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2 years ago

Answer each question below. 18. (16) ZW and ZY are supplementary. If m

Mathematics

2 answers:

BartSMP [9]2 years ago

5 0

Answer:

I'm not answering but i can help you :)

Step-by-step explanation:

Supplementary angles are two angles whose measures add up to 180°.

The two angles of a linear pair, like ∠1 and ∠2 in the figure below, are always supplementary.

But, two angles need not be adjacent to be supplementary. In the next figure, ∠3 and ∠4 are supplementary, because their measures add to 180°.

Example 1:

Two angles are supplementary. If the measure of the angle is twice the measure of the other, find the measure of each angle.

Let the measure of one of the supplementary angles be a .

Measure of the other angle is 2 times a.

So, measure of the other angle is 2a.

If the sum of the measures of two angles is 180°, then the angles are supplementary.

So, a+2a=180°

Simplify.

3a=180°

To isolate a, divide both sides of the equation by 3.

3a3=180°3 a=60°

The measure of the second angle is,

2a=2×60° =120°

So, the measures of the two supplementary angles are 60° and 120°.

Example 2:

Find m∠P and m∠Q if ∠P and ∠Q are supplementary, m∠P=2x+15, and m∠Q=5x−38.

The sum of the measures of two supplementary angles is 180°.

So, m∠P+m∠Q=180°

Substitute 2x+15 for m∠P and 5x−38 for m∠Q.

2x+15+5x−38=180°

Combine the like terms. We get:

7x−23=180°

Add 23 to both sides. We get:

7x=203°

Divide both sides by 7.

7x7=203°7

Simplify.

x=29°

To find m∠P, substitute 29 for x in 2x+15.

2(29)+15=58+15

Simplify.

58+15=73

So, m∠P=73°.

To find m∠Q, substitute 29 for x in 5x−38.

5(29)−38=145−38

Simplify.

145−38=107

So, m∠Q=107°.

lianna [129]2 years ago

4 0

Is there anymore words?

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The vertex for this case would be:

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Step-by-step explanation:

For this case we have the following equation given:

$y= x^2 -6x +14$

We can complete the square like this:

$y= x^2 -6x +(\frac{6}{2})^2 +14 -(\frac{6}{2})^2$

And solving we have:

$y= x^2 -6x +9 + 14 -9$

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$y-5 = (x-3)^2$

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There are several ways two triangles can be congruent.

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In $\mathbf{\triangle AOL}$ and $\mathbf{\triangle BOK}$ (see attachment), we have the following observations

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Using corresponding theorem,

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Neko [114]

Answer:

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Consider all options:

A. Side length and perimeter of 1 face is a linear relationship, because the graph of the function $y=kx$ is a straight line.

B. Perimeter of 1 face and area of 1 face is not a linear relationship, because the graph of this relationship is a quadratic parabola with equation $y=kx^2$ .

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D. Area of 1 face and surface area is a linear relationship, because the graph of the function $y=kx$ is a straight line.

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inna [77]

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Answer:

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Step-by-step explanation:

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