Given: △abc is a right triangle there are 3 shaded squares with sides a, b, and c, respectively. a, b, and c are also the lengths of the sides of the right triangle, such that the area of the square with side a is a2 and the area of the square with side b is b2 and the area of the square with side c is c2. prove: a2 + b2 = c2 (pythagorean theorem) proving which of the following will prove the pythagorean theorem? a when you subtract the area of the smallest square from the medium square the difference equals the area of the largest square. b the sides of a right triangle are also the sides of squares. c m∠a+m∠b=m∠c d the area of the two smaller squares will add up to the area of the largest square.

a2 + b2 = c2 because a + b = c

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

how do heroes resolve conflicts in beowulf?

question 15 options:

they sue each other in court

they get a mediator

the seek counseling

they get even or take revenge

step-by-step explanation: