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3 years ago

2.

Mathematics

2 answers:

Svetllana [295]3 years ago

8 0

He works 40 hours a week.

SIZIF [17.4K]3 years ago

5 0

The answer is 40 if you divide 480 by 12 you’ll get 40

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Please help!! Will mark brainilest, thank you in advance. :))

qaws [65]

Answer:

See image below:) :)

Step-by-step explanation:

6 0

3 years ago

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kaleb has to sell thirty - six chocolate bars to get a prize . If each box contains four chocolate bars how many boxes does he n

irakobra [83]

Answer: 9 boxes

Step-by-step explanation:

6 0

3 years ago

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You want to buy a camera. The retail price is $75. The camera is on sale for 15% off . What is the sale price of the camera?

ludmilkaskok [199]

Answer:

$63.75

Step-by-step explanation:

15% off

Get percentage of retail price

1-0.15 = .85

0.85 = 85%

The sale price is 85% of the retail price

Multiply

0.85 * 75

$63.75

7 0

2 years ago

12x - 2y = -1<br>+ 4x + 6y= -4<br>

svp [43]

Answer:

x = -7/40 , y = -11/20

Step-by-step explanation:

Solve the following system:

{12 x - 2 y = -1 | (equation 1)

4 x + 6 y = -4 | (equation 2)

Subtract 1/3 × (equation 1) from equation 2:

{12 x - 2 y = -1 | (equation 1)

0 x+(20 y)/3 = (-11)/3 | (equation 2)

Multiply equation 2 by 3:

{12 x - 2 y = -1 | (equation 1)

0 x+20 y = -11 | (equation 2)

Divide equation 2 by 20:

{12 x - 2 y = -1 | (equation 1)

0 x+y = (-11)/20 | (equation 2)

Add 2 × (equation 2) to equation 1:

{12 x+0 y = (-21)/10 | (equation 1)

0 x+y = -11/20 | (equation 2)

Divide equation 1 by 12:

{x+0 y = (-7)/40 | (equation 1)

0 x+y = -11/20 | (equation 2)

Collect results:

Answer: {x = -7/40 , y = -11/20

6 0

3 years ago

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

3 years ago

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