All categories

3 years ago

Any assistance would be great!

Mathematics

2 answers:

vladimir1956 [14]3 years ago

4 0

The domain is the input values, which would also be X values.

{ x |x= -5, -3, 1, 2,6}

frozen [14]3 years ago

4 0

For this case we have by definition that the domain of a function y = f (x) is the set of all the values that the variable x takes, for which the function is defined. They are also represented by the starting set.

It is observed in the figure, that the domain is:

${x | x = -3,2, -5,1,6}\\{x | x = -5, -3,1,2,6}$

Answer:

Option A

You might be interested in

I am being asked to calculate and plot residuals and I don't know how my x values are 13.8, 18, 16,7, 18, 0.7, 21.9, 9.2, 19.5,

Igoryamba

Answer:

(13.8,1.42) (18,3.7)(16.7,3.21) and so on REMEMBER ITS (X,Y)

Step-by-step explanation:

3 0

3 years ago

The midpoint of AB is M(-1, -2)M(−1,−2). If the coordinates of AA are (1, 3)(1,3), what are the coordinates of BB?

Sholpan [36]

of the quotient of the rectangular prism of green green and green green yellow green eeewww

4 0

3 years ago

Please help with this ASAP

uranmaximum [27]

Answer:

7w^3/5v

Step-by-step explanation:

42v^3w^4/30v^4w

42 = 6 * 7

30 = 6*5 , factor

6*7v^3w^4/6*5v^4w

6 is common, cancel

7v^3w^4/5v^4w

simplify w^4/w = w^3

v^3/v^4 = 1/v

therefore, 7w^3/5v is the answer

7 0

3 years ago

A tourist in France wants to visit 8 different cities. If the route is randomly selected, what is the probability that she will

Harlamova29_29 [7]

Answer:

Probability p( selecting 8 cities alphabetically) = 2.48×10^-5

Step-by-step explanation:

Number of possible ways of choosing 8 cities=Permutation P(n,k) = n!/(n-k)!

P(8,8)= 8!/(8-8)! = 8! = 40,320

Probability (selecting 8 cities alphabetically) = 1/40320 = 2.48×10^-5

7 0

4 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 $a_i=0$ .

5 0

3 years ago