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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

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 %

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Lelechka [254]

Answer:

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

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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