What two formulas are necessary to find the area of a segment of a circle? i will name brainliest!!!!!!!

Mathematics

2 answers:

TEA [102]3 years ago

6 0

Answer:

When finding the area of a sector, you are actually finding a fractional part of the area of the entire circle. The fraction is determined by the ratio of the central angle of the sector to the "entire central angle" of 360 degrees. [fractional part = ; where n = central angle]

Step-by-step explanation:

NeTakaya3 years ago

5 0

The necessary formulas to get the area of a circle’s segment are the area of the circle’s sector and the triangle it is bounded by.

Hence the formula for the segment’s area is:Area of Segment = Area of sector – area of the triangle.

The formula of the area of a circle’s sector is:Area of Sector = $\frac{central-angle}{2}$ x (radius of circle)²

While the formula for the area of a triangle is:Area of Triangle = $\frac{1}{2}$ x base of triangle x height of triangle

You might be interested in

A perfume is to be mixed in the ratio of 3 drops of pure essence to 7 drops of alcohol. How many drops of pure essence should be

Lemur [1.5K]

the anwser is 24 drops of essentail oil

7 0

3 years ago

Find the slope of the line between points A and B. Simplify your answer.

Taya2010 [7]

Answer:

-2

Step-by-step explanation:

From the given graph, the point A(1, -4) and B(-1, 0)

Slope (m) = (y2 - y1) / (x2 - x1)

x1 = 1, y1 = -4, x2 = -1 and y2 = 0

Slope (m) = (0 - (-4))/(-1 - 1)

= 4/-2

Slope = -2

Thank you.

6 0

3 years ago

Read 2 more answers

Someone please help me I’ve been trying all day!<br><br> NO LINKS , answers only <br> I MEAN IT!!!!

DedPeter [7]

Answer:

Step-by-step explanation:

You want to set up a system. We are given some points fo we can set up inputs and outputs. F(0) = 5, given by point (0,5). and F(3)=135, given by point (3,135). The exponential function general equation is F(x)= a(b)^x. plug in out functions into the general equation to get a(b)^0=5 and a(b)^3=135. Then you simplify and solve for a. In the equation a*b^0=5, anything to the power of 0 is 1. so a*1=5. Therefore a=5. we plug in the a to the second equation to find b. 5*b^3=135. b^3=27. b=3 from simplifying. Therefore, the equation should be 5(3)^x. THATS IT!

3 0

3 years ago

Solve:<br><br> a.x3+x2-8x - 12x+2<br> b. x3-4x2-3x + 18x-3<br> c. x2 + 4x + 4<br> d. x2 - 6x + 9

NeTakaya

<h2>Steps:</h2>

So firstly, this function is asking us to divide f(x) by g(x) so let's set that up as such:

$(\frac{f}{g})(x)=\frac{x^2-x-6}{1}\div \frac{x-3}{x+2}$

Next, remember that <u>dividing by a number is the same as multiplying by its reciprocal.</u> To find the reciprocal of a number, flip the numerator and denominator around. With this info, flip the second fraction to it's reciprocal and change the sign to multiplication:

$(\frac{f}{g})(x)=\frac{x^2-x-6}{1}\times \frac{x+2}{x-3}$

Next, we are going to factor x² - x - 6. Firstly, what two terms have a product of -6x² and a sum of -x? That would be -3x and 2x. Replace -x with 2x - 3x:

$(\frac{f}{g})(x)=\frac{x^2+2x-3x-6}{1}\times \frac{x+2}{x-3}$