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Mathematics, 29.01.2020 10:03 ralewis2931

Pls . need asap. calculus question.

Pls . need asap. calculus question.

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Answer from: ryan23allen

$\displaystyle \frac{sin(\sqrt{x})}{4x}$

General Formulas and Concepts:

Algebra I

Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$ Exponential Rule [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

Derivatives

Derivative Notation

Basic Power Rule:

f(x) = cxⁿ f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Integrals

Integration Rule [Fundamental Theorem of Calculus 2]: $\displaystyle \frac{d}{dx}[\int\limits^x_a {f(t)} \, dt] = f(x)$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle g(t) = \int\limits^{\sqrt{x}}_1 {\frac{sin(t)}{2t}} \, dt$

Step 2: Differentiate

Fundamental Theorem of Calculus 2 [Derivative Rule - Chain Rule]: $\displaystyle g'(x) = \frac{sin(\sqrt{x})}{2\sqrt{x}} \cdot \frac{d}{dx}[\sqrt{x}]$ Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle g'(x) = \frac{sin(\sqrt{x})}{2\sqrt{x}} \cdot \frac{d}{dx} \bigg[ x^\bigg{\frac{1}{2}} \bigg]$ Basic Power Rule: $\displaystyle g'(x) = \frac{sin(\sqrt{x})}{2\sqrt{x}} \cdot \frac{1}{2}x^\bigg{\frac{1}{2} - 1}$ Simplify [Exponential Rule - Rewrite]: $\displaystyle g'(x) = \frac{sin(\sqrt{x})}{2\sqrt{x}} \cdot \frac{1}{2x^\bigg{\frac{1}{2}}}$ Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle g'(x) = \frac{sin(\sqrt{x})}{2\sqrt{x}} \cdot \frac{1}{2\sqrt{x}}$ Multiply: $\displaystyle g'(x) = \frac{sin(\sqrt{x})}{4x}$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

Book: College Calculus 10e

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Answer from: Quest

1gm = 1000mg so 0.1gm = 1000mg

first, convert the weight to kg

160lb * 1kg/2.2lb = 72.7kg

next, the dose

72.7kg * 100mg/kg = 7270mg

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Answer from: Quest

step-by-step explanation:

answers available in this download

(pearson%2c%25202012)%2520bbs.pdf& usg=aovvaw1obmtxnfghfgxhx0urh_je

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