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2 years ago

A doctor asks a nurse to give a

Mathematics

2 answers:

AlekseyPX2 years ago

3 0

Answer:

Hey there!

For this question, we write a proportion.

$\frac{40}{0.6}=\frac{250}{x}$

$40x=150$

$x=3.75$

Thus, the nurse to give the patient exactly 3.75 ml of the solution.

Let me know if this helps :)

vova2212 [387]2 years ago

3 0

Answer:

3.75 ml of the solution.

Step-by-step explanation:

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2 years ago

Find f. f ''(x) = −2 + 36x − 12x2, f(0) = 8, f '(0) = 18 f(x) =

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Integrate f ''(x) = −2 + 36x − 12x2 with respect to x:

f '(x) = -2x + (36/2)x^2 - (12/3)x^3 + c. Find c by letting x = 0 and using f(0)=8.

Then f '(0) = -2x + 18x^2 - 4x^3 + c = 18 (which was given).

Then -0 + 0 - 0 + c = 18, so c = 18 and

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2 years ago

Hailey paid $13\$13 $13 dollar sign, 13 for 137 kg1\dfrac3{7} \text{ kg} 1 7 3 kg 1, start fraction, 3, divided by, 7, end fra

Colt1911 [192]

Answer:

$3.90 was the cost of salami per kilogram.

Step-by-step explanation:

The question is incomplete, so the complete question is below:

$Hailey\ paid\ \$13\ for\ 1\ 3/7\ kg\ of\ sliced\ salami.\ \\What\ was\ the\ cost\ per\ kilogram\ of\ salami?$

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Now, to get the cost of salami per kilogram we use unitary method:

If, the cost of $\frac{10}{3}$ kilogram of salami = $13.

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2 years ago

A tank contains 500 gallons of salt-free water. A brine containing 0.25 lb of salt per gallon runs into the tank at the rate of

Neporo4naja [7]

Answer:

0.0198 lbs per gallon

Step-by-step explanation:

amount of salt free water = 500 gallons

salt rate in = 1 gal/min

salt rate out = 1 gal/min

amount o salt in brine = 0.25 lb per gallon

Let the amount of salt in the tank be A(t) at any time t.

$\frac{dA(t)}{dt} =salt rate in - salt rate out$

salt rate in = 0.25 x 1 = 0.25

salt rate out = $\frac{A(t)\times 1}{500}$

The differential equation is given by

$\frac{dA(t)}{dt} =1 - \frac{A(t)\times 1}{500}$

where, A(0) = 0

So, the equation becomes

$\frac{dA(t)}{dt} + \frac{A(t)}{500} = 1$

Here the integrating factor is $e^{\frac{dt}{500}}=e^{\frac{t}{500}}$

The solution of the above differential equation is given by

$A(t)\times e^{\frac{t}{500}} = \int e^{\frac{t}{500}}dt$

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where, C is the integrating constant.

$A(t)=500+Ce^{-\frac{t}{500}}$

Put, A(0) = 0

C = - 500

$A(t)=500\left ( 1-e^{-\frac{t}{500} \right )$

As concentration is defined as

Concentration = Quantity / Volume

$C(t)=\frac{A(t)}{500}$

$C(t)=1-e^{\frac{-t}{500}}$

Now concentration at t = 10 min

Put, t = 10 min

$C(10)=1-e^{\frac{-10}{500}}$

C (10) = 0.0198 lbs per gallon

Thus, teh concentration after 10 min is 0.0198 lbs per gallon.

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2 years ago

Cindy found a collection of baseball cards in her attic worth $8,000. the collection is estimated to increase in value by 1.5% p

Sergio [31]

A]
Exponential function is given by the form:
y=a(b)ˣ
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b=growth factor
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2 years ago