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IgorLugansk [536]
3 years ago
11

What is x in .80x+30=1.20x+14

Mathematics
1 answer:
Sonja [21]3 years ago
3 0

Answer:

x =  40

Step-by-step explanation:

.80x + 30 = 1.20x + 14

<u>-.80x        = -.80x        </u>

          30 = .4x  + 14

        <u>  -14 =         -14</u>

           16 = .4x

            16/.4  = .4/.4x

              40    =  x

       

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Remove unnecessary parenthesis.

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3 years ago
The top and bottom margins of a poster are each 12 cm and the side margins are each 8 cm. The area of printed material on the po
grin007 [14]

Answer:

Dimensions of printed poster are

length is 32 cm

width is 48 cm


Step-by-step explanation:

Let's assume

length of printed poster is x cm

width of printed poster is y cm

now, we can find area of printed poster

so, area of printed poster is

=xy

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so, we can set it to 1536

xy=1536

now, we can solve for y

y=\frac{1536}{x}

now, we are given

The top and bottom margins of a poster are each 12 cm and the side margins are each 8 cm

so, total area of poster is

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now, we can plug back y

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now, we have to minimize A

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A'=\frac{d}{dx}\left(\left(x+16\right)\left(\frac{1536}{x}+24\right)\right)

we can use product rule

A'=\frac{d}{dx}\left(x+16\right)\left(\frac{1536}{x}+24\right)+\frac{d}{dx}\left(\frac{1536}{x}+24\right)\left(x+16\right)

=\frac{d}{dx}\left(x+16\right)\left(\frac{1536}{x}+24\right)+\frac{d}{dx}\left(\frac{1536}{x}+24\right)\left(x+16\right)

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A'=-\frac{24576}{x^2}+24

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A'=-\frac{24576}{x^2}+24=0

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Since, x is dimension

and dimension can never be negative

so, we will only consider positive value

x=32

now, we can solve for y

y=\frac{1536}{32}

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length is 32 cm

width is 48 cm


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