What is the greatest common factor of 16,32, and 44?

padilas [110]

Answer:

w =20.3

Step-by-step explanation:

i hope this helps you sorry i was late for your last question.

5 0

3 years ago

Read 2 more answers

Please help me I will give you extra points and the brain thing. (image below) 3/5

JulsSmile [24]

I think the answer is A

4 0

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

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

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:

parallel to the third side. half as long as the third side.

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 △XYZ. Construct the midpoint A of side XZ¯¯¯¯¯¯¯¯.

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

We must show that the line through A and parallel to side XY¯¯¯¯¯¯¯¯ will intersect side YZ¯¯¯¯¯¯¯ 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 YZ¯¯¯¯¯¯¯.

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

Proof of 2. We must show that AB=12XY.

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

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

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

So, 2AB=XY and AB=12XY. ⧫

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

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

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

Example 2

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

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

Lesson Summary

8 0

2 years ago

HELP PLS and quick find the volume

Ray Of Light [21]

Answer:

V =96 pi

Step-by-step explanation:

The volume of a cone is given by

V = 1/3 pi r^2 h

Where r is the radius and h is the height

V = 1/3 pi (6)^2 *8

V = 1/3 pi (36)8

V =96 pi

5 0

3 years ago

A 14-foot ladder is placed against a vertical wall of a building, with the bottom of the ladder standing on level ground 9 fee

timama [110]

Answer:

10.7 feet

Step-by-step explanation:

The ladder, the ground and the wall form the shape of a right angled triangle as shown in the image below.

The hypotenuse of the triangle is 14 feet (length of ladder)

The base of the triangle is 9 feet long (the distance from the base of the ladder to the wall)

We need to find the height of the triangle. We can apply Pythagoras rule:

$hyp^2 = a^2 + b^2$

where hyp = hypotenuse

a = base of the triangle

b = height of the triangle

Therefore:

$14^2 = 9^2 + b^2\\\\196 = 81 + b^2\\\\b^2 = 196 - 81 = 115\\\\b = \sqrt{115} \\\\b = 10.7 feet$

The wall reaches 10.7 feet high.

3 0

3 years ago