Graph the direct variation equation. <br> Y=-1/2x

Zinaida [17]

<h3>Original Equation:</h3>

$-\frac{1}{3}+2z=-\frac{5}{6}$

<h3>Steps:</h3>

<em>*To solve for a variable, isolate the desired variable onto one side.</em>

Firstly, we want to add 1/3 to each side however -5/6 and 1/3 do not share the same denominator, and we want them to have that and we can do that by finding the LCD, or lowest common denominator. To find the LCD, list the multiples of 6 and 3 and the lowest multiple that they share is their LCD. In this case, their LCD is 6. Multiply -1/3 by 2/2:

$-\frac{1}{3}\times \frac{2}{2}=-\frac{2}{6}\\\\-\frac{2}{6}+2z=-\frac{5}{6}$

Now that we have common denominators, add both sides by 2/6:

$2z=-\frac{3}{6}\\\\2z=-\frac{1}{2}$

Next, you want to cancel out 2 to isolate z. Usually, one would divide both sides by 2, however remember that <u>dividing by a number is the same as multiplying by it's reciprocal.</u> To find a number's reciprocal, flip the numerator and denominator around. In this case, since 2 is a <em>whole number</em> this means that the denominator is 1. In this case: 2/1 would become 1/2. Multiply both sides by 1/2:

$\frac{1}{2}\times 2z= -\frac{1}{2}\times \frac{1}{2}\\\\z=-\frac{1}{4}$

<h3>Final Answer:</h3>

<u>Your final answer is z = -1/4.</u>

7 0

3 years ago

Start at (12, 3). Move 2 units right and 7 units up. Where do you end up?

Korolek [52]

Answer: (10, 10)

Step-by-step explanation: well if you say start at (12, 3) move 7 units up then 3 units right then it would make it easier to figure it out cus as you can see start at 3 and go up 7 it would stop at 10 then if you go over 2 from 12 it would stop at 10

6 0

3 years ago

Use the drop-down menus to complete the proof. by the unique line postulate, you can draw only one segment,

Doss [256]

The reflection of BC over I is shown below.

<h3>What is reflection?</h3>

A reflection is a mapping from a Euclidean space to itself that is an isometry with a hyperplane as a set of fixed points; this set is known as the reflection's axis (in dimension 2) or plane (in dimension 3).
A figure's mirror image in the axis or plane of reflection is its image by reflection.

See the attached figure for a better explanation:

1. By the unique line postulate, you can draw only one line segment: BC

Since only one line can be drawn between two distinct points.

2. Using the definition of reflection, reflect BC over l.

To find the line segment which reflects BC over l, we will use the definition of reflection.

3. By the definition of reflection, C is the image of itself and A is the image of B.

Definition of reflection says the figure about a line is transformed to form the mirror image.
Now, the CD is the perpendicular bisector of AB so A and B are equidistant from D forming a mirror image of each other.

4. Since reflections preserve length, AC = BC

In Reflection the figure is transformed to form a mirror image.
Hence the length will be preserved in case of reflection.

Therefore, the reflection of BC over I is shown.

Know more about reflection here:

brainly.com/question/1908648

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The question you are looking for is here:

C is a point on the perpendicular bisector, l, of AB. Prove: AC = BC Use the drop-down menus to complete the proof. By the unique line postulate, you can draw only one segment, Using the definition of, reflect BC over l. By the definition of reflection, C is the image of itself and is the image of B. Since reflections preserve , AC = BC.