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Inessa05 [86]
3 years ago
7

 For what value of x is the double product of the binomials x+2 and x–2 is 16 less than the sum of their squares?

Mathematics
2 answers:
solong [7]3 years ago
4 0

x = 2 + 2 - 2 = 4

hope it helps! :D

oee [108]3 years ago
4 0
\bf \begin{cases}
x+2\\\\
x-2
\end{cases}
\begin{cases}\qquad 
product\\
----------\\
(x+2)(x-2)\to x^2-4\leftarrow \textit{difference of squares anyway}\\\\
\textit{double that product}\\
----------\\
\boxed{2(x^2-4)}
\\\\\\
\textit{sum of their squares}\\
----------\\
(x+2)^2+(x-2)^2\\\\
\textit{16 less than that}\\
----------\\
\boxed{(x+2)^2+(x-2)^2-16}
\end{cases}

so.. for what values of "x", are those two expressions, equal to each other then?

well \bf 2x^2-8=x^2\underline{+2x}+4+x^2\underline{-2x}+4-16
\\\\\\
2x^2-8=2x^2+8-16\implies 
\begin{array}{llll}
2x^2-8&=&2x^2-8\\
same&=&same
\end{array}

for a system of equations, the solutions are where the graph converge, or intersect

well, in this case, the left-hand-side expression is exactly the same as the right-hand-side expression
 
so.. if you were to graph the one on the left-hand-side, it's just a parabola shifted vertically down by 8units

if you were to graph the right-hand-side expression, it will be exactly the same, just the same parabola right on top of the other, pancaked on one another

when both expressions are equal, the system has "infinite number of solutions", which is another way to say, for all values of "x", there's a solution

so.. for what values for "x" are those two expressions equal? for all values of "x"  or (-\infty,+\infty)\quad or\quad x\in \mathbb{R}, if you like an interval or set notation
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Step-by-step explanation:

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A rectangular box is to have a square base and a volume of 12 ft3. If the material for the base costs $0.17/ft2, the material fo
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Let the dimensions of the box be x, y and z

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Area of the base =x^2

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Substituting z=\frac{12}{x^2}

C(x) =0.17x^2+0.13x^2+0.10(4x)(\frac{12}{x^2})\\C(x)=0.3x^2+\frac{4.8}{x} \\C(x)=\dfrac{0.3x^3+4.8}{x}

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C'(x)=\dfrac{0.6x^3-4.8}{x^2}\\Setting \:C'(x)=0\\0.6x^3-4.8=0\\0.6x^3=4.8\\x^3=4.8\div 0.6\\x^3=8\\x=\sqrt[3]{8}=2

Recall: z=\frac{12}{x^2}=\frac{12}{2^2}=3\\

Therefore, the dimensions that minimizes the cost of the box are:

(a)Length =2 feet

(b)Width =2 feet

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