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3 years ago

How to convert 18 gallons/1 hour to cups and minutes

1 answer:

8 0

So there are 2 things you need to convert, the first is the gallons to cups and the second is the hours to minutes. so I like to set it up with numerators and denominators and then multiply the ratios. All same terms will cross out and be opposite of each other.

(18 gal/1 hour) x (1 hour / 60 minutes), now we have (0.3 gallons/1 minute) we just have to change gallons to cups

(0.3 gallons/ 1 minute) x (4 quarts/1 gallon) x (2 pints/1 quart) x (2 cups/1 pint)

this comes out 4.8 cups/1 minute

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How to write 0.875 as a fraction in simplest form

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Now, lets first know this:-

0.875 = 875/1000

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875/125 = 7
1000/125 = 8

7/8

Answer: 7/8

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3 years ago

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The area of a circle (A) is given by the formula A=r^2 where r is the circle's radius. The formula to find r is______ . If (A)=5

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Answer: (A) and (A)

Step-by-step explanation:

$A=\pi \cdot r^2\\\\\dfrac{A}{\pi}=r^2\\\\\sqrt{\dfrac{A}{\pi}}=r\ \rightarrow \ \bigg(\dfrac{A}{\pi}\bigg)^{\dfrac{1}{2}}\\\\\\\\r=\sqrt{\dfrac{A}{\pi}}\\\\.\ =\sqrt{\dfrac{54\cdot 7}{22}}\\\\.\ =\sqrt{\dfrac{27\cdot 7}{11}}\\\\.\ =\sqrt{\dfrac{189}{11}}\\\\.\ =\sqrt{17.1818}\\\\.\ =4.15$

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3 years ago

It is a timed test i will give brainleist

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3 years ago

Find the derivative: y={ (3x+1)cos(2x) } / e^2x

DochEvi [55]

Answer:

$\displaystyle y' = \frac{3cos(2x) -2(3x + 1)[sin(2x) + cos(2x)]}{e^{2x}}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Factoring
Exponential Rule [Dividing]: $\displaystyle \frac{b^m}{b^n} = b^{m - n}$
Exponential Rule [Powering]: $\displaystyle (b^m)^n = b^{m \cdot n}$

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

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

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

Trig Derivative: $\displaystyle \frac{d}{dx}[cos(u)] = -u'sin(u)$

eˣ Derivative: $\displaystyle \frac{d}{dx}[e^u] = u'e^u$

Step-by-step explanation:

Step 1: Define

$\displaystyle y = \frac{(3x + 1)cos(2x)}{e^{2x}}$

Step 2: Differentiate

[Derivative] Quotient Rule: $\displaystyle y' = \frac{\frac{d}{dx}[(3x + 1)cos(2x)]e^{2x} - \frac{d}{dx}[e^{2x}](3x + 1)cos(2x)}{(e^{2x})^2}$
[Derivative] [Fraction - Numerator] eˣ derivative: $\displaystyle y' = \frac{\frac{d}{dx}[(3x + 1)cos(2x)]e^{2x} - 2e^{2x}(3x + 1)cos(2x)}{(e^{2x})^2}$
[Derivative] [Fraction - Denominator] Exponential Rule - Powering: $\displaystyle y' = \frac{\frac{d}{dx}[(3x + 1)cos(2x)]e^{2x} - 2e^{2x}(3x + 1)cos(2x)}{e^{4x}}$
[Derivative] [Fraction - Numerator] Product Rule: $\displaystyle y' = \frac{[\frac{d}{dx}[3x + 1]cos(2x) + \frac{d}{dx}[cos(2x)](3x + 1)]e^{2x} - 2e^{2x}(3x + 1)cos(2x)}{e^{4x}}$
[Derivative] [Fraction - Numerator] [Brackets] Basic Power Rule: $\displaystyle y' = \frac{[(1 \cdot 3x^{1 - 1})cos(2x) + \frac{d}{dx}[cos(2x)](3x + 1)]e^{2x} - 2e^{2x}(3x + 1)cos(2x)}{e^{4x}}$
[Derivative] [Fraction - Numerator] [Brackets] (Parenthesis) Simplify: $\displaystyle y' = \frac{[3cos(2x) + \frac{d}{dx}[cos(2x)](3x + 1)]e^{2x} - 2e^{2x}(3x + 1)cos(2x)}{e^{4x}}$
[Derivative] [Fraction - Numerator] [Brackets] Trig derivative: $\displaystyle y' = \frac{[3cos(2x) -2sin(2x)(3x + 1)]e^{2x} - 2e^{2x}(3x + 1)cos(2x)}{e^{4x}}$
[Derivative] [Fraction - Numerator] Factor: $\displaystyle y' = \frac{e^{2x}[(3cos(2x) -2sin(2x)(3x + 1)) - 2(3x + 1)cos(2x)]}{e^{4x}}$
[Derivative] [Fraction] Simplify [Exponential Rule - Dividing]: $\displaystyle y' = \frac{3cos(2x) -2sin(2x)(3x + 1) - 2(3x + 1)cos(2x)}{e^{2x}}$
[Derivative] [Fraction - Numerator] Factor: $\displaystyle y' = \frac{3cos(2x) -2(3x + 1)[sin(2x) + cos(2x)]}{e^{2x}}$

Topic: AP Calculus AB/BC

Unit: Derivatives

Book: College Calculus 10e

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3 years ago

What is x in x - 4 (x - 3) + 7 = 6 - (x - 4)

WINSTONCH [101]

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