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2 years ago

Analyze the asymptotes of the function.

Mathematics

1 answer:

koban [17]2 years ago

5 0

Answer:

vertical asymptote at x=-1

horizontal asymptote at y=0

Step-by-step explanation:

$y=\frac{1}{x+1}$

To find vertical asymptote we set the denominator =0 and solve for x

x+1=0 (subtract 1 from both sides)

x=-1

So, vertical asymptote at x=-1

To find horizontal asymptote we look at the degree of both numerator and denominator

there is no variable at the numerator , so degree of numerator =0

degree of denominator =1

When the degree of numerator is less than the degree of denominator

then horizontal asymptote at y=0

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Help with this math problem

marishachu [46]

see explanations

Step-by-step explanation:

Formula for mid-point is;

$midpoint=((x_1+x_2)/2,(y_1+y_2)/2)$

1.Given S(-3,4), U(-3,0) and T(5,0) then finding the mid-points

SU = [(-3+-3)/2,(4+0)/2] =(-3,2)

TU= [(-3+5)/2,(0+0)/2] =(1,0)

ST=[(-3+5)/2,(4+0)/2]=(1,2)

2.Mid-point formula for this case given W(-5,1) and Z(3,3) will be;

Midpoint = [ (x₁+x₂)/2,(y₁+y₂)/2]

Midpoint= [(-5+3)/2 ,(1+3)/2]

Midpoint=(1,2)

3. Finding the length of segments and the midpoint of the segments

Formula for length of segment, d, is given as;

For C(-1,5) and D(5,1)

$d=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2} \\\\d=\sqrt{(1-5)^2+(5--1)^2} =\sqrt{-4^2+6^2} =\sqrt{16+36} =\sqrt{52} =7.21$

Midpoint of the segment CD will be

[(-1+5)/2,(5+1)/2] = (2,3)

For E (-3,-6) and F(5,0)

$d=\sqrt{(0--6)^2+(5--3)^2} =\sqrt{6^2+8^2} =\sqrt{36+64} =\sqrt{100} =10$

The midpoint for segment EF will be;

[(-3+5)/2 ,(-6+0)/2] =(1,-3)

For L(-4,4) and M(-1,-5)

$d=\sqrt{(-5--4)^2+(-1-4)^2} \\d=\sqrt{-1^2+-5^2} \\\\d=\sqrt{1+25}=\sqrt{26} =5.1$

Learn More

Midpoint of a segment :brainly.com/question/10424143

Distance of a segment :brainly.com/question/11591660

Keywords :midpoint, midpoint formula, segment length, line segment

#LearnwithBrainly

4 0

3 years ago

Help plz

Yuki888 [10]

A) V * (2/5) = I
B) V = I + 42

Substituting A into B
V = (2/5) V + 42
3/5 V = 42
Victor has 70 tools
Ilya has 28 tools

together they have 98 tools

8 0

2 years ago

Read 2 more answers

n ΔABC, AB = 10 and BC = 5. Which expression is always true? A. 5 < AC < 10 B. AC = 5 C. 5 < AC < 15 D. AC = 10

Nataly [62]

Answer:

A. 5 < AC < 10

Step-by-step explanation:

If ∆ABC is a right angled triangle, we use the Pythagoras formula:

c² = a² + b²

Where c = longest side

When given sides AB, AC and BC, the formula becomes:

AB² = AC² + BC²

Where AB = Longest side

In the question,

AB = 10 and BC = 5.

10² = AC² + 5²

AC² = 10² - 5²

AC² = 100 - 25

AC² = 75

AC = √75

AC = 8.6602540378

Therefore, the expression that is always true = A. 5 < AC < 10

5 0

2 years ago

Please help anybody no links or else I report<br> Due in one min

Kaylis [27]

0.7 = 7/10 = 14/20

The missing numbers are 10 and 14

3 0

2 years ago

Read 2 more answers

Let f(x)=x2+6x−16 . Enter the x-intercepts of the quadratic function in the boxes. and

gizmo_the_mogwai [7]

Answer:

x = - 8, x = 2

Step-by-step explanation:

Given

f(x) = x² + 6x - 16

To find the x- intercepts equate f(x) to zero, that is

x² + 6x - 16 = 0 ← in standard form

(x + 8)(x - 2) = 0 ← in factored form

Equate each factor to zero and solve for x

x + 8 = 0 ⇒ x = - 8

x - 2 = 0 ⇒ x = 2

8 0

2 years ago