Will give brainliest in this unit, you calculated the surface area of solid figures composed of polygons, such as rectangles and triangles. but imagine finding the surface area of a cylinder—a solid figure that has several curved surfaces. a cylinder is made up of two circular bases and a curved (lateral) surface in between. develop a formula for finding the surface area of the outside of a cylinder based on what you know about the circumference of a circle and the areas of circles and rectangles. (hint: think about rolling a rectangular piece of paper into the shape of a cylinder.) discuss your approach and why it works.

The area for a circle, is A = πr^2

There is a top and bottom circle so you would multiply that by 2 to get 2πr^2

Now to find the area of the lateral surface you need to find the circumference of the circle ( which would be the length of a rectangle)

The formula for circumference is 2πr, then you need to multiply that by the height so you would have 2πrh ( this is equivalent to length x width).

Now add both equations together to get total surface area:

Surface area = 2πr^2 + 2πrh

You can use the formula [2(πr^2)+(2πrh) to find the surface area of a cylinder. First, you find the area of the circles, so 2(πr^2). You multiply it by two because there are two circles. Then, you have to find the area of the lateral surface. You can do this by finding the area of one of the circles and then multiply it by the height of the cylinder, so (2πrh). Finally, you just add both of those numbers to get the surface area of a cylinder. This works because when you find the surface area, you just find the area of what you can see, not the inside too. This equation does that and only finds the surface area of the cylinder.

Step-by-step explanation:

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2