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2 years ago

25 POINTS!!!! PLEASE EXPLAIN TO ME STEP BY STEP!

2 answers:

8 0

-x-y-z=-8
-x-y-z=-8

Let's solve for x.−x−y−z=−8
Step 1: Add y to both sides.−x−y−z+y=−8+y−x−z=y−8
Step 2: Add z to both sides.−x−z+z=y−8+z−x=y+z−8
Step 3: Divide both sides by -1.−x−1=y+z−8−1x=−y−z+8
Answer: x=−y−z+8

Feliz [49]2 years ago

3 0

Easy peassy
so basically what you would try to do is eliminate one type of variable from 2 different equations
look, we have
2x+2z=4
that already has y eliminated
so we just need to eliminate y from another equation

so

we can eliminate the y from the 2nd equation
multiply the first equation by 4 and add to 2nd equation

-4x-4y-4z=-32
-4x+4y+5z=7 +
-8x-0y+1z=-25

-8x+z=-25

we have
-8x+z=-25 and
2x+2z=4

we can elminate the z's
multiply first (top) equation by -2 and add to bottom equaton
16x-2z=50
2x+2z=4 +
18x=54

18x=54
divide both sides by 18
x=3

sub back
2x+2z=4
2(3)+2z=4
6+2z=4
minus 6 both sides
2z=-2
divide 2
z=-1

sub back

-x-y-z=-8
-3-y-(-1)=-8
-3-y+1=-8
-2-y=-8
add 2 to both sides
-y=-6
times -1
y=6

x=3
y=6
z=-1
(x,y,z)
(3,6,-1)

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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

The area of a sail is 40 1/2 square feet. The base and the height of the sail are equal. What is the height of the sail (in feet

grin007 [14]

Both base and height of the sail is 40 1/2 feet.

A sail is triangular in shape. Finding the area of a triangle is multiplying its base and height and dividing the product by 2.

A = 1/2 bh

Area = 40 1/2 or 40.50 sq. ft
Since base and height are equal.
40.50 x 2 = 81
bh = 81
bh = 81 ÷ 2
b = 40 1/2 ft
h = 40 1/2 ft

3 0

3 years ago

Read 2 more answers

What factors were used to create 4 6 9 14 49

Butoxors [25]

The factors are 2 (two)

7 0

2 years ago

PROBABILITY QUESTION!!! PLEASE HELP

Evgen [1.6K]

Answer:

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Step-by-step explanation:

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4 0

2 years ago