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3 years ago

What is the missing length?

Mathematics

2 answers:

Alika [10]3 years ago

8 0

The missing length is 51.5, I hope this helps :)

scoray [572]3 years ago

7 0

Answer:

Length=12mi

Step-by-step explanation:

Area=123.6mi^3

Width=10.3mi

Length=u

Formula: Area=LxW

Area÷W=u

123.6÷10.3=12

Length=12mi

Hope this helps :)

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3 years ago

Write a polynomial function that meets the given conditions. Answers may vary. Degree 2 polynomial with zeros of 2√ 13 and -2√13

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Answer:

The polynomial function is $x^{2} - 52$

Step-by-step explanation:

A polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = (x - x_{1})*(x - x_{2})$ .

In this problem:

The roots are $x_{1} = 2\sqrt{13}$ and $x_{2} = -2\sqrt{13}$

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3 years ago

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Answer:

Step-by-step explanation:

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2 years ago

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3 years ago

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$x^2 + 11x - 4 = 0$

i.e. now we're looking for the roots of the polynomial. To find them, we can use the following formula:

$x_{1,2} = \frac{-b\pm\sqrt{b^2-4ac}}{2}$

where $x_{1,2}$ is a compact way to indicate both solutions $x_1$ and $x_2$ , while $a,b,c$ are the coefficients of the quadratic equation, i.e. we consider the polynomial $ax^2+bx+c$ .

So, in your case, we have $a=1,\ \ b=11,\ \ c=-4$

Plug those values into the formula to get

$x_{1,2} = \frac{-11\pm\sqrt{121+16}}{2} = \frac{-11\pm\sqrt{137}}{2}$

So, the two solutions are

$x_1 = \frac{-11+\sqrt{137}}{2}$

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