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Fudgin [204]
4 years ago
11

At time (t) hours after taking the cough suppressant hydrocodone bitartrate, the amount, A, in mg, remaining in the body is give

n by A = 14(0.8)^t.. What percent of the drug leaves the body each hour?. How much of the drug is left in the body 2 hours after the dose is administered? How long is it until only 1 mg of the drug remains in the body?
Mathematics
1 answer:
Mice21 [21]4 years ago
5 0
1)we calculate the amount of the drug is left in the body 2 hours after the dose is administered.
t=2
A(t)=14(08)^t
A(t)=14(0.8)²=14(0.64)=8.96

Answer: after of two hours the amount of drug left in te body is 8.96 mg.

2)we calculate the time required to stay in the body 1 mg of drug.

A(t)=1
14(0.8)^t=1   
ln [14(0.8)^t]=ln1
ln14+tln(0.8)=0
tln(0.8)=-ln 14
t=-ln 14 / ln (0.8)
t=11.8267...≈11.83

Answer: the time required is 11.83 hours  (≈11 hours, 49 minutes, 48 seconds).

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What is the perimeter of a polygon with vertices at (−2, 1) , ​ (−2, 4) ​, (2, 7) , ​ (6, 4) ​, and (6, 1) ​?
alekssr [168]
First we need the distances of the sides of the polygon, because perimeter = sum of all sides.
d =  \sqrt{{(y2 - y1)}^{2} + {(x2 - x1)}^{2} }
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d3 \: = \sqrt{{(7 - 4)}^{2} + {(2 - 6)}^{2} } \\  =  \sqrt{{(3)}^{2} +  {( - 4)}^{2} } =  \sqrt{9 + 16}  \\  =  \sqrt{25} \:  = 5
d4 = \sqrt{{(4 - 1)}^{2} + {(6 - 6)}^{2} } \\  =  \sqrt{ {(3)}^{2} +  {(0)}^{2}  }  =  \sqrt{9} \:  = 3
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3 years ago
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worty [1.4K]

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Step-by-step explanation:

<em>First, we need to take our given numbers</em>

<em>-12x+5x-x</em>

<em>Now let's start with the start of the equation</em>

<em>-12x + 5x = -7x</em>

<em>Now we take -7x and -x</em>

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3 years ago
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In the figure shown, triangle RST undergoes reflections across two line. R''S''T’’ is the final image of RST
soldi70 [24.7K]
<h3>Answer: D.  y = -x</h3>

======================================================

Explanation:

Point T is at (-2,1)

When we reflect it over the line x = 0, aka the y axis, we use the rule (x,y) \to (-x,y) so T(-2,1) becomes T ' (2, 1). The x coordinate flipped in sign, while the y coordinate stays the same.

Then the final transformation is reflecting over y = -x using the rule (x,y) \to (-y, -x). Therefore, the point (2,1) moves to (-1, -2) which is where T'' is located in the diagram.

You apply the same two transformations for the points R and S to get R'' and S'' respectively.

Note: A composition of two reflections, where the lines of reflection aren't parallel, form a rotation. In this case, we have a 90 degree counterclockwise rotation when going from triangle RST to triangle R''S''T''.

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