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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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A piece of wire of length L will be cut into two pieces, one piece to form a square and the other piece to form an equilateral t
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Answer:

  (a)  square: L; triangle: 0.

  (b)  square: L·(-16+12√3)/11; triangle: L·(27-12√3)/11

Step-by-step explanation:

<u>Strategy</u>: First we will write each area in terms of its perimeter. Then we will find the total area in terms of the amount devoted to the square. Differentiating will give a way to find the minimum total area.

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In terms of its perimeter p, the area of a square is ...

  A_square = p^2/16

In terms of its perimeter p, the area of an equilateral triangle is ...

  A_triangle = p^2/(12√3)

Then the total area of the two figures whose total perimeter is L with "x" devoted to the square is ...

  A_total = x^2/16 + (L-x)^2/(12√3)

__

(a) We know when polygons are regular, the one with the most area for the least perimeter is the one with the most sides. Hence, the total area is maximized when all of the wire is devoted to the square.

__

(b) The derivative of A_total with respect to x is ...

  dA/dx = x/8 -(L-x)/(6√3)

This will be zero when ...

  x/8 = (L-x)/(6√3)

  x(6√3) = 8L -8x

  x(8 +6√3) = 8L

  x = L·8/(6√3 +8) = 8L(6√3 -8)/(64-108)

  x = L·(12√3 -16)/11

The total area is minimized when L·(12√3 -16)/11 is devoted to the square, and the balance is devoted to the triangle.

5 0
3 years ago
Jerome tries to find the value of the expression 3 - 6 + 5 by first applying the commutative property.He rewrites the expression
Afina-wow [57]
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Hopefully this helps :)
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AveGali [126]
Since it is an equality that means the left side of the equation (x/8) has to be equal to the right side of the equation (28/32). One way you can solve for x is to isolate it. Since it’s an equality like we said you have to do to the left side the same that you would do to the right side to isolate x. So if I multiply by 8 on the left I need to also multiply by 8 on the right. So 8* (x/8) becomes just x since 8x/8 becomes 8/8 * x which is 1 *x. On the right side we do 8 *(28/32) which becomes 28 * 8/32 which becomes 7. So x = 7.
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Answer:

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Step-by-step explanation:

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plug in the equation x^2+2x-5=0

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