Given: ∆AMK, MP ⊥ AK , MP = 10 m∠A = 72º, m∠PMK = 50° Find AM, MK, AK
1 answer:
Answer:
Step-by-step explanation:
The diagram of triangle AMK is shown on the attached photo. To determine AM, we would apply trigonometric ratio since triangle AMP is a right angle triangle.
Sin# = opposite/hypotenuse
Sin 72 = 10/AM
AMSin72 = 10
AM = 10/Sin72 = 10/0.9511
AM = 10.51
To determine MK,
Cos# = adjacent/hypotenuse
Cos 50 = 10/MK
MKCos50 = 10
MK = 10/Cos50 = 10/0.6428
MK = 15.6
AK = AP + KP
Tan# = opposite/adjacent
Tan 72 = 10/AP
AP tan 72 = 10
AP =10/tan72 = 10/ 3.0777 = 3.25
Tan 50 = KP/10
KP = 10tan50
KP= 10× 1.1918 = 11.918
Therefore,
AK = 3.25 + 11.918 = 15.168
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Answer:
- h ≥ 0
- C = 0; concentration eventually decays to nothing
- (0, 0) is the only intercept in the domain. It means the concentration in the bloodstream is zero at the time the drug is injected.
- 1.95 hours
Step-by-step explanation:
1. a. The domain of the function in the context of this problem is h ≥ 0. The maximum value of h will correspond to the time at which the concentration is considered to be negligible. If that time is when the concentration decays to 1% of its peak value, then perhaps the suitable domain is 0 ≤ h ≤ 180.
b. The equation of the vertical asymptote of this function is where the denominator of C(h) is zero, that is ...
h^3 +8 = 0
h = ∛(-8)
h = -2 . . . . the equation of the vertical asymptote
This is not a concern for a medical professional because it is outside the domain of the function.
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2. As h gets large the value of the function approaches 2/h, which is to say the horizontal asymptote is C = 0. It means the concentration of medication in the blood eventually decays to zero.
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3. C(0) = 0 is the only intercept in the domain of the function. (h, C(h)) = (0, 0)
In the context of this problem, it means the blood concentration of medication is zero at the time of the injection.
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4. About 1.95 hours after injection, a maximum concentration of the drug occur in the bloodstream. This value is easily found using a graphing calculator.
Taking the derivative of the function gives you ...
C'(h) = -2(h^4 +5h^3 -16h -20)/(h^3 +8)^2
This has two real zeros and two complex zeros. The positive real zero is near h = 1.94527, about 1.95.
That is, the concentration in the bloodstream reaches a maximum about 1.95 hours after injection into the muscle.
