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2 years ago

I need help on this pls

Mathematics

1 answer:

Varvara68 [4.7K]2 years ago

6 0

Answer:

Step-by-step explanation:

a). Given function is y = $\frac{1}{2}x^3-2x+1$

At x = -3,

y = $\frac{1}{2}(-3)^3-2(-3)+1$

y = -13.5 + 6 + 1

y = -6.5

For x = -2,

y = $\frac{1}{2}(-2)^3-2(-2)+1$

y = -4 + 4 + 1

y = 1

For x = 2

y = $\frac{1}{2}(2)^2-2(2)+1$

y = 4 - 4 + 1

y = 1

Therefore, table for the given function will be,

x -3 -2 -1 0 1 2

y -6.5 1 2.5 1 -0.5 1

b). From the table given in part (a) we find the points lying on the graph are (-2, 1), (-1, 2.5) and (2, 1).

These points justify the graph (C)

Therefore, graph (C) will be the answer.

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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 $a_i=0$ .