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Assoli18 [71]
2 years ago
10

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 <br>(3,-1), (3,9) and (-2,4)<br>​
mash [69]
Y=-x+2

OR if you are looking for point slope:

y+1=-1(x-3)
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2 years ago
1v1 lol i'm a girl the code is za9nja
Naya [18.7K]

Answer:

i thought you was a guy ?

Step-by-step explanation:

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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)

<span>y\approx6-(1.68029746+2.63940507809)

</span><span>\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}</span>
3 0
3 years ago
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