The answer is true. Originally, the Olympic Games consisted of fewer events than they do now. The original events included wrestling, discuss throw, javelin throw, running, chariot racing and some others. Much recent addition to the Olympic Games include table tennis and martial arts.

3 0

3 years ago

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When a person has an alcoholic beverage, what percent of the alcohol contained is absorbed by the stomach?

geniusboy [140]

A.
About 20% (:

Good Luck!

3 0

3 years ago

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Differentiate the following functions (i) x(1+x)^3

statuscvo [17]

Answer:

$\displaystyle y' = (1 + x)^2(4x + 1)$

General Formulas and Concepts:

Algebra I

Terms/Coefficients
Functions
Function Notation
Factoring

Calculus

Derivatives

Derivative Notation

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

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

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

Explanation:

Step 1: Define

Identify

y = x(1 + x)³

Step 2: Differentiate

Product Rule [Derivative Rule - Chain Rule]: $\displaystyle y' = \frac{d}{dx}[x] \cdot (1 + x)^3 + x \cdot \frac{d}{dx}[(1 + x)^3] \cdot \frac{d}{dx}[1 + x]$
Derivative Property [Addition/Subtraction]: $\displaystyle y' = \frac{d}{dx}[x] \cdot (1 + x)^3 + x \cdot \frac{d}{dx}[(1 + x)^3] \cdot (\frac{d}{dx}[1] + \frac{d}{dx}[x])$
Basic Power Rule: $\displaystyle y' = x^{1 - 1} \cdot (1 + x)^3 + x \cdot 3(1 + x)^{3 - 1} \cdot (0 + x^{1 - 1})$
Simplify: $\displaystyle y' = (1 + x)^3 + 3x(1 + x)^2$
Factor: $\displaystyle y' = (1 + x)^2 \bigg[ (1 + x) + 3x \bigg]$
Combine like terms: $\displaystyle y' = (1 + x)^2(4x + 1)$

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

Unit: Derivatives

Book: College Calculus 10e

5 0

3 years ago

(

Montano1993 [528]

10 its literraly right thereಠ_ಠ

6 0

3 years ago