All categories

2 years ago

20. Lee uses Bill Payer from his bank account to pay his internet bill. Lee pays the same amount each month.

1 answer:

3 0

It's the same amount each month, for a year. So:
total money ÷ times money was taken out= how much has been taken out.

$600÷12=50

You might be interested in

The number. Find the nume.

aliya0001 [1]

Answer: 8x+3x=12x+9

The number is -9.

Step-by-step explanation: Combine like terms.

11x=12x+9

Subtract 12x from the other side. 11x-12x= -1x

-1x+9

Divide by -1 to isolate x.

-1/-1 cancels out.

9/-1= -9

x=-9

Plug in -9 to check your answer if you want to double check.

3 0

2 years ago

Simplify. 3 1 8 8 OA) 2 1 B) 2 C) OD) 8

lawyer [7]

Hello Love!! ♡　

»»————- $\red{Answer\:}$ ————-««

$\Huge{\boxed{\frac{1}{4}}}$

☆♬○♩●♪✧♩ $\bold{EXPLANATION\;}$ ♩✧♪●♩○♬☆

$\mathfrak{apply~the~fraction~rule~:}~\frac{a}{c}~ -~\frac{b}{c} = \frac{a~-~b}{c}$

$= \frac{3-1}{8}$

$\mathfrak{subtract~the~numbers~:}~ 3 - 1=2$

$=\frac{2}{8}$

$\mathfrak{cancel~the~numbers~:}~\frac{2}{8} = \frac{1}{4}$

$=\frac{1}{4}$

°:⋆ₓₒ　Hope It Helps. . .　ₓₒ⋆:°

$Answer~:$

$\mathfrak{Jace}$

6 0

2 years ago

Read 2 more answers

(2x^2-5x+1)+(x^2-x-4)

maksim [4K]

Answer:

3x^2-6x -3

Step-by-step explanation:

(2x^2-5x+1)+(x^2-x-4)

we sum all coefficient according to the exponent of the variable

2x^2-5x+1+x^2-x-4 = 3x^2-6x -3

5 0

2 years ago

Answer fast please! I need an answer before 9:00!

dolphi86 [110]

Answer:

1536 cm^2 be the correct answ

4 0

3 years ago

Read 2 more answers

Find the limit

Lana71 [14]

Step-by-step explanation:

<h3>Appropriate Question :-</h3>

Find the limit

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

On substituting directly x = 1, we get,

$\rm \: = \: \sf \dfrac{1-2}{1 - 1}-\dfrac{1}{1 - 3 + 2}$

$\rm \: = \sf \: \: - \infty \: - \: \infty$

which is indeterminant form.

Consider again,

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

can be rewritten as

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 3x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 2x - x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( x(x - 2) - 1(x - 2))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ {(x - 2)}^{2} - 1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 2 - 1)(x - 2 + 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)(x - 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)}{x(x - 2)}\right]$

$\rm \: = \: \sf \: \dfrac{1 - 3}{1 \times (1 - 2)}$

$\rm \: = \: \sf \: \dfrac{ - 2}{ - 1}$

$\rm \: = \: \sf \boxed{2}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right] = 2 \: }}$

$\rule{190pt}{2pt}$

7 0

2 years ago

Read 2 more answers