Show that A(t)=300−250e0.2−0.02t satisfies the differential equation ⅆAⅆt=6−0.02A with initial condition A(10)=50 .

Given the expression

Factor out 0.02 from the xpression to have:

Separate the variables;

Integrate both sides of the eequation to have:

Apply the initial condition to get the value of C

Substitute the constant into the function A(t) to get the required proof

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Step-by-step explanation:

dA/dt = 6 − 0.02A

dA/dt = -0.02 (A − 300)

Separate the variables.

dA / (A − 300) = -0.02 dt

Integrate.

ln(A − 300) = -0.02t + C

Solve for A.

A − 300 = Ce^(-0.02t)

A = 300 + Ce^(-0.02t)

Use initial condition to find C.

50 = 300 + Ce^(-0.02 × 10)

50 = 300 + Ce^(-0.2)

-250 = Ce^(-0.2)

C = -250e^(0.2)

A = 300 − 250e^(0.2) e^(-0.02t)

A = 300 − 250e^(0.2 − 0.02t)

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

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step-by-step explanation: