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3 years ago

What is quotient of 6 1/4 divided by 5/8

Mathematics

1 answer:

lara31 [8.8K]3 years ago

7 0

Answer: The quotient = 10

Step-by-step explanation: A quotient is the result you get when a number is used to divide another number. e . g.

6¹/₄ ÷ ⁵/₈ ( where 5/8 is the divisor )

First convert 6¹/₄ to improper fraction

= 25/4 ÷ 5/8

= 25/4 × 5/8

= 10

The quotient = 10.

NOTE: The 5/8 changed to 8/5 because when dividing fractions, immediately you applied the multiplication symbol, the denominator of the divisor becomes the numerator and vice versa

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Answer:

$\displaystyle y' = \frac{5x - 2xy^2}{2y(x^2 - 3y)}$

General Formulas and Concepts:

Calculus

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

Implicit Differentiation

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle 5x^2 - 2x^2y^2 + 4y^3 - 7 = 0$

Step 2: Differentiate

Implicit Differentiation: $\displaystyle \frac{dy}{dx}[5x^2 - 2x^2y^2 + 4y^3 - 7] = \frac{dy}{dx}[0]$
Rewrite [Derivative Property - Addition/Subtraction]: $\displaystyle \frac{dy}{dx}[5x^2] - \frac{dy}{dx}[2x^2y^2] + \frac{dy}{dx}[4y^3] - \frac{dy}{dx}[7] = \frac{dy}{dx}[0]$
Rewrite [Derivative Property - Multiplied Constant]: $\displaystyle 5\frac{dy}{dx}[x^2] - 2\frac{dy}{dx}[x^2y^2] + 4\frac{dy}{dx}[y^3] - \frac{dy}{dx}[7] = \frac{dy}{dx}[0]$
Basic Power Rule [Product Rule, Chain Rule]: $\displaystyle 10x - 2 \Big( \frac{d}{dx}[x^2]y^2 + x^2\frac{d}{dx}[y^2] \Big) + 12y^2y' - 0 = 0$
Basic Power Rule [Chain Rule]: $\displaystyle 10x - 2 \Big( 2xy^2 + x^22yy' \Big) + 12y^2y' - 0 = 0$
Simplify: $\displaystyle 10x - 4xy^2 - 4x^2yy' + 12y^2y' = 0$
Isolate y' terms: $\displaystyle -4x^2yy' + 12y^2y' = 4xy^2 - 10x$
Factor: $\displaystyle y'(-4x^2y + 12y^2) = 4xy^2 - 10x$
Isolate y': $\displaystyle y' = \frac{4xy^2 - 10x}{-4x^2y + 12y^2}$
Simplify: $\displaystyle y' = \frac{5x - 2xy^2}{2y(x^2 - 3y)}$