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Dimas [21]
1 year ago
7

Solve for x in the equation x squared minus 12 x + 59 = 0. x = negative 12 plus-or-minus StartRoot 85 EndRoot x = negative 6 plu

s-or-minus StartRoot 23 EndRoot i x = 6 plus-or-minus StartRoot 23 EndRoot i x = 12 plus-or-minus StartRoot 85 EndRoot
Mathematics
2 answers:
jonny [76]1 year ago
5 0

Answer:

On edge it's C

Step-by-step explanation:

x= 6 ±\sqrt{23} i

fgiga [73]1 year ago
4 0

Value of x is: x=6\pm\sqrt{23}i

Option B is correct option.

Step-by-step explanation:

We need to solve the equation x^2-12x+59=0 and find value of x.

Using quadratic formula:

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

Putting values of a, b and c and finding the value of x

a=1, b=-12 and c=59

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x=\frac{-(-12)\pm\sqrt{(-12)^2-4(1)(59)}}{2(1)}\\x=\frac{12\pm\sqrt{144-236}}{2}\\x=\frac{12\pm\sqrt{-92}}{2}\\x=\frac{12\pm\sqrt{92}i}{2}\\We\,\,know\,\sqrt{-1} \,\,is\,\,i\\x=\frac{12\pm2\sqrt{23}i}{2}\\x=\frac{2(6\pm\sqrt{23}i)}{2}\\x=6\pm\sqrt{23}i

So, value of x is: x=6\pm\sqrt{23}i

Option B is correct option.

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1 year ago
Find the quantity q that maximizes profit if the total revenue, R(q), and total cost, C(q) are given in dollars by R(q)=3q−0.002
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Answer:

The value of q that maximize the profit is q=200 units

Step-by-step explanation:

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we have

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The vertex represent a maximum

The x-coordinate of the vertex represent the value of q that maximize the profit

The y-coordinate of the vertex represent the maximum profit

using a graphing tool

Graph the quadratic equation

The vertex is the point (200,-120)

see the attached figure

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The value of q that maximize the profit is q=200 units

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