18. The shaded portion on the grid represents

Isosceles triangle LMN is graphed with vertices L(0, 1), M(3, 5), and N(6,1). What is the slope of side LN?

Vaselesa [24]

L(0, 1)
N(6, 1)

slope formula;
m = y2 - y1/ x2 - x1

m = 1 - 1/ 6 - 1

m = 0/5

m = 0

hope this helped, God bless!

9 0

3 years ago

Read 2 more answers

Suppose that triangle TUV is an isosceles with the base VU.

vekshin1

Set them equal to each other and solve
3x + 39 = 4x +34
-x=-5
X=5
Plug it in
3(5) + 39 = 54
4(5) +34 = 54
For the final angle ,
54 + 54 = 108
180-108 =72
The angle measures are 54,54, and 72

3 0

2 years ago

2. Check the boxes for the following sets that are closed under the given

son4ous [18]

The properties of the mathematical sequence allow us to find that the recurrence term is 1 and the operation for each sequence is

a) Subtraction

b) Addition

c) AdditionSum

d) in this case we have two possibilities

* If we move to the right the addition

* If we move to the left the subtraction

The sequence is a set of elements arranged one after another related by some mathematical relationship. The elements of the sequence are called terms.

The sequences shown can be defined by recurrence relations.

Let's analyze each sequence shown, the ellipsis indicates where the sequence advances.

a) ... -7, -6, -5, -4, -3

We can observe that each term has a difference of one unit; if we subtract 1 from the term to the right, we obtain the following term

-3 -1 = -4

-4 -1 = -5

-7 -1 = -8

Therefore the mathematical operation is the subtraction.

b) $0. \sqrt{1}. \sqrt{4}, \sqrt{9}, \sqrt{16}, \sqrt{25}$ ...

In this case we can see more clearly the sequence when writing in this way

$0, \sqrt{1^2}. \sqrt{2^2}, \sqrt{3^2 } . \sqrt{4^2} , \sqrt{5^2}$

each term is found by adding 1 to the current term,

$\sqrt{(0+1)^2} = \sqrt{1^2} \\\sqrt{(1+1)^2} = \sqrt{2^2}\\\sqrt{(2+1)^2} = \sqrt{3^2}\\\sqrt{(5+1)^2} = \sqrt{6^2}$

Therefore the mathematical operation is the addition

c) $... \frac{-10}{2}. \frac{-8}{2}, \frac{-6}{2}, \frac{-4}{2}. \frac{-2}{2}. ...$

The recurrence term is unity, with the fact that the sequence extends to the right and to the left the operation is

To move to the right add 1

- $\frac{-10}{2} + 1 = \frac{-10}{2} - \frac{2}{2} = \frac{-8}{2}\\\frac{-8}{2} + \frac{2}{2} = \frac{-6}{2}$

To move left subtract 1

$\frac{-2}{2} - 1 = \frac{-4}{2}\\\frac{-4}{2} - \frac{2}{2} = \frac{-6}{2}$

Using the properties the mathematical sequence we find that the recurrence term is 1 and the operation for each sequence is

a) Subtraction

b) Sum

c) Sum

d) This case we have two possibilities

If we move to the right the sum

If we move to the left we subtract

Learn more here: brainly.com/question/4626313

5 0

2 years ago

Write 6 ratinol numbers between - 3 and - 4<br><br>

WARRIOR [948]

Answer:

-3.1, -3.2, -3.3, -3.4, -3.5, -3.6

Step-by-step explanation:

7 0

2 years ago

En un triangulo ABC, el angulo B mide 64° y el angulo C mide 72°. La bisectriz interior CD corta a la altura BH y a la bisectriz

nadezda [96]

Answer:

The difference between the greatest and the smallest angle of the triangle PBQ is 98°.

Step-by-step explanation:

The question is:

In a triangle ABC, angle B measures 64 ° and angle C measures 72°. The inner bisector CD intersects the height BH and the bisector BM at P and Q respectively. Find the difference between the greatest and the smallest angle of the triangle PBQ.

Solution:

Consider the triangle ABC.

The measure of angle A is:

angle A + angle B + angle C = 180°

angle A = 180° - angle B - angle C

= 180° - 72° - 64°

= 44°

It is provided that CD and BM are bisectors.

That:

angle BCP = angle PCH = 36°

angle CBQ = angle QBD = 32°

angle BHC = 90°

Compute the measure of angle HBC as follows:

angle HBC = 180° - angle BHC + angle BCH

= 180° - 90° - 72°

= 18°

Compute the measure of angle BPC as follows:

angle BPC = 180° - angle PCB + angle CBP

= 180° - 18° - 36°

= 126°

Then the measure of angle BPQ will be:

angle BPQ = 180° - angle BPC

= 180° - 126°

angle BPQ = 54°

Compute the measure of angle PBQ as follows:

angle PBQ = angle B - angle QBD - angle HBC

= 64° - 32° - 18°

angle PBQ = 14°

Compute the measure of angle BQP as follows:

angle BQP = 180° - angle PBQ - angle BPQ

= 180° - 14° - 54°

angle BQP = 112°

So, the greatest and the smallest angle of the triangle PBQ are:

angle BQP = 112°

angle PBQ = 14°

Compute the difference:

<em>d</em> = angle BQP - angle PBQ

= 112° - 14°

= 98°

Thus, the difference between the greatest and the smallest angle of the triangle PBQ is 98°.

4 0

2 years ago