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3 years ago

5

Linear equations 3(x+1)=3x+3

1 answer:

vaieri [72.5K]3 years ago

3 0

4x + 2 - 3x + 5 - 2(x + 5) = (2x - 5) + (3x + 4)
4x + 2 - 3x + 5 - 2x - 10 = 2x - 5 + 3x + 4
4x - 3x - 2x + 5 + 2 - 10 = 2x + 3x - 5 + 4
- x - 3 = 5x - 1
-x - 3 = 5x - 1 is the line above recopied
+x = +x
   -3 = 6x - 1
<span>   +1 =    + 1 </span>
   - 2 = 6x
   - 2/6 = 6x/6
   - 1/3 = x, the answer

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3 years ago

A box with a square base and closed top is required to have a volume of 27,000 cm3

julia-pushkina [17]

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h=30cm

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Step-by-step explanation:

Let x be the dimension of the sides and h the dimension of height;

$V=x^2h$

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$Area=2x^2+4xh$

Find area as a function of x:

$V=x^2h=27000\\\\h=\frac{27000}{x^2}\\\\Area=2x^2+4x(\frac{27000}{x^2})\\\\=2x^2+\frac{108000}{x}$

To minimize Area, we find the first derivative of Area:

$A=2x^2+\frac{108000}{x}\\\\A\prime=4x+\frac{108000}{x^2}=0\\\\\frac{4x^3-108000}{x^2}=0\\\\4x^3=108000\\\\x=30$

#find the second derivative to verify the minimum:

$A\prime=4x+\frac{108000}{x^2}\\A\prime\prime=4x+\frac{108000}{x^3}=124, x=30$

Substitute x in the volume equation to find h:

$h=\frac{27000}{x^2}\\\\=\frac{27000}{30^2}\\\\h=30$

Hence, the dimensions of the box that minimize the amount of material used h=30cm and x=30cm

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