In a plane, points p and q are 20 inches apart. if point r is randomly chosen from all the points in the plane that are 20 inches from p, what is the probability that r is closer to p than it is to q?
a. 00
b. 1414
c. 1313
d. 1212
e. 23

E.

Step-by-step explanation:

Consider a circle having centre P and radius 20 inches,

So, the area of the circle,

Also, suppose points Q and R are on the circumference,

i.e. PQ = PR = 20 inches,

If QR = 20 inches,

So, triangle PQR is an equilateral triangle,

⇒ m∠RPQ = 60°,

Now, suppose R' is another point on the circle,

Such that, ΔPQR ≅ Δ PQR',

⇒ m∠QPR' = 60°,

Thus, minor angle, m∠RPR' = m∠RPQ + m∠QPR' = 60° + 60° = 120°,

⇒ major angle, m∠RPR' = 360° - 120° = 240°,

So, the area of the circle where a point on the circumference is closer to P than it is to Q

Hence, the probability that a point is closer to P than it is to Q =

i.e. OPTION E is correct.

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

the equation is not included but the answer is (1.50, 2.00, 2.75)

step-by-step explanation: