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Eduardwww [97]
3 years ago
5

The table below shows the values of y for different values of x:

Mathematics
1 answer:
mr Goodwill [35]3 years ago
4 0

Answer:

B. y=8x

Step-by-step explanation:

Take the numbers and idk..

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Charlie is at a small airfield watching for the approach of a small plane with engine trouble. He sees the plane at an angle of
Lera25 [3.4K]
Check the picture.

Let the given points represent the following:

F : Charlie's feet
E: Charlie's eyes
P: location of the plane
G: the foot of the perpendicular drawn from P to the ground
A: a point on PG, such that |GA|=|FE|=5.2 feet.

Clearly EAP is a right triangle, with m(AEP)=32°, side PA=1,700-5.2=1694.8 feet.

<span>the ground distance from Charlie to the plane is |FG|=|EA|

from right angle trigonometry, we know that:

</span>tan32= \frac{PA}{AE}
<span>
</span>0.625= \frac{1694.8}{AE}
<span>
</span>AE= \frac{1694.8}{0.625}=2711.68 feet
<span>

Answer: 2711.68 feet</span>

3 0
3 years ago
Della bought a tree seedling that was 2 1/4 feet tall. During the first year, it grew 1 1/6 Feet. After two years, it was 5 feet
Margarita [4]

Answer:

Step-by-step explanation:

Alright, lets get started.

Della bought a tree seeding that was 2\frac{1}{4} feet tall means \frac{9}{4} feet.

First year it grew 1\frac{1}{6} feet means it grew \frac{7}{6} feet.

It means after 1 year, the height of tree will be = \frac{9}{4}+\frac{7}{6}=\frac{41}{12}

After 2 years, the height of tree is = 5 feet

So, it grew in second year = 5-\frac{41}{12}

So, it grew in second year = \frac{60-41}{12}=\frac{19}{12}

So, it grew in second year=1\frac{7}{12} feet.   :  Answer

Hope it will help :)

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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