five roots by assigning k = 0, 1, 2, 3, and 4
When K = 0, 1st root = 2 + i sin (cos 56 + i sin 56)k = 1, 2nd root = 2 (cos 128 + i sin 128)k = 2, 3rd root = 2 (cos 200 + i sin 200)k = 3, 4th root = 2 (cos 272 + i sin 272)k = 4, 5th root = 2 (cos 344 + i sin 344)

The fifth root is 5th root = 2 (cos 344 + i sin 344)

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What is the value of x? x = 23 x = 35 x = 58 x = 93

Akimi4 [234]

we know that

m∠SQR+x= $180$ -------> by supplementary angles

m∠SQR= $180-x$

The sum of the internal angles of a triangle is equal to $180$ degrees

m∠SQR+m∠QSR+m∠SRQ= $180$

substitute the values in the formula

m∠SQR+ $35$ + $58$ = $180$

m∠SQR= $87$ degrees

<u>Find the value of x</u>

m∠SQR= $180-x$

x= $180$ -m∠SQR

x= $180-87$

x= $93$ degrees

therefore

<u>the answer is</u>

the value of x is $93$ degrees

5 0

3 years ago

Read 2 more answers

All vectors are in Rn. Check the true statements below:

Oduvanchick [21]

Answer:

A), B) and D) are true

Step-by-step explanation:

A) We can prove it as follows:

$Proy_{cv}y=\frac{(y\cdot cv)}{||cv||^2}cv=\frac{c(y\cdot v)}{c^2||v||^2}cv=\frac{(y\cdot v)}{||v||^2}v=Proy_{v}y$

B) When you compute the product Ax, the i-th component is the matrix of the i-th column of A with x, denote this by Ai x. Then, we have that $||Ax||=\sqrt{(A_1 x)^2+\cdots (A_n x)^2}$ . Now, the colums of A are orthonormal so we have that (Ai x)^2=x_i^2. Then $||Ax||=\sqrt{(x_1)^2+\cdots (x_n)^2}=||x||$ .

C) Consider $S=\{(0,2),(2,0)\}\subseteq \mathbb{R}^2$ . This set is orthogonal because $(0,2)\cdot(2,0)=0(2)+2(0)=0$ , but S is not orthonormal because the norm of (0,2) is 2≠1.

D) Let A be an orthogonal matrix in $\mathbb{R}^n$ . Then the columns of A form an orthonormal set. We have that $A^{-1}=A^t$ . To see this, note than the component $b_{ij}$ of the product $A^t A$ is the dot product of the i-th row of $A^t$ and the jth row of $A$ . But the i-th row of $A^t$ is equal to the i-th column of $A$ . If i≠j, this product is equal to 0 (orthogonality) and if i=j this product is equal to 1 (the columns are unit vectors), then $A^t A=I$

E) Consider S={e_1,0}. S is orthogonal but is not linearly independent, because 0∈S.

In fact, every orthogonal set in R^n without zero vectors is linearly independent. Take a orthogonal set $\{u_1,u_2\cdots u_p\}$ and suppose that there are coefficients a_i such that $a_1u_1+a_2u_2\cdots a_nu_n=0$ . For any i, take the dot product with u_i in both sides of the equation. All product are zero except u_i·u_i=||u_i||. Then $a_i||u_i||=0$ then $a_i=0$ .