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ICE Princess25 [194]
3 years ago
9

What is the value of n in the equation –1/2(2n + 4) + 6 = –9 + 4(2n + 1)?

Mathematics
1 answer:
ioda3 years ago
7 0

Answer:

<h2>n = 1</h2>

Step-by-step explanation:

-\dfrac{1}{2}(2n-4)+6=-9+4(2n+1)\qquad\text{use the distributive property}\\\\\left(-\dfrac{1}{2}\right)(2n)+\left(-\dfrac{1}{2}\right)(-4)+6=-9+(4)(2n)+(4)(1)\\\\-n-2+6=-9+8n+4\qquad\text{combine like terms}\\\\-n+(-2+6)=8n+(-9+4)\\\\-n+4=8n-5\qquad\text{subtract 4 from both sides}\\\\-n=8n-9\qquad\text{subtract}\ 8n\ \text{from both sides}\\\\-9n=-9\qquad\text{divide both sides by (-9)}\\\\n=1

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Find all the missing sides or angles in each right triangles
astra-53 [7]
In previous lessons, we used the parallel postulate to learn new theorems that enabled us to solve a variety of problems about parallel lines:

Parallel Postulate: Given: line l and a point P not on l. There is exactly one line through P that is parallel to l.

In this lesson we extend these results to learn about special line segments within triangles. For example, the following triangle contains such a configuration:

Triangle <span>△XYZ</span> is cut by <span><span>AB</span><span>¯¯¯¯¯¯¯¯</span></span> where A and B are midpoints of sides <span><span>XZ</span><span>¯¯¯¯¯¯¯¯</span></span> and <span><span>YZ</span><span>¯¯¯¯¯¯¯</span></span> respectively. <span><span>AB</span><span>¯¯¯¯¯¯¯¯</span></span> is called a midsegment of <span>△XYZ</span>. Note that <span>△XYZ</span> has other midsegments in addition to <span><span>AB</span><span>¯¯¯¯¯¯¯¯</span></span>. Can you see where they are in the figure above?

If we construct the midpoint of side <span><span>XY</span><span>¯¯¯¯¯¯¯¯</span></span> at point C and construct <span><span>CA</span><span>¯¯¯¯¯¯¯¯</span></span> and <span><span>CB</span><span>¯¯¯¯¯¯¯¯</span></span> respectively, we have the following figure and see that segments <span><span>CA</span><span>¯¯¯¯¯¯¯¯</span></span> and <span><span>CB</span><span>¯¯¯¯¯¯¯¯</span></span> are midsegments of <span>△XYZ</span>.

In this lesson we will investigate properties of these segments and solve a variety of problems.

Properties of midsegments within triangles

We start with a theorem that we will use to solve problems that involve midsegments of triangles.

Midsegment Theorem: The segment that joins the midpoints of a pair of sides of a triangle is:

<span>parallel to the third side. half as long as the third side. </span>

Proof of 1. We need to show that a midsegment is parallel to the third side. We will do this using the Parallel Postulate.

Consider the following triangle <span>△XYZ</span>. Construct the midpoint A of side <span><span>XZ</span><span>¯¯¯¯¯¯¯¯</span></span>.

By the Parallel Postulate, there is exactly one line though A that is parallel to side <span><span>XY</span><span>¯¯¯¯¯¯¯¯</span></span>. Let’s say that it intersects side <span><span>YZ</span><span>¯¯¯¯¯¯¯</span></span> at point B. We will show that B must be the midpoint of <span><span>XY</span><span>¯¯¯¯¯¯¯¯</span></span> and then we can conclude that <span><span>AB</span><span>¯¯¯¯¯¯¯¯</span></span> is a midsegment of the triangle and is parallel to <span><span>XY</span><span>¯¯¯¯¯¯¯¯</span></span>.

We must show that the line through A and parallel to side <span><span>XY</span><span>¯¯¯¯¯¯¯¯</span></span> will intersect side <span><span>YZ</span><span>¯¯¯¯¯¯¯</span></span> at its midpoint. If a parallel line cuts off congruent segments on one transversal, then it cuts off congruent segments on every transversal. This ensures that point B is the midpoint of side <span><span>YZ</span><span>¯¯¯¯¯¯¯</span></span>.

