What is the formula for the area of a circle?

A sprinkler sprays a section of 200 degrees with a radius of water reaching 5 meters.

beks73 [17]

Answer: The area that it covers is 78.53 square meters

Step-by-step explanation:

Hi, to answer this question we have to apply the next formula:

Area of a circle (A): π r²

Where:

r = radius (in our case is 5 meters)

So, replacing with the values given:

A= π (5)² = π 25 =78.53 square meters

In conclusion, the area that it covers is 78.53 square meters

Feel free to ask for more if needed or if you did not understand something.

7 0

2 years ago

line segment AB is a directed segment with endpoints A(2,2) and B(14,14). In two or more complete sentences, explain how you wou

Anit [1.1K]

Answer: Section formula

Step-by-step explanation:

Given

The line segment AB with endpoints $A(2,2)$ and $B(14,14)$ is divided by a point X in the ratio 1:2

To verify that X divides the line segment in the ratio 1:2, we will use section formula. Section formula tells the coordinates of the point which divide the line segment in the certain ratio or vice-versa.

We will assume that X divides AB in k:1 and insert the values in the section formula. If the ratio k:1 comes out to be 1:2, then it verifies the claim.

8 0

3 years ago

Find the largest possible area of a rectangle with base on x-axis and vertices on curve y = 4 - x^2

astraxan [27]

The region bounded by the parabola and the x-axis is symmetric, so any inscribed rectangle whose base lies in the x-axis will have its base extend symmetrically around the origin, i.e. if the base has length $2a$ , with $a>0$ , then its base is the line segment connecting the points $(-a,0)$ and $(a,0)$ .

The height of such a rectangle will then by $4-(\pm a)^2=4-a^2$ .

The area of such a rectangle is then a function of $a$ :

$A(a)=2a(4-a^2)=8a-2a^3$

Differentiating with respect to $a$ gives

$A'(a)=8-6a^2$

which has critical points at

$8-6a^2=0\implies a=\pm\dfrac2{\sqrt3}$

We omit the negative root. Checking the sign of the second derivative at the positive critical point (it's negative) confirms that $a=\dfrac2{\sqrt3}$ is the site of a local maximum.

This means the largest area of this rectangle is

$A\left(\dfrac2{\sqrt3}\right)=\dfrac{32}{3\sqrt3}$