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Y_Kistochka [10]
3 years ago
10

The following information was reported in an article in a newspaper. A sheriff reported that investigators found 7.5 kilograms o

f cocaine in a car belonging to a man who lives in Virginia. This is a significant amount of cocaine, having a street value of about $1.4 million dollars. Assume that a typical dose contains half a gram of pure cocaine. How much would a dose cost
Mathematics
1 answer:
ruslelena [56]3 years ago
5 0

Answer:

  $93.33 per dose

Step-by-step explanation:

 $1.4×10⁶/(7.5 kg) × (1 kg)/(1000 g) × (0.5 g)/(1 dose)

  = (1.4×10⁶×0.5)/(7.5×10³) $/dose = 700/7.5 $/dose ≈ $93.33/dose

__

Converting dollars per kilogram to dollars per dose involves units conversion from kilograms to doses. We have done it in two steps, first to grams, then to doses.

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Given that G=ab find the percentage increase in G when both a and b<br> increase by 10%
monitta

The percentage change in G is 21 %

<h3>What is Percentage change ?</h3>

Percentage change is defined as the increase or decrease in the value as compared to the original value multiplied by 100.

It is given that

G = ab

when a is increased by 10% the new a will be = 1.1 a

When b is increased by 10% the new b will be 1.1 b

So,

G' = 1.1a *1.1 b

G' = 1.21 ab

G' = 1.21

(G' - G)*100/G = (1.21-1)*100/1

The percentage change is 21 %

To know more about  percentage change

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3 0
2 years ago
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Lana71 [14]

Answer:

Im not sure but, i would see what each corner is each.

6 0
2 years ago
Please help me with these, oh sweet jesus
Lelechka [254]

Answer:

77.  \cot^{6} x = \cot^{4} x \csc^{2}x - \cot^{4} xProved

78.  \sec^{4}x \tan^{2} x = \sec^{2}x [\tan^{2}x + \tan^{4}x ] Proved

79. \cos^{3} x\sin^{2} x = [\sin^{2}x - \sin^{4}x] \cos x Proved.

80. \sin^{4}x - \cos^{4}x = 1 - 2\cos^{2}x + 2 \cos^{4} x Proved.

Step-by-step explanation:

77. Left hand side

= \cot^{6} x

= \cot^{4} x \times \cot^{2} x

= \cot^{4}x [\csc^{2}x - 1]  

{Since we know, \csc^{2} x - \cot^{2}x = 1}

= \cot^{4} x \csc^{2}x - \cot^{4} x  

= Right hand side (Proved)

78. Left hand side

= \sec^{4}x \tan^{2} x

= \sec^{2} x [1 + \tan^{2}x] \tan^{2} x  

{Since \sec^{2}x - \tan^{2}x = 1}

= \sec^{2}x [\tan^{2}x + \tan^{4}x ]

= Right hand side (Proved)

79. Left hand side  

= \cos^{3} x\sin^{2} x

= \cos x[1 - \sin^{2} x] \sin^{2} x

{Since \sin^{2}x + \cos^{2} x = 1}

= [\sin^{2}x - \sin^{4}x] \cos x

= Right hand side

80. Left hand side  

= \sin^{4}x - \cos^{4}x

= [\sin^{2}x + \cos^{2}x]^{2} - 2\sin^{2} x \cos^{2}x

{Since \sin^{2}x + \cos^{2} x = 1}

= 1 - 2\cos^{2} x[1 - \cos^{2}x ]

= 1 - 2\cos^{2}x + 2 \cos^{4} x

= Right hand side. (Proved)

7 0
3 years ago
In the fallowing diagram, it is shown that mEHG is a right angle, mEHF = x+ 1 and mGHF = 5x+5. Find the value of x, the numerica
GalinKa [24]

Answer:

x = 14

EHF = 15 degree  GHF = 75 degree  

Step-by-step explanation:

(x+1) + (5x+5) = 90

(x+5x)+ (1+5) = 90

6x + 6 = 90

6x = 90-6

6x = 84

x = 84/6

x = 14

EHF + GHF = 90 degree

EHF = x+1

       = 14 + 1 = 15

       = 15 degree

GHF = 5x+5

        = 14 x  5 (+5) = 70 + 5

        = 75 degree

EHF = 15 degree  GHF = 75 degree  

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2 years ago
A is a prorportional three dimensional model of an object
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Scale factor mark brainliest plz
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