All categories

3 years ago

Y=(1/2)^x +2. What is the horizontal asymptote of this function?

1 answer:

6 0

$\lim_{x\to -\infty} \left(\frac{1}{2}\right)^x+2=\infty+2=\infty\\
\lim_{x\to \infty} \left(\frac{1}{2}\right)^x+2=0+2=2\\$

There is one-sided horizontal asymptote $y=2$

You might be interested in

45% of 80 show your work

Elan Coil [88]

.45 times 80

8 times 4 =32
8 times 0.5 =4

Just add these together to get 36.

4 0

2 years ago

Read 2 more answers

PLEASE HELP! SO CONFUSED! 25 POINTS!!

MrRa [10]

Answer:

Step-by-step explanation:

(f • g)(x)

= f(x) × g(x)

= (x + 4)(3x² - 7)

each term in the second factor is multiplied by each term in the first factor , that is

x(3x² - 7) + 4(3x² - 7) ← distribute both parenthesis

= 3x³ - 7x + 12x² - 28

= 3x³ + 12x² - 7x - 28 ← in standard form

7 0

1 year ago

What is the name for an 8-sided polygon

Mazyrski [523]

An 8 sided polygon is, an Octagon

8 0

2 years ago

Read 2 more answers

A cement walkway of uniform width has been built around an in-ground rectangular pool. the area of the walkway is 1344 square fe

Elza [17]

Answer:

6 feet.

Step-by-step explanation:

Dimensions of the Pool =80 feet long by 20 feet wide

Area of the walkway =1344 square feet.

If the width of the walkway=w

Length of the Larger Rectangle =80+2w
Width of the Larger Rectangle =20+2w

Area of the Walkway =Area of the Larger Rectangle - Area of the Pool

$1344=(80+2w)(20+2w)-(80*20)\\1344=80*20+160w+40w+4w^2-80*20\\4w^2+200w=1344\\4w^2+200w-1344=0\\We factorize\\4(w^2+50w-336)=0\\4(w^2-6w+56w-336)=0\\w(w-6)+56(w-6)=0\\(w-6)(w+56)=0\\w-6=0$ or $ w+56=0\\w=6$ ft or $ w=-56$ \\Therefore, the width of the walkway , w=6 feet.$

6 0

2 years ago

Evaluate the integral by interpreting it in terms of areas Draw a picture of the region the integral

denis23 [38]

Answer:

Step-by-step explanation:

The picture is below of how to separate this into 2 different regions, which you have to because it's not continuous over the whole function. It "breaks" at x = 2. So the way to separate this is to take the integral from x = 0 to x = 2 and then add it to the integral for x = 2 to x = 3. In order to integrate each one of those "parts" of that absolute value function we have to determine the equation for each line that makes up that part.

For the integral from [0, 2], the equation of the line is -3x + 6;

For the integral from [2, 3], the equation of the line is 3x - 6.

We integrate then:

$\int\limits^2_0 {-3x+6} \, dx+\int\limits^3_2 {3x-6} \, dx$ and

$-\frac{3x^2}{2}+6x\left \} {{2} \atop {0}} \right. +\frac{3x^2}{2}-6x\left \} {{3} \atop {2}} \right.$ sorry for the odd representation; that's as good as it gets here!

Using the First Fundamental Theorem of Calculus, we get:

(6 - 0) + (-4.5 - (-6)) = 6 + 1.5 = 7.5

5 0

2 years ago