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3 years ago

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A coach took 3/1/2 hours to travel a distance of 240 km from Kuala Lumpur to Cameron Highlands. It travelled part of the journey

at an average speed of 60km/h and the rest of the journey at 75km/h. Calculate the distance it travelled at 60km/h.

1 answer:

Marina CMI [18]3 years ago

5 0

Step-by-step explanation:

He travelled 108 km in speed of 65km/h.

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Answer:

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3 years ago

Help with num 3 please. thanks

Alja [10]

Answer:

a) $\displaystyle \frac{dy}{dx} \bigg| \limits_{x = 0} = -1$

b) $\displaystyle \frac{dy}{dx} \bigg| \limits_{x = \frac{\pi}{2}} = -1$

General Formulas and Concepts:

<u>Pre-Calculus</u>

Unit Circle

<u>Calculus</u>

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

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 [Quotient Rule]: $\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

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

Trigonometric Differentiation

Logarithmic Differentiation

Step-by-step explanation:

a)

<u>Step 1: Define</u>

<em>Identify</em>

$\displaystyle y = ln \bigg( \frac{1 - x}{\sqrt{1 + x^2}} \bigg)$

<u>Step 2: Differentiate</u>

Logarithmic Differentiation [Chain Rule]: $\displaystyle \frac{dy}{dx} = \frac{1}{\frac{1 - x}{\sqrt{1 + x^2}}} \cdot \frac{d}{dx}[\frac{1 - x}{\sqrt{1 + x^2}}]$
Simplify: $\displaystyle \frac{dy}{dx} = \frac{-\sqrt{x^2 + 1}}{x - 1} \cdot \frac{d}{dx}[\frac{1 - x}{\sqrt{1 + x^2}}]$
Quotient Rule: $\displaystyle \frac{dy}{dx} = \frac{-\sqrt{x^2 + 1}}{x - 1} \cdot \frac{(1 - x)'\sqrt{1 + x^2} - (1 - x)(\sqrt{1 + x^2})'}{(\sqrt{1 + x^2})^2}$
Basic Power Rule [Chain Rule]: $\displaystyle \frac{dy}{dx} = \frac{-\sqrt{x^2 + 1}}{x - 1} \cdot \frac{-\sqrt{1 + x^2} - (1 - x)(\frac{x}{\sqrt{x^2 + 1}})}{(\sqrt{1 + x^2})^2}$
Simplify: $\displaystyle \frac{dy}{dx} = \frac{-\sqrt{x^2 + 1}}{x - 1} \cdot \bigg( \frac{x(x - 1)}{(x^2 + 1)^\bigg{\frac{3}{2}}} - \frac{1}{\sqrt{x^2 + 1}} \bigg)$
Simplify: $\displaystyle \frac{dy}{dx} = \frac{x + 1}{(x - 1)(x^2 + 1)}$

<u>Step 3: Find</u>

Substitute in <em>x</em> = 0 [Derivative]: $\displaystyle \frac{dy}{dx} \bigg| \limit_{x = 0} = \frac{0 + 1}{(0 - 1)(0^2 + 1)}$
Evaluate: $\displaystyle \frac{dy}{dx} \bigg| \limits_{x = 0} = -1$

b)

<u>Step 1: Define</u>

<em>Identify</em>

$\displaystyle y = ln \bigg( \frac{1 + sinx}{1 - cosx} \bigg)$

<u>Step 2: Differentiate</u>

Logarithmic Differentiation [Chain Rule]: $\displaystyle \frac{dy}{dx} = \frac{1}{\frac{1 + sinx}{1 - cosx}} \cdot \frac{d}{dx}[\frac{1 + sinx}{1 - cosx}]$
Simplify: $\displaystyle \frac{dy}{dx} = \frac{-[cos(x) - 1]}{sin(x) + 1} \cdot \frac{d}{dx}[\frac{1 + sinx}{1 - cosx}]$
Quotient Rule: $\displaystyle \frac{dy}{dx} = \frac{-[cos(x) - 1]}{sin(x) + 1} \cdot \frac{(1 + sinx)'(1 - cosx) - (1 + sinx)(1 - cosx)'}{(1 - cosx)^2}$
Trigonometric Differentiation: $\displaystyle \frac{dy}{dx} = \frac{-[cos(x) - 1]}{sin(x) + 1} \cdot \frac{cos(x)(1 - cosx) - sin(x)(1 + sinx)}{(1 - cosx)^2}$
Simplify: $\displaystyle \frac{dy}{dx} = \frac{-[cos(x) - sin(x) - 1]}{[sin(x) + 1][cos(x) - 1]}$

<u>Step 3: Find</u>

Substitute in <em>x</em> = π/2 [Derivative]: $\displaystyle \frac{dy}{dx} \bigg| \limit_{x = \frac{\pi}{2}} = \frac{-[cos(\frac{\pi}{2}) - sin(\frac{\pi}{2}) - 1]}{[sin(\frac{\pi}{2}) + 1][cos(\frac{\pi}{2}) - 1]}$
Evaluate [Unit Circle]: $\displaystyle \frac{dy}{dx} \bigg| \limit_{x = \frac{\pi}{2}} = -1$

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

Unit: Differentiation

Book: College Calculus 10e

4 0

3 years ago

The final balance in an account that has a monthly fee and automatic monthly deposits is given by the expression. The initial ac

trapecia [35]

150x represents the monthly deposit, and -4.99 represents the monthly fee.

6 0

4 years ago

Read 2 more answers

What is the base of the trapezoid with an area of 150 a height of 10 and a base of 14?

slavikrds [6]

The formula for the area of a trapezoid is:

A= (a+b)/2*h

A=150
h=10
b=14
a=?

So we need to solve for a. Let's plug in the values that we do know...

A= (a+b)/2*h
150=(a+14)/2*10
150=(a/2+14/2)*10
150=(a/2+7)*10
Divide both sides by 10
15=a/2+7
subtract 7 from both sides
8=a/2
multiply both sides by 2
16=a

The other base is 16

3 0

3 years ago

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A store has a display case with cherry, peach,

STatiana [176]

Answer:

56 grape

Step-by-step explanation:

160 fruit chews

25% are cherry, 40% are peach.....25 + 40 = 65%....so 65% of the fruit chews are NOT grape....meaning (100 - 65) = 35 % ARE grape.

so 35% of all the fruit chews are grape

35% of 160 = 0.35 * 160 = 56 are grape <===

5 0

3 years ago