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3 years ago

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If the probability that it will rain tomorrow is 1/5, what is the probability that it will not rain tomorrow? A 4/5 B 3/5 C 2/5

D 2/10

2 answers:

kipiarov [429]3 years ago

7 0

A. 4/5

P(A)’ = 1 - P(A)

1 - 1/5 = 4/5

Natali5045456 [20]3 years ago

3 0

Answer:

A

Step-by-step explanation:

Because if the chance of rain is 1/5 you just got to think because the remainder is 4/5 so all together it is 5/5

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4 x 3/5 as a mixed number

algol [13]

<span>The mixed number is</span><span> = 2 <span>2/5.</span></span>

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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

PLEASE p-l-e-a-s-e help, there are 3 math questions, thank you!

Serjik [45]

You use Pythagorean theorem to solve this problem.a^2+b^2=c^2
A.53*53=45*45+28*28
2809=2025+784
2809=2809

B.85*85=13*13+84*84
7725=169+7056
7725=7725
C.85*85=36*36+77*77
7725=1296+5929
7725=7725
D.65*65=16*16+61*61
4225=256+3721
4225=3977

The answer is D

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4 years ago

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Mazyrski [523]

You just have to look at the table that give you it was angle 31 so then you would go to angle and go across the line to where it says cos then you right the number which would be 0.8572

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3 years ago

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telo118 [61]

The answer is 615.44

3 0

2 years ago