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jekas [21]
1 year ago
6

A wire 15 cm long is cut into two pieces. The longer piece is 3.cmidonger than the shorter piece.Find the length of the shorter

piece of wirecm
Mathematics
1 answer:
Andre45 [30]1 year ago
7 0

A piece of wire is cut into two pieces. That means each part would be assigned an unknown variable. Let one part be x and the other part be y.

That means,

x+y=15\operatorname{cm}

If the longer part is x, and the longer part is 3cm longer than the shorter part, then we would have the following;

\begin{gathered} x+y=15 \\ x=y+3 \\ \text{This is because x is 3cm longer,} \\ So\text{ the length of x would be y+3} \end{gathered}

We can now refine the equation as follow;

\begin{gathered} \text{Where;} \\ x=y+3 \\ x+y=15 \\ y+3+y=15 \\ 2y+3=15 \\ \text{Subtract 3 from both sides;} \\ 2y+3-3=15-3 \\ 2y=12 \\ \text{Divide both sides by 2;} \\ \frac{2y}{2}=\frac{12}{2} \\ y=6 \\ \text{When;} \\ x+y=15 \\ x+6=15 \\ \text{Subtract 6 from both sides;} \\ x+6-6=15-6 \\ x=9 \end{gathered}

ANSWER:

The length of the shorter piece of wire is 6cm

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Answer:

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Step-by-step explanation:

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Let call  " x " side of the base then as it is square area is A₁ = x²

Sides areas are 4 each one equal to x * h  (where h is the high of the box)

And volume of the box is   13,5 cm³  = x²*h

Then   h  =  13,5/x²

Side area is :  A₂ =  x* 13,5/x²

A(b)  = A₁  + A₂

Total area of the box as functon of x is:

A(x)  = 2*x²  + 4* 13,5 / x

And finally cost of the box is

C(x)  = 10*2*x²   +  2.50*4*13.5/x

C(x)  = 20*x²  +  135/x

Taking derivatives on both sides of the equation:

C´(x)  =  40*x   -  135*/x²

C´(x)  = 0     ⇒      40*x   -  135*/x² = 0    ⇒  40*x³ = 135

x³  = 3.375

x  = 1,5 cm

And   h  =  13,5/x²     ⇒   h  =  13,5/ (1,5)²

h = 6 cm

C(min)  = 20*x²  + 135/x

C(min)  = 45  +  90

C(min)  = 135 $

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Answer:

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Step-by-step explanation:

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