The quantity of the medicine left in the patient's system after 2 hours is 18 mg.
Consider the function:
where m is the number of times this growth/decay occurs, a = initial amount, and r = fraction by which this growth/decay occurs.
It should be noted that only 60% of the medicine of the previous hour is left n the patient's system every hour.
Thus, the model of the scenario,
where D is the dosage at any hour n.
From the model above with n equal to 2 then D becomes 18.
Hence, The quantity of the medicine left in the patient's system after 2 hours is 18 mg.
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The glass needed to frame the poster is 1500 square inches.
Given,
Matt is framing his favorite poster.
The poster is 30 incheswide and 50 inches long.
We need to find out how many square inches of glass will he need toframe the poster.
It is given by:
Area = Length x Width
We have,
Length = 30 inches
Width = 50 inches
We see that it is a rectangle.
Area of rectangle:
= Length x Width
= 30 x 50
= 1500 square inches.
Thus the glasses needed to frame the poster is 1500 square inches.
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(B) f(t) = 4(t − 1)2 + 3; the minimum height of the roller coaster is 1 meter from the ground
(C) f(t) = 4(t − 1)2 + 2; the minimum height of the roller coaster is 2 meters from the ground
(D) f(t) = 4(t − 1)2 + 2; the minimum height of the roller coaster is 1 meter from the ground
The function is a parabola, and the problem asks to transform the equation into f(t)=a(x-h)2 + k
Given f(t) = 4t2 -8t +7
= (4t2 - 8t + 4) + 7 - 4
=4 (t2 - 2t + 1) + 3
= 4 (t-1) 2 +3
This removes C and D from the viable choices.
Differentiating the f(t),
f’(t) = 8t – 8, the maximum/minimum value occurs at f’(t) = 0
0 = 8t – 8
t = 1
determining if maximum or minimum, f”(t) > 0 if minimum, f”(t) < 0 maximum
f”(t) = 8 > 0, therefore minimum
f(1) =4(1)^2 – 8(1) +7
= 3
Therefore, minimum height is 3.
Answer:
T= 4 s (second option).
Answer:
440
Step-by-step explanation:
Prime factors of 40 = 2 x 2 x 2 x 5
Prime factors of 220 = 2 x 2 x 5 x 11
Common factors = 2² x 5
Prime factors of 40 (excluding common factors with 220) = 2
Prime factors of 220 (excluding common factors with 40) = 11
Multiply the common factors with remaining factors
2² x 5 x 2 x 11
= 440
Answer:
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Step-by-step explanation: