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Ivanshal [37]
3 years ago
5

Two numbers multiply to be -7 and add to be -6. What are the numbers? Example; 2,3

Mathematics
1 answer:
rewona [7]3 years ago
3 0

<u>Given</u>:

Let the two numbers be x and y.

Two numbers multiply to be -7 and add to be -6.

This can be written in equation as,

xy=-7 and

x+y=-6

<u>Value of the two numbers:</u>

Let us determine the value of the two numbers using substitution method.

Substituting y=-6-x in the equation xy=-7, we get;

x(-6-x)=-7

Simplifying, we get;

       -6x-x^2=-7

   x^2+6x-7=0

(x+7)(x-1)=0

x=-7, 1

Thus, the values of x are x = 1,-7

When x = 1 , the equation x+y=-6 becomes y=-7

When x = -7, the equation x+y=-6 becomes y=1

Therefore, the two numbers are 1 and -7

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Steve likes to entertain friends at parties with "wire tricks." Suppose he takes a piece of wire 60 inches long and cuts it into
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Answer:

a) the length of the wire for the circle = (\frac{60\pi }{\pi+4}) in

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Step-by-step explanation:

If one piece of wire for the square is y; and another piece of wire for circle is (60-y).

Then; we can say; let the side of the square be b

so 4(b)=y

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Area of the square which is L² can now be said to be;

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On the otherhand; let the radius (r) of the  circle be;

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r = \frac{60-y}{2\pi }

Area of the circle which is πr² can now be;

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     =( \frac{60-y}{4\pi } )^2

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A = A_S+A_C

   = \frac{y^2}{16} +(\frac{60-y}{4\pi } )^2

For the smallest possible area; \frac{dA}{dy}=0

∴ \frac{2y}{16}+\frac{2(60-y)(-1)}{4\pi}=0

If we divide through with (2) and each entity move to the opposite side; we have:

\frac{y}{18}=\frac{(60-y)}{2\pi}

By cross multiplying; we have:

2πy = 480 - 8y

collect like terms

(2π + 8) y = 480

which can be reduced to (π + 4)y = 240 by dividing through with 2

y= \frac{240}{\pi+4}

∴ since y= \frac{240}{\pi+4}, we can determine for the length of the circle ;

60-y can now be;

= 60-\frac{240}{\pi+4}

= \frac{(\pi+4)*60-240}{\pi+40}

= \frac{60\pi+240-240}{\pi+4}

= (\frac{60\pi}{\pi+4})in

also, the length of wire for the square  (y) ; y= (\frac{240}{\pi+4})in

The smallest possible area (A) = \frac{1}{16} (\frac{240}{\pi+4})^2+(\frac{60\pi}{\pi+y})^2(\frac{1}{4\pi})

= 126.0223095 in²

≅ 126.02 in² ( to two decimal places)

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