What is the domain of the function represented by the graph

Mathematics

1 answer:

Dimas [21]3 years ago

5 0

Answer:

Step-by-step explanation:

Domain of the function is the whole R

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How do I solve 1/2 in minus 3 equals 5

Mandarinka [93]

Answer:

n = 16

Step-by-step explanation:

$\dfrac{1}{2}n-3=5\\\\\dfrac{1}{2}n=8 \qquad\text{add 3}\\\\n=16 \qquad\text{multiply by 2}$

___

In general, you identify the terms containing the variable. Then figure out what is being done to the variable. Here, it is multiplied by 1/2 and then 3 is subtracted. You undo these operations in reverse order.

Undo the subtraction of 3 by adding 3. Undo the multiplication by 1/2 by multiplying by its inverse, 2. Of course, the rules of equality tell you that anything you do to one side of the equation must also be done to the other side.

You can consider it as "unwrapping the package". You start at the outside, the last operation applied, and work your way in.

4 0

3 years ago

Given the points A(-1,-3) and B(5,2). A. Write the equation of the straight line AB. Another line MN is parallel to AB and passe

Alexus [3.1K]

Find slope of AB where A = (-1,-3) and B = (5,2):

$\text {Slope = } \dfrac{Y_2-Y_1}{X_2-X_1}$

$\text {Slope = } \dfrac{2+3}{5+1} = \dfrac{5}{6}$

Write equation with the slope found:

$y = mx + b$

$y = \dfrac{5}{6} x + b$

Find y-intercept:

$y = \dfrac{5}{6} x + b$

$2 = \dfrac{5}{6} (5) + b$

$2 = \dfrac{25}{6} + b$

$b = 2 - \dfrac{25}{6}$

$b = -\dfrac{13}{6}$

Form the equation:

$y = \dfrac{5}{6} x - \dfrac{13}{6}$

$\boxed {\boxed {\bf \text {Answer: Line AB : }y = \dfrac{5}{6} x - \dfrac{13}{6}}}$


Find Slope of Line MN where MN is parallel to AB:

$\text {Slope of MN = Slope of AB = } \dfrac{5}{6}$

Write equation with the slope found:

$y = mx + b$

$y = \dfrac{5}{6} x + b$

Find y-intercept given that the line passed through (6, -2):

$y = \dfrac{5}{6} x + b$

$-2 = \dfrac{5}{6} (6) + b$

$-2 =5 + b$

$b = -2 - 5$

$b = -7$

$
\boxed {\boxed {\bf \text {Answer: Line MN } : y = \dfrac{5}{6} x -7 }}$

8 0

4 years ago

The question is in the attached file

sergeinik [125]

Answer:

Step-by-step explanation:

using

x × y = $\sqrt{x^2+y^2}$ , then