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Sindrei [870]
2 years ago
5

Um I’m kinda confused and I need some help

Mathematics
1 answer:
mel-nik [20]2 years ago
8 0

Answer:

with what

Step-by-step explanation:

what do you need help with if there was photo we can't see it

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trasher [3.6K]
The answer would be standard deviation.
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3 years ago
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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.  

5 0
3 years ago
What is the inverse of y=2x squared -4
Bumek [7]
y=2x^2-4\ \ \ |+4\\\\2x^2=y+4\ \ \ |:2\\\\x^2=\dfrac{y+4}{2}\\\\x=\sqrt{\dfrac{y+4}{2}}

y^{-1}=\sqrt{\dfrac{x+4}{2}}=\dfrac{\sqrt{x+4}}{\sqrt2}=\dfrac{\sqrt{2(x+4)}}{2}=\dfrac{\sqrt{2x+8}}{2}
6 0
3 years ago
The solution is <br><br> u = ?
Tpy6a [65]

Answer:

u = (b/hn)

Step-by-step explanation:

The inverse!!

3 0
1 year ago
You use the Internet to purchase 9 tickets for a concert The tickets are $22 each you have to play a handling fee of $3 per tick
Naddik [55]

Answer: $206

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6 0
2 years ago
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