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3 years ago

Please quick ???!!!!!!

Mathematics

1 answer:

harina [27]3 years ago

3 0

The third one I think I’m sorry if I get this wrong

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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

4 years ago

For each graph below, state whether it represents a function.

valina [46]

Graph 1 and 3 are the only functions

6 0

3 years ago

sue stacked one box onto another. the bottom had the height of 2 1/3 feet and the top box had the height of 3 2/3 feet. how tall

Scilla [17]

Height of bottom box : - $2 \frac{1}{3} => 7/3$
Height of top box : - $3 \frac{2}{3} => 11/3$

Thus, height of two boxes =

$\frac{7}{3} + \frac{11}{3} = \frac{18}{3} => 6ft$

Happy to help chodini!

3 0

3 years ago

Read 2 more answers

In which direction must the graph of f(x) = |x| be shifted to produce the graph of g(x) = |x + 4|?

Tamiku [17]

Answer:

Left on the x-axis

Step-by-step explanation:

The opposite of the coefficient's amount of units is how much the graph will be moved.

Table comparisons

g(x)=|x+4|

x 1 -2 -4

y 5 2 0

Underlined is the x-int of the equation.

f(x)=|x|

x 1 0 -1 3

y 1 0 1 3

5 0

4 years ago

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Rewrite the equation: y-3 = -1/2(x + 2), in slope-intercept form.

exis [7]

Y equals 3x-3 is the answer

3 0

3 years ago