Can a real number be both rational and irrational?

yes

step-by-step explanation:

answer:

given the linear inequality :

add 2y to both sides we get;

simplify:

add 12 to both sides of an equation;

simplify:

divide both sides by 2 we get;

or

the related equation for the inequality equation [1] is;

y-intercepts defined as the graph crosses the y-axis.

substitute x =0 in [2] to solve for y;

y = 6

x-intercepts defined as the graph crosses the x-axis.

substitute value y = 0 to solve for x;

subtract 6 from both sides we get

multiply both sides by 2 we get;

or

x = -12

now, plot these point (0, 6) and (-12, 0) on graph as shown below and shade the graph below the solid line.

2 equivalent expressions to represent the area of a rectangle and the top is n-3 and the side is 2

area of the rectangle = length * width

given length = n - 3

width = 2

replace it in the area

so area of the rectangle = (n-3) *2

= 2(n-3)

one of the expression is 2(n-3)

to get another expression we distribute 2 inside the parenthesis

2(n-3) is 2n - 6

so another expression is 2n - 6

two equivalent expression are

2(n - 3)   , 2n - 6

20/27

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step-by-step explanation:

1 simplify brackets

-\frac{8}{9}\times \frac{5}{6}−

​9

​

​8

​​ ×

​6

​

​5

​​

2 use this rule: \frac{a}{b}\times \frac{c}{d}=\frac{ac}{bd}

​b

​

​a

​​ ×

​d

​

​c

​​ =

​bd

​

​ac

​​

-\frac{8\times 5}{9\times 6}−

​9×6

​

​8×5

​​

3 simplify 8\times 58×5 to 4040

-\frac{40}{9\times 6}−

​9×6

​

​40

​​

4 simplify 9\times 69×6 to 5454

-\frac{40}{54}−

​54

​

​40

​​

5 simplify \frac{40}{54}

​54

​

​40

​​   to \frac{20}{27}

​27

​

​20

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