Find the center and the radius of the circle given the equation of a circle
below.

(x - 2)^2 + (y – 5)^2 = 49

A: Center: (2,5) & Radius=7

B: Center: (5,2) & Radius=49

C: Center: (-2,-5) & Radius=49

D: Center: (-5,-2) & Radius= 7

Option A) Center: (2,5) & Radius=7 is the correct one.

Step-by-step explanation:

The general form of the equation of the circle is given as :

⇒ (x-h)² + (y-k)² = r²

where,

Therefore, you need to compare the given equation with the general equation of the circle.

The given equation of circle is (x-2)² + (y-5)² = 49

From the given equation, it can be determined that

h = 2 and y = 5 and r² = 49

The center (h,k) of the circle is (2,5).

To find the radius r :

⇒ r² = 49

⇒ r = 7

The radius of the circle is 7.

∴ Option A) Center: (2,5) & Radius=7 is the correct one.

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

answerthe vertex form

step-by-step explanation: