What is 5/12 minus 3/20

1 answer:

6 0

$\frac{5}{12} - \frac{3}{20}$

First, find the LCD of the two fractions.
Multiples of 12: 12, 24, 36, 48, 60.
Multiples of 20: 20, 40, 60
The LCD is 60.
Second, change the denominators to be the same as the LCD.
$\frac{5x5}{12x5} - \frac{3x3}{20x3}$
Third, 5 × 5 = 25 / 12 × 5 = 60 / 3 × 3 = 9 / 20 × 3 = 60
$\frac{25}{60} - \frac{9}{60}$
Fourth, combine the denominators.
$\frac{25-9}{60}$
Fifth, subtract 25 - 9 to get 16.
$\frac{16}{60}$
Sixth, find the GCF of 16 and 60.
The GCF is 4.
Seventh, divide the numerator by the GCF.
$16$ ÷ $4 = 4$
Eighth, divide the denominator by the GCF.
$60$ ÷ $4 = 15$
Ninth, rewrite the fraction in simplest form.
$\frac{4}{15}$

Answer as fraction: $\frac{4}{15}$
Answer as decimal: 0.2667

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Jessie works at least 36 hours a week. She works as a math

Olegator [25]

Answer:

The feasible region is unbounded which is the shaded portion in the graph.

Step-by-step explanation:

Given that, the number of hours Jessie works as a teacher and tutor is y and x respectively, which can't be negative.

So, $x\ge0\;\cdots(i)$

y\ge0\;\cdots(ii)

She works at least 36 hours a week.

$\Rightarrow x+y\ge 36$

$\Rightarrow y\ge 36-x\;\cdots (iii)$

She works as a teacher for al least twice the number of hours she works as a tutor.

$\Rightarrow y\ge2x\;\cdots(iv)$

The boundary of the inequalities from equations (i), (ii), (iii), and (iv) are

$x=0, y=0, y=36-x$ and $y=2x$ .

The feasible region is the shaded portion which is the unbounded region as shown in the graph. Any point P(x,y) from the feasible region will satisfies all the constraints.