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3 years ago

Is this a function ?

Mathematics

1 answer:

Brrunno [24]3 years ago

5 0

Answer:

no i hope this helped

Step-by-step explanation:

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For all the values of x,<br> f(x)=x²+3 g(x)=2x-1 <br> a) find fg(2) <br> b) find gf(2)

erik [133]

Answer:

12 and 13

Step-by-step explanation:

(a)

To evaluate f(g(2)), evaluate g(2) then use the value obtained to evaluate f(x)

g(2) = 2(2) - 1 = 4 - 1 = 3, then

f(3) = 3² + 3 = 9 + 3 = 12

--------------------------------------------

(b)

To evaluate g(f(2)), evaluate f(2) the use the value obtained to evaluate g(x)

f(2) = 2² + 3 = 4 + 3 = 7, then

g(7) = 2(7) - 1 = 14 - 1 = 13

5 0

3 years ago

What is 10x as much as 70?

sleet_krkn [62]

9514 1404 393

Answer:

700

Step-by-step explanation:

You figure this out by multiplying 70 by 10:

10×70 = 700

5 0

3 years ago

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 $a_i=0$ .