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iogann1982 [59]
3 years ago
5

Suppose f(x) is a function such that if p < q, f(p) < f(q). Which statement best describes f(x)?

Mathematics
2 answers:
quester [9]3 years ago
4 0

Answer:

option B is correct.

Step-by-step explanation:

we are given that for any p<q

f(p)<f(q)

this clearly implies that f is an increasing function.

Now we know that if f is an increasing function then -f is always an decreasing function and vice-versa.

so here -f(x) will be an decreasing function.

Let us consider a example f(x)=x then f(x) is clearly an increasing function.

and -f(x)= -x is an decreasing function. also it is an odd function but not an even function.

so option B holds.



Oxana [17]3 years ago
4 0
p \ \textless \  q, f(p) \ \textless \  f(q) means that the function f(x) is increasing. -f(x) is therefore decreasing.

Increasing/decreasing functions can't be even, but can be odd, so it's B.
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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
Find the solution of the system of equations.
avanturin [10]

Answer:

3

Step-by-step explanation:

The number is 3.

Step-by-step explanation:

Finding the Number

To find the number, we have to translate the problem above to an algebraic equation. The algebraic equation refers to the statement of the equality of two algebraic expressions.

Equation:

Let "x" be the number.

4 is divided by a number - 4/x

3 divided by the number decreased by 2 - 3/x-2

"4 is divided by a number is equal to 3 divided by the number decreased by 2"

4/x = 3/x-2

Solution:

Cross multiply.

4/x = 3/x-2

x(3) = 4(x-2)

3x = 4x - 8

(Combine similar terms.)

3x - 4x = - 8

- x = - 8

- x/- 1 = - 8/- 1

x = 8

Final Answer:

8

Checking:

4/x = 3/x-2

4/8 = 3/8-2

1/2 = 3/6

1/2 = 1/2 ✔

7 0
3 years ago
-3 / 4 + (-2 / 7) - (- 5 ½)
Igoryamba
Answer:
125/28
4.46428571
4 13/28
There all right
Pls mark me brainliest
6 0
3 years ago
Read 2 more answers
Please me ASAP will choose brainliest
pogonyaev
The Least Common Denomiator Is 12
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8 0
3 years ago
Read 2 more answers
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malfutka [58]
?-54-42=?
I hoped this helped
4 0
4 years ago
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