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lana [24]
2 years ago
13

Find x helpppppppppppp​

Mathematics
1 answer:
lidiya [134]2 years ago
3 0

|2x - 1|  =  | - x - 2|

Thus ;

2x - 1 =  - x - 2

Add both sides x

2x + x - 1 =  - x + x - 2

3x - 1 =  - 2

Add both sides 1

3x - 1 + 1 =  - 2 + 1

3x =   - 1

Divide both sides by 3

\frac{3x}{3}  =  \frac{ - 1}{3}  \\

x =  -  \frac{1}{3}  \\

<h2>Or </h2>

2x - 1 =  - ( - x - 2)

2x - 1 = x + 2

Subtract both sides x

2x  - x  - 1 = x - x + 2

x - 1 = 2

Add both sides 1

x - 1 + 1 = 2 + 1

x = 3

There u go ...

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A rectangular storage container with an open top is to have a volume of 24 cubic meters. The length of its base is twice the wid
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Answer:

419.25

Step-by-step explanation:

The calculation of the cost of materials for the cheapest such container is shown below:-

We assume

Width = x

Length = 2x

Height = h

where, length = 2 \times width

Base area = lb

= 2x^2

Side area = 2lh + 2bh

= 2(2x)h + 2(x)h

= 4xh + 2xh

Volume = 24 which is lbh = 24

h = \frac{24}{2x^2} \\\\ h = \frac{12}{x^2}

Now, cost is

= 13(2x^2) + 9(4xh + 2xh)\\\\ = 13(2x^2) + 9(4x + 2x)\times \frac{12}{x^2} \\\\ = 26x^2 + \frac{648}{x}

now we have to minimize C(x)

So, we need to compute the C'(x)

= 52x - \frac{648}{x^2}

C"(x)  = 52x - \frac{1,296}{x^3}

now for the critical points, we will solve the equation C'(x) = 0

= 52x - \frac{648}{x^2} = 0\\\\ x = \frac{648}{52}^{\frac{1}{3}}

C" = ((\frac{648}{52} ^{\frac{1}{3} } = 52 + \frac{1296}{(\frac{648}{52})^\frac{1}{3} )^3}\\\\ = 52 + \frac{1296}{\frac{648}{52} } >0

So, x is a point of minima that is

= (\frac{648}{52} )^\frac{1}{3}

Now, Base material cost is

= 13(2x^2)\\\\ = 26(\frac{648}{52} )^\frac{2}{3}

= 139.75

Side material cost is

= \frac{648}{x} \\\\ = \frac{648}{(\frac{648}{52})^\frac{1}{3}  }

= 279.50

and finally

Total cost is

= 139.75 + 279.50

= 419.25

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