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pochemuha
2 years ago
15

A certain drug dosage calls for 17 mg per kg per day and is divided into two doses (1 every 12 hours). If a person weighs 255 po

unds, how many milligrams of the drug should he receive every 12 hours? Round your dosage to the tenth of a milligram.
Mathematics
1 answer:
vekshin12 years ago
3 0

Answer:

985.15mg

Step-by-step explanation:

Calculation for how many milligrams of the drug should he receive every 12 hours

First step is to calculate the weight in kg

255 lbs = 255/2.2 = 115.9 kg

Second step is multiply mg/kg in order to get total medicine per day

Dose per day=115.9kg × 17 mg

Dose per day= 1,970.3

Third step is to calculate how many milligrams of the drug should he receive by dividing the day dose by 2

Dose milligrams=1,970.3/2

Dose milligrams=985.15mg

Therefore how many milligrams of the drug should he receive every 12 hours will be 985.15mg

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Step-by-step explanation:

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B

Step-by-step explanation:

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Eight years ago, Clyde was 7 years old. Which equation represents Clyde age now?
vichka [17]

Answer:

D.

Step-by-step explanation:

Clyde's present age is 7+8, because he has aged 8 years since he was 7.

So, the equation would be C=7+8.

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If you subtract 8 from both sides of the equation C=7+8, you get C-8=7, which is D.

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3 years ago
Verify each trigonometric equation by substituting identities to match the right hand side of the equation to the left hand side
lorasvet [3.4K]

Answer:

Step-by-step explanation:

1.

cot x sec⁴ x = cot x+2 tan x +tan³x

L.H.S = cot x sec⁴x

       =cot x (sec²x)²

       =cot x (1+tan²x)²     [ ∵ sec²x=1+tan²x]

       =  cot x(1+ 2 tan²x +tan⁴x)

       =cot x+ 2 cot x tan²x+cot x tan⁴x

        =cot x +2 tan x + tan³x        [ ∵cot x tan x =\frac{ \textrm{tan x }}{\textrm{tan x}} =1]

       =R.H.S

2.

(sin x)(tan x cos x - cot x cos x)=1-2 cos²x

 L.H.S =(sin x)(tan x cos x - cot x cos x)

          = sin x tan x cos x - sin x cot x cos x

           =\textrm{sin x cos x }\times\frac{\textrm{sin x}}{\textrm{cos x} } - \textrm{sinx}\times\frac{\textrm{cos x}}{\textrm{sin x}}\times \textrm{cos x}

           = sin²x -cos²x

           =1-cos²x-cos²x

           =1-2 cos²x

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3.

1+ sec²x sin²x =sec²x

L.H.S =1+ sec²x sin²x

         =1+\frac{{sin^2x}}{cos^2x}                       [\textrm{sec x}=\frac{1}{\textrm{cos x}}]

         =1+tan²x                        [\frac{\textrm{sin x}}{\textrm{cos x}} = \textrm{tan x}]

         =sec²x

        =R.H.S

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\frac{\textrm{sinx}}{\textrm{1-cos x}} +\frac{\textrm{sinx}}{\textrm{1+cos x}} = \textrm{2 csc x}

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-tan²x + sec²x=1

L.H.S=-tan²x + sec²x

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        =\frac{1}{cos^2x} -\frac{sin^2x}{cos^2x}

        =\frac{1- sin^2x}{cos^2x}

        =\frac{cos^2x}{cos^2x}

        =1

     

       

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(2,-2) and (4,1).<br> Y=
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Answer:

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