Sphere P has a radius of 5 inches. Find the area of the cross section formed by the intersection of a plane 3 cm above the center of sphere P and parallel to its equator.
O None of the other answers are correct
O 16pi in
O4pi in
O 7pi in
O 9pi in

pi*16 in^2

Step-by-step explanation:

In the image, we can see a triangle rectangle.

Where one cathetus is equal to 3 inches, the hypotenuse is equal to the radius of the sphere, 5 inches, and the other cathetus is the radius of the cross-section of which we can find the area.

Using the Pythagorean theorem, we will have that:

x^2 + (3in)^2 = (5in)^2

x = √( (5in)^2 - (3in)^) = 4 in

Then the radius of the cross-section is 4 inches.

And we know that the area of a circle of radius R is:

A = pi*R^2

Then the area of the cross-section will be:

A = pi*(4in)^2 = pi*16 in^2

Then the correct option is the first one.

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!