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faltersainse [42]
3 years ago
5

Find the value of x. PLEASE HELP!!!!

Mathematics
1 answer:
Diano4ka-milaya [45]3 years ago
4 0

Answer:

Step-by-step explanation:

Because of the right angle (90 deg) in red, and the horizontal line on the bottom axis,

we know that :

49 + (1+5x) = 90

solving for x delivers x=8

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Please help!!!!!!!!!!!!​
daser333 [38]

Answer:

orochimaru vessle

Step-by-step explanation:

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2 years ago
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During the trip, the Morales family drove on several different highways. Gabriela and Luis noticed that sometimes the mile marke
Elenna [48]

<u><em>Answer:</em></u>

The slope of the line is 77

<u><em>Explanation:</em></u>

We know the speed is constant and that the mile-marker location over time can be represented by a line. This means that the equation is linear.

<u>The slope can, therefore, be calculated as follows:</u>

slope = \frac{y_2-y_1}{x_2-x_1}

where (x₁ , y₁) and (x₂ , y₂) are two points on the line

We are given that the two points (1, 123) and (3, 277) are two points in the line

<u>Substitute with them in the above equation to get the slope as follows:</u>

slope = \frac{277-123}{3-2}=77

Hope this helps :)

3 0
3 years ago
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>
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3 years ago
What is an equation that relates weight on earth to weight on the moon
dalvyx [7]
Weight is equal to the mass of the object multiplied by the force of gravity. The force of gravity on earth is greater then the force of the moon's gravity causing a smaller weight. The equation used to find an objects weight is W=mg. W is the weight, m is the mass, and g is the gravity's acceleration. <span />
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A large bag contains 12/15 pound of granola.How many 1/3 pound bags can be filled with this amoumt of granola?How much granola i
mylen [45]
1/3 is 5/15 so you can fill 5/15 +5/15 (2bags) and you will have 2/15 pounds left
5 0
3 years ago
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