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natali 33 [55]
4 years ago
15

What is y=x+2 graphed

Mathematics
1 answer:
Travka [436]4 years ago
4 0

The graphing is shown below

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What number must be distributed in the following equation?
allochka39001 [22]

Answer:

-6

Step-by-step explanation: -6 is your answer because it’s outside of the parenthesis

8 0
3 years ago
Read 2 more answers
A function is given. (a) Find all the local maximum and minimum values of the function and the value of x at which each occurs.
Allisa [31]

Answer:

a)

x = -1/√3 = -0.58    is a local maximum

x = 1/√3  =  0.58     is a local minimum

b)

(-1/√3 , 1/√3 ) DECREASING

(-∞, -1/√3) U (1/√3, ∞)    INCREASING

Step-by-step explanation:

To answer all of the questions we must obtain the derivative of the fucntion:

If

U(x) = 4(x^3 - x)

then

U'(x) = 4(3x^2 - 1)

U''(x) = 4(3*2 x) = 24 x

U'''(x) = 24

The local maxima and minima of the function U(x) can be found when U'(x) = 0

this occurs when :

3x^2 = 1

that is:

x = ±1/√3

We will know if they are a minimum or a maximum evaluating this points on the second derivative (you can look for it as <u><em>Second derivative test</em></u>), if the result is positive the point corresponds to a minimum and if it is negative it will be a maximum.

Here it is easy to determine wheather is a maximum or a minimum because the second derivative is 24x, therefore if x is negative or positive so the second derivative will be.

a)

x = -1/√3 = -0.58    is a local maximum

x = 1/√3  =  0.58     is a local minimum

b)

the intervals at which the function is increasing { decreasing } is given when the first derivative is positive { negative }

The first derivative will be positive when:

3x^2 > 1

|x| > 1/√3  -->  x > 1/√3   and    x < -1/√3

The first derivatice will be negative when:

3x^2 < 1

|x| < 1/√3  -->  x < 1/√3   and    x > -1/√3

Therefore the intervals are:

(-1/√3 , 1/√3 ) DECREASING

(-∞, -1/√3) U (1/√3, ∞)    INCREASING

<u><em>** The attached image is a plot of the fucntion where you can see part of the intervals and the local maximum and minimum</em></u>

8 0
3 years ago
For the function f defined by f(x)=5x−12 find f(−x),−f(x), and −f(−x).
Slav-nsk [51]
f(x) = 5x-12

f(-x) = 5(-x)-12 = -5x-12

-f(x) = -(5x-12) = -5x+12

-f(-x) = -(5(-x)-12) = -(-5x-12) = 5x+12

This is probably asking you to consider the symmetry of f(x). If f(x) is equal to f(-x), then f is symmetric about the origin. If f(x) is equal to -f(x), then f is symmetric about the y-axis.
3 0
3 years ago
List the fractions from least to greatest 14/25 29/50 53/100 13/20 3/5
Inga [223]
14/25 = 56/100
29/50 = 58/100
53/100
13/20 = 65/100
3/5 = 60/100

 order least to greatest:
53/100, 14/25, 29/50, 3/5, 13/20

8 0
3 years ago
A 2D vector can have a component equal to zero even when its magnitude is nonzero.
lys-0071 [83]

Answer:

T F F F T F T

Step-by-step explanation:

A 2D vector can have a component equal to zero even when its magnitude is nonzero. TRUE.

it will be actually a 1d vector, but it's still a 2d vector too. this happens when the vector it's aligned to one of the axis.

The direction of a vector can be different in different coordinate systems. FALSE

The direction of a vector is independent of any coordinate system.

A 2D vector can have a magnitude equal to zero even when one of its components it nonzero. FALSE.

because the only way \sqrt{x^{2} +y^{2} } =0 is that both x and y are equal to zero

The magnitude of a vector can be different in different coordinate systems. FALSE.

The magnitude stays the same in every coordinate system

It is possible to multiply a vector by a scalar. TRUE.

it changes only it's magnitude.

It is possible to add a scalar to a vector. FALSE.

you can´t sum elements of diferent spaces. R, R^{2}, etc

The components of a vector can be different in different coordinate systems.TRUE.

The components of a vector are just a way for identifying the vector, in a determinate coordinate system, the way i call those components will change as I change the name I call every point in the plane.

5 0
3 years ago
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