Answer:
D.
Step-by-step explanation:
Debido a restricciones de extensión y la características del ejercicio, recomendamos leer la explicación de esta pregunta para mayores detalles sobre la adición de números <em>enteros</em>.
<h3>¿Cuáles son los resultados de cada suma?</h3>
En este ejercicio tenemos un grupo de sumas con números <em>enteros</em> <em>positivos</em> y <em>negativos</em>, en las cuales se prueba la capacidad del estudiante para realizar varias operaciones en serie (adición, sustracción) y comprender las diferencias entre números <em>positivos</em>, <em>negativos</em> y <em>neutros</em>. Ahora procedemos a determinar el resultado de cada una de las expresiones:
20 + 50 + 30 + 7 = 107
30 + 5 + 2 = 37
- 200 - 50 - 70 - 8 = - 328
- 500 + 100 - 20 + 50 = - 370
10 - 5 = 5
20 + 50 - 25 - 10 = 35
- 100 + 20 = - 80
- 30 + 5 + 4 - 20 + 8 = - 33
- 258 + 8 = - 250
- 10 + 20 + 520 - 100 + 8 = 438
- 20 - 5 - 42 + 3 = - 64
1000 - 200 + 50 + 30 - 45 + 75 - 87 + 90 + 50 - 100 + 50 - 10 = 903
- 400 + 500 - 200 - 50 + 48 + 8 - 47 - 50 = - 191
300 + 20 - 50 + 30 - 84 + 35 - 7 + 20 - 40 + 10 - 45 + 65 + 8 - 55 = 207
800 + 50 - 69 + 8 - 35 + 85 - 54 + 40 + 85 + 74 - 32 - 8 + 65 - 27 = 982
Para aprender más sobre sumas: brainly.com/question/1456841
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Here are the steps
1: Put the compass on Q and make the width equal to the distance from Q to L. Extend line LM towards the left side of L and draw an arc hitting the line segment on the left side of L
2. <span> Without changing the width and position of the compass, draw an arc between L and M.
3. Without changing the width of the compass, put the compass on the point of intersection of the arc and line LM (left side of L). Draw an arc above line LM.
4. Without changing the width of the compass, put the compass on the point of intersection of the arc and line LM (right side of L). Draw an arc above line LM.
5. Use a straight edge to make a line from the intersection of the two arcs above line LM to Q intersecting through L along the way. </span>
Ok
first of all, for q(x)/p(x)
if the degree of q(x) is less than the degree of p(x),then the horizontal assemtote is 0
then simplify
any factors you factored out is now a hole, remember them
to find the vertical assemtotes of a function, set the SIMPLIFIED denomenator equal to 0 and solve
so
y=(x-5)/(x^2-1)
q(x)<p(x)
horizontal assemtote is y=0
no factors to simplify so no holes
set denomenator to 0 to find vertical assemtote
x^2-1=0
(x-1)(x+1)=0
x-1=0
x=1
x+1=0
x=-1
the horizontal assemtotes are x=1 and -1
Answer:
The equation does not match the form of any conic section.
Not a Conic Section
Step-by-step explanation:
