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3 years ago

How an object reflected in the x axis and y axis on the cartesian

1 answer:

8 0

if a point is reflected on the x axis, the y coordinate changes. If an object is reflected on the y axis the x coordinate changes

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Write 100% as a decimal and as a fraction <br>middle school not high school.<br>9/3/20 ↑

balandron [24]

Answer:

1.0 and 1/1

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

The following table gives three different relations.

Drupady [299]

Answer:

First relation: It's a function. No order pair need to be changed.

Second relation: It's a function. No order pair need to be changed.

Third relation: It's not a function. The ordered pair (5,8) needs to be changed.

Step-by-step explanation:

A function may not have two y-values assigned to the same x-value.

Each x-element of the first relation has only one y-element associated with it. So we conclude that it's a function.

Each x-element of the second relation has only one y-element associated with it. So we conclude that it's a function.

The points (5,8) and (5,2) shows that there are two y-elements (y=8 and y=2) associated with the value x=5. So we conclude that it's not a function.

8 0

4 years ago

Express f(x) = |x-2| +|x+2| in the non-modulus form. Hence, sketch the graph of f.

alexgriva [62]

Recall that

$|x|=\begin{cases}x&\text{if }x\ge0\\-x&\text{if }x$

There are three cases to consider:

(1) When $x+2$ , we have $|x+2|=-(x+2)$ and $|x-2|=-(x-2)$ , so

$|x-2|+|x+2|=-(x-2)-(x+2)=-2x-4$

(2) When $x+2\ge0$ and $x-2$ , we get $|x+2|=x+2$ and $|x-2|=-(x-2)$ , so

$|x-2|+|x+2|=-(x-2)+(x+2)=4$

(3) When $x-2\ge0$ , we have $|x+2|=x+2$ and $|x-2|=x-2$ , so

$|x-2|+|x+2|=(x-2)+(x+2)=2x$

So

$|x-2|+|x+2|=\begin{cases}-2x-4&\text{if }x$

4 0

3 years ago

Find the first partial derivatives of the function f(x,y,z)=4xsin(y−z)

Amanda [17]

Answer:

$f_x(x,y,z)=4\sin (y-z)$

$f_x(x,y,z)=4x\cos (y-z)$

$f_z(x,y,z)=-4x\cos (y-z)$

Step-by-step explanation:

The given function is

$f(x,y,z)=4x\sin (y-z)$

We need to find first partial derivatives of the function.

Differentiate partially w.r.t. x and y, z are constants.

$f_x(x,y,z)=4(1)\sin (y-z)$

$f_x(x,y,z)=4\sin (y-z)$

Differentiate partially w.r.t. y and x, z are constants.

$f_y(x,y,z)=4x\cos (y-z)\dfrac{\partial}{\partial y}(y-z)$

$f_y(x,y,z)=4x\cos (y-z)$

Differentiate partially w.r.t. z and x, y are constants.

$f_z(x,y,z)=4x\cos (y-z)\dfrac{\partial}{\partial z}(y-z)$

$f_z(x,y,z)=4x\cos (y-z)(-1)$

$f_z(x,y,z)=-4x\cos (y-z)$

Therefore, the first partial derivatives of the function are $f_x(x,y,z)=4\sin (y-z), f_x(x,y,z)=4x\cos (y-z)\text{ and }f_z(x,y,z)=-4x\cos (y-z)$ .

4 0

4 years ago

If Mateo converts his anxiety score into a z-score, he can both indicate how many standard deviation units his score is from the

sertanlavr [38]

Answer:

The answer is "Option a"

Step-by-step explanation:

In the given scenario, the z-score value is used as the score which is used to shows that how many standard deviation units in this score were uses as the from mean and whenever this score is above or below by the mean, therefore the given statement is true.

5 0

3 years ago