An isosceles right triangle has legs of equal length if the hypotenuse is 10 cm long find the length of each leg

Since this is a right triangle, we know we have to use the Pythagorean Theorem: A² + B² = C² , where A and B = the legs of the triangle (which are shorter than the hypotenuse) and C = the hypotenuse.

In this isosceles triangle, A = B (because the legs are the same size) and C is 10cm longer than A and B.

Thus we get the following,

A = x

B = x

C = x + 10

We simply plug in these variables into the equation and solve for x:

x² + x² = (x+10)²

2x² = x² + 20x + 100

x² - 20x - 100 = 0

We solve for x by completing the square:

(x² - 20x + 100) - 100 = 100

(x-10)² -100 = 100

(x-10)² = 200

x - 10 = √(200)

x - 10 = 10 √2

x = 10 + 10√2 = 24

Finally,

A ≈ 24 (or 10 + 10√2)

B ≈ 24 (or 10 + 10√2)

C ≈ 24 + 10 ≈ 34 (or 10 + 10√2 + 10 = 20 + 10√2)

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

need points to ask question

step-by-step explanation: