Add [1/-4 3/5] [-2/6 -2/4]

25/138

Step-by-step explanation:

1 Convert 4\frac{3}{5}4

5

3

 to improper fraction. Use this rule: a \frac{b}{c}=\frac{ac+b}{c}a

c

b

=

c

ac+b

.

\frac{1}{-(\frac{4\times 5+3}{5})}(-\frac{2}{6}-\frac{2}{4})

−(

5

4×5+3

)

1

(−

6

2

−

4

2

)

2 Simplify  4\times 54×5  to  2020.

\frac{1}{-(\frac{20+3}{5})}(-\frac{2}{6}-\frac{2}{4})

−(

5

20+3

)

1

(−

6

2

−

4

2

)

3 Simplify  20+320+3  to  2323.

\frac{1}{-(\frac{23}{5})}(-\frac{2}{6}-\frac{2}{4})

−(

5

23

)

1

(−

6

2

−

4

2

)

4 Simplify  \frac{2}{6}

6

2

  to  \frac{1}{3}

3

1

.

\frac{1}{-(\frac{23}{5})}(-\frac{1}{3}-\frac{2}{4})

−(

5

23

)

1

(−

3

1

−

4

2

)

5 Simplify  \frac{2}{4}

4

2

  to  \frac{1}{2}

2

1

.

\frac{1}{-(\frac{23}{5})}(-\frac{1}{3}-\frac{1}{2})

−(

5

23

)

1

(−

3

1

−

2

1

)

6 Find the Least Common Denominator (LCD) of \frac{1}{3},\frac{1}{2}

3

1

,

2

1

. In other words, find the Least Common Multiple (LCM) of 3,23,2.

LCD = 66

7 Make the denominators the same as the LCD.

-\frac{1\times 2}{3\times 2}-\frac{1\times 3}{2\times 3}−

3×2

1×2

−

2×3

1×3

8 Simplify. Denominators are now the same.

-\frac{2}{6}-\frac{3}{6}−

6

2

−

6

3

9 Join the denominators.

\frac{-2-3}{6}

6

−2−3

10 Simplify  -\frac{1}{3}-\frac{1}{2}−

3

1

−

2

1

  to  -\frac{5}{6}−

6

5

.

\frac{1}{-(\frac{23}{5})}\times \frac{-5}{6}

−(

5

23

)

1

×

6

−5

11 Move the negative sign to the left.

-\frac{1}{\frac{23}{5}}\times \frac{-5}{6}−

5

23

1

×

6

−5

12 Invert and multiply.

-\frac{5}{23}\times \frac{-5}{6}−

23

5

×

6

−5

13 Use this rule: \frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}

b

a

×

d

c

=

bd

ac

.

-\frac{5\times -5}{23\times 6}−

23×6

5×−5

14 Simplify  5\times -55×−5  to  -25−25.

-\frac{-25}{23\times 6}−

23×6

−25

15 Simplify  23\times 623×6  to  138138.

-\frac{-25}{138}−

138

−25

16 Move the negative sign to the left.

-(-\frac{25}{138})−(−

138

25

)

17 Remove parentheses.

\frac{25}{138}

138

25

distance covered by pamela in 9 l of fuel = 99km

let us apply unitary method here.

distance covered by pamela in 1 l of fuel =99/9 =11 km

now let us find distance covered by pamela if she has 12 l of fuel.

distance covered in 12 l = distance covered in 1 l of fuel * 12

we know that distance covered by here in 1 l is 11 km.

so distance covered in 12 l of fuel = 11*12 = 132 km.

answer: bearing in mind that a rectangle has a perimeter of length + length + width + width.

step-by-step explanation:

p = perimeter

l = length

x=width

p = 2l + 2x

p = 2(l+x)     < ==== perimeter of the base of the box.