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scoray [572]
4 years ago
7

Juanita has 3 rectangular cards that are 8 inches by 9 inches. How can she arrange the cards, without overlapping, to make one l

arger polygon with the smallest possible perimeter? How will the area of the polygon compare to the combined area of the 3 cards?

Mathematics
1 answer:
ivann1987 [24]4 years ago
6 0

Answer:

Step-by-step explanation:

Given  

  • 3 rectangular cards
  • each card is 8 inches by 9 inches

The way that she can arrange the cards, without overlapping, to make one larger polygon with the smallest possible perimeter is that:

Put 3 cards together according to the length

So  the smallest possible perimeter is:

8 + 8 + 8 + 8 + 8 + 8 + 9 + 9 = 66 inches as you can see in the attached photo.

The area of the polygon is equal to the combined area of the 3 cards without overlapping or extending.

Hope it can find you well.

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The three roots of x^3 + 7x^2 + 12x = 0 is 0,-3 and -4

<u>Solution:</u>

We have been given a cubic polynomial.

x^{3}+7 x^{2}+12 x=0

We need to find the three roots of the given polynomial.

Since it is a cubic polynomial, we can start by taking ‘x’ common from the equation.

This gives us:

x^{3}+7 x^{2}+12 x=0

x\left(x^{2}+7 x+12\right)=0   ----- eqn 1

So, from the above eq1 we can find the first root of the polynomial, which will be:

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Now, we need to find the remaining two roots which are taken from the remaining part of the equation which is:

x^{2}+7 x+12=0

we have to use the quadratic equation to solve this polynomial. The quadratic formula is:

x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

Now, a = 1, b = 7 and c = 12

By substituting the values of a,b and c in the quadratic equation we get;

\begin{array}{l}{x=\frac{-7 \pm \sqrt{7^{2}-4 \times 1 \times 12}}{2 \times 1}} \\\\{x=\frac{-7 \pm \sqrt{1}}{2}}\end{array}

<em><u>Therefore, the two roots are:</u></em>

\begin{array}{l}{x=\frac{-7+\sqrt{1}}{2}=\frac{-7+1}{2}=\frac{-6}{2}} \\\\ {x=-3}\end{array}

And,

\begin{array}{c}{x=\frac{-7-\sqrt{1}}{2}} \\\\ {x=-4}\end{array}

Hence, the three roots of the given cubic polynomial is 0, -3 and -4

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