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2 years ago

What are the domain, range, and asymptote of h(x) = (0.5)x – 9?

Mathematics

1 answer:

Rzqust [24]2 years ago

8 0

Let's solve for h

hx = 0.5x - 9

Step 1: Divide both sides by x.

hx/x = 0.5x - 9/ x

Answer = h = 0.5x - 9/ x

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Find the equation of the circle which passes through points (3,-1), (3,9) and (-2,4)

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Y=-x+2

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y+1=-1(x-3)

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2 years ago

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2 years ago

Several students from boxerville middle stood in a circle, evenly spaced, to play a game. If the 6th student was directly across

myrzilka [38]

Answer:

There were 14 students in the circle

Step-by-step explanation:

* Lets explain how to solve the problem

- The students stood in a circle

∴ The line joining each two opposite students is the diameter

of the circle

- The 6th student was directly across from the 13th student

∴ The 6th student is opposite to the 13th student

- Lets count how many students between 6th student and

13th student

∵ 7th , 8th , 9th , 10th , 11th , 12th students between the 6th

student and 13th student

∴ There are 6 students between them

- Lets count how many students between 13th student and

6th student

∵ 14th , 1st , 2nd , 3rd , 4th , 5th students between the 13th

student and 6th student

∴ There are 6 students between them

∵ There are 6 students between 6th and 13th

∵ There are 6 students between 13th and 6th

∵ There are 2 students in 6th and 13th

∴ The number of the students = 6 + 6 + 2 = 14

∴ There were 14 students in the circle

8 0

3 years ago

Solve -3(2x-3)+10x=3q

scoray [572]

Answer:

4x+9=3q

Step-by-step explanation:

-3(2x-3)+10x=3q

-6x+9+10x=3q

4x+9=3q

7 0

3 years ago

Read 2 more answers

A normal window is constructed by adjoining a semicircle to the top of an ordinary rectangular window, (see figure ) The perimet

Dima020 [189]

Let's find the perimeter of the window.

The bottom side is $x$ . The left and right sides make $2y$ .
The perimeter of a circle is $2\pi r$ , so the perimeter of a semicircle must be $\pi r$ , The radius is $\frac{1}2x$ , so that gives $\frac{1}2\pi x$ for the curve. All of that is equal to 12.

$x+2y+\frac{1}2\pi x=12$

We only want to use one variable to create the area formula, so let's solve for $y$ .

$2y=12-x-\frac{1}2\pi x$

$y=6-\frac{1}2x-\frac{1}4\pi x$

Now that we have a value for $y$ in terms of $x$ , we can find the area in terms of $x$ .

The area of the rectangle is going to be $xy$ , which then becomes

$A_r=x(6-\frac{1}2x-\frac{1}4\pi x$

$A_r=6x-\frac{1}2x^2-\frac{1}4\pi x^2$

The area of the semicircle is going to be $\frac{1}2\pi r^2$ .

Since $r=\frac{1}2x$ , $A_{sc}=\frac{1}2\pi (\frac{1}2x)^2$ .

$A_{sc}=\frac{1}2\pi \frac{1}4x^2$

$A_{sc}=\frac{1}8\pi x^2$

Now let's add the areas of the rectangle and semicircle.

$A=A_r+A_{sc}$

$A=6x-\frac{1}2x^2-\frac{1}4\pi x^2+\frac{1}8\pi x^2$

$A=6x-\frac{1}2x^2-\frac{1}8\pi x^2$

If you wanted to factor out $\frac{1}8$ like you did, this would become

$\boxed{A(x)=\frac{1}8(48x=4x^2-\pi x^2)}$

Now what we want to do is find what $x$ is when $A$ is at its highest point, Once we have the value for $x$ we can also find the value for $y$ , of course.

Let's put our equation in the general form of a quadratic.

$A(x)=(-\frac{1}2-\frac{1}8\pi )x^2+6x$

Now we can use the vertex formula $x=\frac{-b}{2a}$ .
( $a$ and $b$ refer to $ax^2+bx+c$ .)

$x=\frac{-6}{2(-\frac{1}2-\frac{1}8\pi)}$

$x=\frac{-6}{-\frac{1}4\pi -1}$

$x=\frac{-24}{-\pi -4}$

$\boxed{x=\frac{24}{\pi +4}}$

Now let's plug that in for $y=6-\frac{1}2x-\frac{1}4\pi x$ .

Since our final answers are in decimal form and not exact form, we can make our lives a little easier here and just use $x\approx3.36059492$ .

$y\approx6-\frac{1}2(3.36059492)-\frac{1}4\pi(3.36059492)$

 $y\approx6-(1.68029746+2.63940507809)$

 $\boxed{y\approx1.68029746191}$

Let's take our answers for $x$ and $y$ and round to 2 decimal places.

$\boxed{x\approx 3.36\ ft}$

$\boxed{y\approx 1.68\ ft}$

3 0

3 years ago