In triangle ABC, angle B = 90 degrees. Semicircles are constructed on sides AB, AC, and BC, as shown below. Show that the total area of the shaded region is equal to the area of triangle ABC.

Explanation:

The area of a semicircle is given by ...

  A = πr^2/2

where r is the radius. Here, we're given diameters, so in terms of diameter, the area of a semicircle is ...

  A = π(d/2)^2/2 = (π/8)d^2

__

The area of the semicircle with diameter AC is ...

 white area = (π/8)AC^2

The area of the semicircle with diameter BC is ...

  left semicircle area = (π/8)BC^2

And the area of the semicircle with diameter AB is ...

  right semicircle area = (π/8)AB^2

__

We can use the relationship between the areas to find the shaded area:

  triangle area + left semicircle area + right semicircle area =

     white area + shaded area

Then the shaded area is ...

  shaded area = triangle area + left semicircle area + ...

     right semicircle area - white  area

__

Filling in the values for area from above, we have ...

  shaded area = triangle area+ (π/8)BC^2 +(π/8)AB^2 -(π/8)AC^2

  shaded area = triangle area + (π/8)(BC^2 +AB^2 -AC^2)

From the Pythagorean theorem, we know that ...

  AC^2 = BC^2 +AB^2

Substituting this into the above equation gives ...

  shaded area = triangle area + (π/8)((Bc^2 +AB^2 -(BC^2 +AB^2))

  shaded area = triangle area + 0 . . . . simplify

  shaded area = triangle area

just find the gcf and solve the gcf's until you get each answer i have a really good feeling somethings missing xd (like the dimensions of the fence) *cough *cough *hint *hint

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