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Romashka [77]
3 years ago
9

Select the two values of x that are roots of this equation

Mathematics
1 answer:
Jet001 [13]3 years ago
7 0
<h2>Hello!</h2>

The answers are:

B.   \frac{-3-\sqrt{29}}{2}

C.   \frac{-3+\sqrt{29}}{2}

<h2>Why?</h2>

We can use the quadratic equation to find the two values of x that are roots of the given equation. We must remember that most of the quadratic equations have two roots, however, we could find quadratic equations with just one root or even with no roots, at least in the real numbers.

Quadratic equation:

\frac{-b+-\sqrt{b^{2}-4ac} }{2a}

So,

From the given equation we have:

a=1\\b=3\\c=-5

Substituting it into the quadratic equation to find the roots, we have:

\frac{-b+-\sqrt{b^{2}-4ac} }{2a}=\frac{-3+-\sqrt{3^{2}-4*1*-5} }{2*1}\\\\\frac{-3+-\sqrt{3^{2}-4*1*-5} }{2*1}=\frac{-3+-\sqrt{9+20} }{2}\\\\\frac{-3+-\sqrt{29}}{2}

So,

x_{1}=\frac{-3-\sqrt{29}}{2}\\\\x_{2}=\frac{-3+\sqrt{29}}{2}

Hence, the correct options are B and C.

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A lake polluted by bacteria is treated with an antibacterial chemical. Aftertdays, thenumber N of bacteria per milliliter of wat
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<u>Extrema values of functions </u>

If the first and second derivative of a function f exists, then f'(a)=0 will produce values for a called critical points. If a is a critical point and f''(a) is negative, then x=a is a local maximum, if f''(a) is positive, then x=a is a local minimum.  

We are given a function (corrected)

N(t) = 20(t^2-lnt^2)+ 30

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(a)

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Simplifying

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(b)

N(1)=20(1^2-2ln1)+ 30

N(1)=50 is a minimum

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(d)

N(15)=20(15^2-2ln15)+ 30

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Step-by-step explanation:

hope this helps

7 0
3 years ago
Read 2 more answers
Answer???
olga nikolaevna [1]
5+6.50+7.21=18.71 30-18.71=$11.29 left
5 0
3 years ago
Read 2 more answers
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