Is 1.252525252525 repeating irrational

HELP PLEASE! What’s the domain and range?

NeX [460]

Answer:

Domain → (-∞, -1)∪(-1, ∞)

Range → (-∞, ∞)

Step-by-step explanation:

Given function is,

$\frac{f(x)}{g(x)}=\frac{x^2+2x-5}{2x+2}$

This function is not defined when the denominator is zero.

2x + 2 = 0

2x = -2

x = -1

Therefore, x = -1 is not in the domain of this function.

And the domain will be,

(-∞, -1)∪(-1, ∞)

For all values of x (except x = -1) we get some output values (value of $\frac{f(x)}{g(x)}$ ).

Therefore, range of the function will be (-∞, ∞).

7 0

3 years ago

Write the equation of the line that passes through (2,3) and (4, 1) in point-slope form. Use (2,3) for (x1, y1).

earnstyle [38]

Answer:

y-2 = -(x-2)

Step-by-step explanation:

To find the slope

m = (y2-y1)/(x2-x1) where (x1,y1) and (x2,y2) are two points on the line

= (1-3)/(4-2)

=-2/2

=-1

The point slope form is

y-y1 = m(x-x1) wher m is the slope and (x1,y1) is a point on the line

y-3 = -1(x-2)

y-2 = -(x-2)

5 0

3 years ago

What is the area of the shaded region in the figure below ? Leave answer in terms of pi and in simplest radical form

ser-zykov [4K]

Answer:

Step-by-step explanation:

That shaded area is called a segment. To find the area of a segment within a circle, you first have to find the area of the pizza-shaped portion (called the sector), then subtract from it the area of the triangle (the sector without the shaded area forms a triangle as you can see). This difference will be the area of the segment.

The formula for the area of a sector of a circle is:

$A_s=\frac{\theta}{360}*\pi r^2$ where our theta is the central angle of the circle (60 degrees) and r is the radius (the square root of 3).

Filling in:

$A_s=\frac{60}{360}*\pi (\sqrt{3})^2$ which simplifies a bit to

$A_s=\frac{1}{6}*\pi(3)$ which simplifies a bit further to

$A_s=\frac{1}{2}\pi$ which of course is the same as

$A_s=\frac{\pi}{2}$

Now for tricky part...the area of the triangle.

We see that the central angle is 60 degrees. We also know, by the definition of a radius of a circle, that 2 of the sides of the triangle (formed by 2 radii of the circle) measure √3. If we pull that triangle out and set it to the side to work on it with the central angle at the top, we have an equilateral triangle. This is because of the Isosceles Triangle Theorem that says that if 2 sides of a triangle are congruent then the angles opposite those sides are also congruent. If the vertex angle (the angle at the top) is 60, then by the Triangle Angle-Sum theorem,

180 - 60 = 120, AND since the 2 other angles in the triangle are congruent by the Isosceles Triangle Theorem, they have to split that 120 evenly in order to be congruent. 120 / 2 = 60. This is a 60-60-60 triangle.

If we take that extracted equilateral triangle and split it straight down the middle from the vertex angle, we get a right triangle with the vertex angle half of what it was. It was 60, now it's 30. The base angles are now 90 and 60. The hypotenuse of this right triangle is the same as the radius of the circle, and the base of this right triangle is $\frac{\sqrt{3} }{2}$ . Remember that when we split that 60-60-60 triangle down the center we split the vertex angle in half but we also split the base in half.

Using Pythagorean's Theorem we can find the height of the triangle to fill in the area formula for a triangle which is

$A=\frac{1}{2}bh$ . There are other triangle area formulas but this is the only one that gives us the correct notation of the area so it matches one of your choices.

Finding the height value using Pythagorean's Theorem:

$(\sqrt{3})^2=h^2+(\frac{\sqrt{3} }{2})^2$ which simplifies to

$3=h^2+\frac{3}{4}$ and

$3-\frac{3}{4}=h^2$ and

$\frac{12}{4} -\frac{3}{4} =h^2$ and

$\frac{9}{4} =h^2$

Taking the square root of both the 9 and the 4 (which are both perfect squares, thankfully!), we get that the height is 3/2. Now we can finally fill in the area formula for the triangle!

$A=\frac{1}{2}(\sqrt{3})(\frac{3}{2})$ which simplifies to