Answer:
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!!!
Answer:
26
2
/3 cm3
Step-by-step explanation:
V = (6
2
3
)(2)(2) = 26
2
3
Can also be solved by counting the fractional cubes.
Given:
To rewrite using distributive property, distribute the 28 to all values inside the parenthesis.
Thus, we have:
ANSWER:
28y - 4
Answer:
Step-by-step explanation:
x + 3y = 24.....multiply by -1
5x + 3y = 36
------------------
-x - 3y = -24 (result of multiplying by -1)
5x + 3y = 36
------------------add
4x = 12
x = 12/4
x = 3 <=======
x + 3y = 24
3 + 3y = 24
3y = 24 - 3
3y = 21
y = 21/3
y = 7 <========
Answer:
(4z-5a)(2xy)
Step-by-step explanation:
8xyz-10axy
Check for similar terms/like terms
x and y are the same.
The hcf or highest common factor is 2.
The z and a are not common in both terms.
Factor.
8xyz-10axy
2xy(4z-5a)
(4z-5a)(2xy)
