Prove in full that the square 9.9cm diameter fits the circle.
Prove here again if you can help id appreciate it

1/2 the points are repeated on another question on my page.
Total including this is 40 points. + Brainliest.
Others unanswered 100 points in 3 questions.

see below

Step-by-step explanation:

We need to find the diameter of the square

We can find this using the Pythagorean theorem

a^2+b^2 = c^2  where the legs are 7 and 7 and the diameter is c

7^2 +7^2 = c^2

49+49 = c^2

98 = c^2

Taking the square root of each side

sqrt(98) = sqrt(c^2)

9.899494937 = c

Since 9.9 is less than 11 which is the diameter of the circle it will never touch the circle.

Since the longest part of the square is less than the diameter of the circle, the square will fit  inside the circle without touching

Diagonal < Diameter

Step-by-step explanation:

The square will fit inside without touching if the diagonal length is less than the diameter

Diagonal = sqrt(7² + 7²)

7sqrt(2)

Diameter = 11

7sqrt(2) < 11

Because

(7sqrt(2))² < 11²

98 < 121

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

i belive those are the ones that are under 90 degreese.