A direct relationship occurs when

Mathematics

1 answer:

amid [387]3 years ago

4 0

A direct relationship requires a change in the same direction of both variables, answer c.

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The model shown was used to find the product of two fractions: A rectangle divided into nine columns of equal size and 4 rows of

Valentin [98]

The equation that represents the array (rectangles and area) multiplication model that sows two grey shaded columns of length one ninth each and three rows with dots of width one fourth each is option a

a) The equation with fractions two ninths times three fourths is equal to six thirty sixths

$\dfrac{2}{9} \times \dfrac{3}{4} = \dfrac{6}{36}$

<h3>What is an array (area) multiplication model?</h3>

An array representation of a multiplication is a rectangular visual order of positioning of rows and columns that indicates the terms of a multiplication equation.

Please find attached the area model to multiply the fractions

The terms of the equation represented by the model are indicated by the two columns of length one ninth each shaded grey and the three rows of width one fourth each covered with dots, such that the equation can be presented as follows;

$\left(\dfrac{1}{9} +\dfrac{1}{9}\right) \times \left(\dfrac{1}{4} +\dfrac{1}{4} + \dfrac{1}{4} \right) = \dfrac{2}{9} \times \dfrac{3}{4} =\dfrac{6}{36}$

The equation that the model represents is therefore;

The equation with fractions two ninths times three fourths is equal to six thirty sixths

Learn more about multiplication models here:

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8 0

2 years ago

For what values of x is f(x) = |x + 1| differentiable? I'm struggling my butt off for this course

pav-90 [236]

By definition of absolute value, you have

$f(x) = |x+1| = \begin{cases}x+1&\text{if }x+1\ge0 \\ -(x+1)&\text{if }x+1$

or more simply,

$f(x) = \begin{cases}x+1&\text{if }x\ge-1\\-x-1&\text{if }x$

On their own, each piece is differentiable over their respective domains, except at the point where they split off.

For x > -1, we have

(x + 1)' = 1

while for x < -1,

(-x - 1)' = -1

More concisely,

$f'(x) = \begin{cases}1&\text{if }x>-1\\-1&\text{if }x$

Note the strict inequalities in the definition of f '(x).

In order for f(x) to be differentiable at x = -1, the derivative f '(x) must be continuous at x = -1. But this is not the case, because the limits from either side of x = -1 for the derivative do not match:

$\displaystyle \lim_{x\to-1^-}f'(x) = \lim_{x\to-1}(-1) = -1$