Since <span><span><span>XA</span><span>¯¯¯¯¯¯¯¯</span></span>≅<span><span>AZ</span><span>¯¯¯¯¯¯¯</span></span></span>, we have <span><span><span>BZ</span><span>¯¯¯¯¯¯¯</span></span>≅<span><span>BY</span><span>¯¯¯¯¯¯¯¯</span></span></span>. Hence, by the definition of midpoint, point B is the midpoint of side <span><span>YZ</span><span>¯¯¯¯¯¯¯</span></span>. <span><span>AB</span><span>¯¯¯¯¯¯¯¯</span></span> is a midsegment of the triangle and is also parallel to <span><span>XY</span><span>¯¯¯¯¯¯¯¯</span></span>.

Proof of 2. We must show that <span>AB=<span>12</span>XY</span>.

In <span>△XYZ</span>, construct the midpoint of side <span><span>XY</span><span>¯¯¯¯¯¯¯¯</span></span> at point C and midsegments <span><span>CA</span><span>¯¯¯¯¯¯¯¯</span></span> and <span><span>CB</span><span>¯¯¯¯¯¯¯¯</span></span> as follows:

First note that <span><span><span>CB</span><span>¯¯¯¯¯¯¯¯</span></span>∥<span><span>XZ</span><span>¯¯¯¯¯¯¯¯</span></span></span> by part one of the theorem. Since <span><span><span>CB</span><span>¯¯¯¯¯¯¯¯</span></span>∥<span><span>XZ</span><span>¯¯¯¯¯¯¯¯</span></span></span> and <span><span><span>AB</span><span>¯¯¯¯¯¯¯¯</span></span>∥<span><span>XY</span><span>¯¯¯¯¯¯¯¯</span></span></span>, then <span>∠<span>XAC</span>≅∠<span>BCA</span></span> and <span>∠<span>CAB</span>≅∠<span>ACX</span></span> since alternate interior angles are congruent. In addition, <span><span><span>AC</span><span>¯¯¯¯¯¯¯¯</span></span>≅<span><span>CA</span><span>¯¯¯¯¯¯¯¯</span></span></span>.

Hence, <span>△<span>AXC</span>≅△<span>CBA</span></span> by The ASA Congruence Postulate. <span><span><span>AB</span><span>¯¯¯¯¯¯¯¯</span></span>≅<span><span>XC</span><span>¯¯¯¯¯¯¯¯</span></span></span> since corresponding parts of congruent triangles are congruent. Since C is the midpoint of <span><span>XY</span><span>¯¯¯¯¯¯¯¯</span></span>, we have <span>XC=CY</span> and <span>XY=XC+CY=XC+XC=2AB</span> by segment addition and substitution.

So, <span>2AB=XY</span> and <span>AB=<span>12</span>XY</span>. ⧫

Example 1

Use the Midsegment Theorem to solve for the lengths of the midsegments given in the following figure.

M, N and O are midpoints of the sides of the triangle with lengths as indicated. Use the Midsegment Theorem to find

<span><span> A. <span>MN</span>. </span><span> B. The perimeter of the triangle <span>△XYZ</span>. </span></span><span><span> A. Since O is a midpoint, we have <span>XO=5</span> and <span>XY=10</span>. By the theorem, we must have <span>MN=5</span>. </span><span> B. By the Midsegment Theorem, <span>OM=3</span> implies that <span>ZY=6</span>; similarly, <span>XZ=8</span>, and <span>XY=10</span>. Hence, the perimeter is <span>6+8+10=24.</span> </span></span>

We can also examine triangles where one or more of the sides are unknown.

Example 2

<span>Use the Midsegment Theorem to find the value of x in the following triangle having lengths as indicated and midsegment</span> <span><span>XY</span><span>¯¯¯¯¯¯¯¯</span></span>.

By the Midsegment Theorem we have <span>2x−6=<span>12</span>(18)</span>. Solving for x, we have <span>x=<span>152</span></span>.

<span> Lesson Summary </span>
8 0
3 years ago
Pls ANWSER pls pls pls I’ll get brainly
Charra [1.4K]

Answer:

C

Step-by-step explanation:

Its C

7 0
2 years ago
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Virty [35]
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Step-by-step explanation:

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What are angles that are equal called?
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I think the word you're looking for is congruent. Congruent angles always have the same angles.
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