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2 years ago

Consider the following sets:

Mathematics

2 answers:

aleksandr82 [10.1K]2 years ago

7 0

***********************m = 2 ****************************

bagirrra123 [75]2 years ago

4 0

U = {points on the coordinate plane}
A = {solutions to the equation y = 2x + 5}
B = {points on the line y = mx}

Value of the slope m so that {2x + 5} ∩ {mx} =Ф
This means that {2x + 5} never intersects with {mx}
To that end m=2 (same slope), if so the 2 linear functions:

y = 2x+5 and y = 2x are PARALLEL

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The components of a vector are Ax = –10 units and Ay = 8 units. What angle does this vector make with the positive x axis? g

Mandarinka [93]

Answer:

This vector make an angle of 141° with the positive x axis

Step-by-step explanation:

Refer the attached figure

Components of vector :

$A_x=-10 \\A_y = 8$

We are supposed to find What angle does this vector make with the positive x axis

$tan \theta =\frac{A_y}{A_x}\\tan \theta =\frac{8}{-10}\\\theta =tan^{-1}(\frac{8}{-10})\\\theta =tan^{-1}(\frac{8}{-10})=-38.65^{\circ}$

Now ,

$\alpha = 180-38.65=141.35^{\circ}$

So, this vector make an angle of 141° with the positive x axis

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2 years ago

What is the net force on an object that has a 25N force pulling it to the right, and two 10N forces pulling it to the left?

Umnica [9.8K]

Answer:

Step-by-step explanation:

25 - (10 x 2) = 5N

4 0

2 years ago

Read 2 more answers

Is the point (-2, 5) a solution to the equation 2x + 5y = 0?

ratelena [41]

Answer:

Step-by-step explanation:

substitute -2 for 'x' and 5 for 'y'

2(-2) + 5(5) = 0?

-4 + 25 = 21, not 0

5 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

Pls help me with this!

joja [24]

Answer:

A = 78.5 cm²

Step-by-step explanation:

Area = π · radius²

if diameter is 10 then radius is 5

A = 5²·π

A = 25 × 3.14

A = 78.5 cm²

6 0

2 years ago