All categories

4 years ago

Given the following linear function sketch the graph of the function and find the domain and range.

Mathematics

1 answer:

barxatty [35]4 years ago

3 0

As with any odd-degree polynomial, ...
the domain is "all real numbers"
the range is "all real numbers"

Here is a graph.

You might be interested in

What is the inverse of the function f(x) = x +3?

KatRina [158]

Next time please add answer choices

f(x) = y = x+3, Consider y=x+3. You just switch x and y:
so y= x+3, becomes x=y+3. Once done, you isolate y:
x=y+3 or y = x-3. This this the inverse function of f(x)

5 0

3 years ago

Mr. Chiu made a dot plot after asking his students a question.

stepladder [879]

Answer:

...

Step-by-step explanation:

Im going to need a question before i can properly answer this.

5 0

3 years ago

What is the equation for the plane illustrated below?

TiliK225 [7]

Answer:

Hence, none of the options presented are valid. The plane is represented by $3 \cdot x + 3\cdot y + 2\cdot z = 6$ .

Step-by-step explanation:

The general equation in rectangular form for a 3-dimension plane is represented by:

$a\cdot x + b\cdot y + c\cdot z = d$

Where:

$x$ , $y$ , $z$ - Orthogonal inputs.

$a$ , $b$ , $c$ , $d$ - Plane constants.

The plane presented in the figure contains the following three points: (2, 0, 0), (0, 2, 0), (0, 0, 3)

For the determination of the resultant equation, three equations of line in three distinct planes orthogonal to each other. That is, expressions for the xy, yz and xz-planes with the resource of the general equation of the line:

xy-plane (2, 0, 0) and (0, 2, 0)

$y = m\cdot x + b$

$m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Where:

$m$ - Slope, dimensionless.

$x_{1}$ , $x_{2}$ - Initial and final values for the independent variable, dimensionless.

$y_{1}$ , $y_{2}$ - Initial and final values for the dependent variable, dimensionless.

$b$ - x-Intercept, dimensionless.

If $x_{1} = 2$ , $y_{1} = 0$ , $x_{2} = 0$ and $y_{2} = 2$ , then:

Slope

$m = \frac{2-0}{0-2}$

$m = -1$

x-Intercept

$b = y_{1} - m\cdot x_{1}$

$b = 0 -(-1)\cdot (2)$

$b = 2$

The equation of the line in the xy-plane is $y = -x+2$ or $x + y = 2$ , which is equivalent to $3\cdot x + 3\cdot y = 6$ .

yz-plane (0, 2, 0) and (0, 0, 3)

$z = m\cdot y + b$

$m = \frac{z_{2}-z_{1}}{y_{2}-y_{1}}$

Where:

$m$ - Slope, dimensionless.

$y_{1}$ , $y_{2}$ - Initial and final values for the independent variable, dimensionless.

$z_{1}$ , $z_{2}$ - Initial and final values for the dependent variable, dimensionless.

$b$ - y-Intercept, dimensionless.

If $y_{1} = 2$ , $z_{1} = 0$ , $y_{2} = 0$ and $z_{2} = 3$ , then:

Slope

$m = \frac{3-0}{0-2}$

$m = -\frac{3}{2}$

y-Intercept

$b = z_{1} - m\cdot y_{1}$

$b = 0 -\left(-\frac{3}{2} \right)\cdot (2)$

$b = 3$

The equation of the line in the yz-plane is $z = -\frac{3}{2}\cdot y+3$ or $3\cdot y + 2\cdot z = 6$ .

xz-plane (2, 0, 0) and (0, 0, 3)

$z = m\cdot x + b$

$m = \frac{z_{2}-z_{1}}{x_{2}-x_{1}}$

Where:

$m$ - Slope, dimensionless.

$x_{1}$ , $x_{2}$ - Initial and final values for the independent variable, dimensionless.

$z_{1}$ , $z_{2}$ - Initial and final values for the dependent variable, dimensionless.

$b$ - z-Intercept, dimensionless.

If $x_{1} = 2$ , $z_{1} = 0$ , $x_{2} = 0$ and $z_{2} = 3$ , then:

Slope

$m = \frac{3-0}{0-2}$

$m = -\frac{3}{2}$

x-Intercept

$b = z_{1} - m\cdot x_{1}$

$b = 0 -\left(-\frac{3}{2} \right)\cdot (2)$

$b = 3$

The equation of the line in the xz-plane is $z = -\frac{3}{2}\cdot x+3$ or $3\cdot x + 2\cdot z = 6$

After comparing each equation of the line to the definition of the equation of the plane, the following coefficients are obtained:

$a = 3$ , $b = 3$ , $c = 2$ , $d = 6$

Hence, none of the options presented are valid. The plane is represented by $3 \cdot x + 3\cdot y + 2\cdot z = 6$ .

8 0

3 years ago

Will mark brainliest please show work

SVEN [57.7K]

Answer:

5/4

Step-by-step explanation:

7 0

3 years ago

Which of the following correctly justifies statement 4 of the two-column proof?

slavikrds [6]

Answer:

D.Transitive property of equality

Step-by-step explanation:

We are given that segment JK is parallel to segment LM

We have to prove $\angle 3=\angle 6$

We have to find which option correctly justifies the statement 4 of the two - column proof.

1.Statement : JK is parallel to segment LM

Reason: Given

2. $\angle 7\cong \angle 6$

Reason: Vertical angles theorem

3. $\angle 3\cong \angle 7$

Reason:Corresponding angles theorem

4. $\angle 3\cong \angle 6$

Reason: Transitive property of equality.

If a=b and b=c then a=c

Hence, option D is true.

3 0

3 years ago

Read 2 more answers