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2 years ago

6

A premature infant weighing 3lbs is ordered to 1 milligram per gram of a drug. How many milligrams should you administer to the

patient?

1 answer:

Alona [7]2 years ago

4 0

Answer:

1360.78 mg

Step-by-step explanation:

3 pounds in grams is 1360.78 grams.

so, 1 mg per gram would mean the baby should recieve 1360.78 mg of the drug.

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1. (5pts) Find the derivatives of the function using the definition of derivative.

andreyandreev [35.5K]

2.8.1

$f(x) = \dfrac4{\sqrt{3-x}}$

By definition of the derivative,

$f'(x) = \displaystyle \lim_{h\to0} \frac{f(x+h)-f(x)}{h}$

We have

$f(x+h) = \dfrac4{\sqrt{3-(x+h)}}$

and

$f(x+h)-f(x) = \dfrac4{\sqrt{3-(x+h)}} - \dfrac4{\sqrt{3-x}}$

Combine these fractions into one with a common denominator:

$f(x+h)-f(x) = \dfrac{4\sqrt{3-x} - 4\sqrt{3-(x+h)}}{\sqrt{3-x}\sqrt{3-(x+h)}}$

Rationalize the numerator by multiplying uniformly by the conjugate of the numerator, and simplify the result:

$f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x} - 4\sqrt{3-(x+h)}\right)\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x}\right)^2 - \left(4\sqrt{3-(x+h)}\right)^2}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16(3-x) - 16(3-(x+h))}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16h}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}$

Now divide this by <em>h</em> and take the limit as <em>h</em> approaches 0 :

$\dfrac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ \displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-x}\left(4\sqrt{3-x} + 4\sqrt{3-x}\right)} \\\\ \implies f'(x) = \dfrac{16}{4\left(\sqrt{3-x}\right)^3} = \boxed{\dfrac4{(3-x)^{3/2}}}$

3.1.1.

$f(x) = 4x^5 - \dfrac1{4x^2} + \sqrt[3]{x} - \pi^2 + 10e^3$

Differentiate one term at a time:

• power rule

$\left(4x^5\right)' = 4\left(x^5\right)' = 4\cdot5x^4 = 20x^4$

$\left(\dfrac1{4x^2}\right)' = \dfrac14\left(x^{-2}\right)' = \dfrac14\cdot-2x^{-3} = -\dfrac1{2x^3}$

$\left(\sqrt[3]{x}\right)' = \left(x^{1/3}\right)' = \dfrac13 x^{-2/3} = \dfrac1{3x^{2/3}}$

The last two terms are constant, so their derivatives are both zero.

So you end up with

$f'(x) = \boxed{20x^4 + \dfrac1{2x^3} + \dfrac1{3x^{2/3}}}$

8 0

2 years ago

patriot [66]

Answer:

Let me take 2,3,3 be 2x, 3x, 3x

Colin = 2x

Natasha = 3x

Angad = 3x

The equation forms to find out the value of x

2x + 3x + 3x = 48

=> 8x = 48

=> x = 48/8

=> x = 6

So share of Colin = 2x = £(2×6) = £12

Share of Natasha = 3x = £(3×6) = £18

share of Angad = 3x = £(3×6) = £18

For proof

£12 + £18 + £18 = £48

Hope it helps

8 0

2 years ago

Which expressions are equivalent to the one below? Check all that apply. 2^5 • 2^x

umka21 [38]

When you multiply exponential expressions that have the same base, you simply add the exponents. This is true for both numbers and variables. When you include other numbers in the multiplication, you simply factor and break the expression into several multiplications.
2^5 * 2^x= 2^(5 + X)

3 0

2 years ago

Helppp Plsss Asap!! Thanks!!

LenKa [72]

the answer is 12 because if you look at the position of the y axis when x equals 2 then look at the y value when x equals 4 you will get the 2 numbers 4 and 16 now subtract 16 by 4 and you get 12.

8 0

3 years ago

Read 2 more answers

Iĺl mark brainlist

N76 [4]

Answer:

C. 16√5

Step-by-step explanation:

a squared plus b squared equals c squared!

7 0

2 years ago