To overcome an infection an anti-biotic is injected into David’s bloodstream. After the injection the anti-biotic in his body decreases at a rate proportional to the amount, Q(t), present at time t. If the initial injection was 8 ccs and 5 ccs remain after 4 hours, estimate how many ccs will remain after 5 hours?

4.45 ccs will remain after 5 hours.

Step-by-step explanation:

The anti-biotic in his body decreases at a rate proportional to the amount, Q(t), present at time t.

(dQ/dt) = - kQ ((Minus sign because it's a rate of reduction)

(dQ/dt) = -kQ

(dQ/Q) = -kdt

 ∫ (dQ/Q) = -k ∫ dt 

Solving the two sides as definite integrals by integrating the left hand side from Q₀ to Q and the Right hand side from 0 to t.

We obtain

In (Q/Q₀) = -kt

(Q/Q₀) = e⁻ᵏᵗ

Q(t) = Q₀ e⁻ᵏᵗ

the initial injection was 8 ccs and 5 ccs remain after 4 hours

Q₀ = 8 ccs,

At t = 4 hours, Q = 5 ccs

5 = 8 e⁻ᵏᵗ

e⁻ᵏᵗ = 0.625

-kt = In (0.625) = -0.47

-4k = 0.47

k = 0.1175 /hour

Q(t) = Q₀ e⁻⁰•¹¹⁷⁵ᵗ

At t = 5 hours, Q = ?

Q = 8 e⁻⁰•¹¹⁷⁵ᵗ

0.1175 × 5 = 0.5875

Q = 8 e(^-0.5875)

Q = 4.45 ccs

Hope this Helps!!!

The surface area would be 768 cm².

Area is a square measurement; therefore to find the ratio of the areas, we square the ratio of the linear measurements:

(5/8)²=5²/8²=25/64

Now we set up a proportion to solve for the missing surface area:

25/64 = 300/x

Cross multiply:
25*x = 300*64
25x = 19200

Divide by 25:
25x/25 = 19200/25
x = 768