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sdas [7]
2 years ago
15

Which of the following is the complete list of roots for the polynomial function f(x)=(x2+6x+8)(x2+6x*13)

Mathematics
1 answer:
anygoal [31]2 years ago
3 0
Factorise x^2+6x+8=0 to get (x+4)(x+2)=0
Solve to get x=-4, x=-2
Factorise x^2+6x+13=0 (I assumed you mistyped and wrote * instead of +)
Use the quadratic formula and you find that there are no real roots so the answer for this is just x=-4, x=-2
Tell me if I made any mistakes
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PEMDAS is used to help us remember the "order of operations." What do we use to help us remember the right triangle trig formula
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SOH CAH TOA

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3 years ago
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Find positive numbers x and y satisfying the equation xyequals12 such that the sum 4xplusy is as small as possible. Let S be the
WITCHER [35]

Answer:

The objective function in terms of one​ number, x is

S(x) = 4x + (12/x)

The values of x and y that minimum the sum are √3 and 4√3 respectively.

Step-by-step explanation:

Two positive numbers, x and y

x × y = 12

xy = 12

S(x,y) = 4x + y

We plan to minimize the sum subject to the constraint (xy = 12)

We can make y the subject of formula in the constraint equation

y = (12/x)

Substituting into the objective function,

S(x,y) = 4x + y

S(x) = 4x + (12/x)

We can then find the minimum.

At minimum point, (dS/dx) = 0 and (d²S/dx²) > 0

(dS/dx) = 4 - (12/x²) = 0

4 - (12/x²) = 0

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To just check if this point is truly a minimum

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4 0
3 years ago
21. In 4 + In(4x - 15) = In(5x + 19)
Alexxx [7]

Answer:

x=-30;\quad \:I\ne \:0

Step-by-step explanation:

In\cdot \:4+In\left(4x-15\right)=In\left(5x+19\right)

\:4+In\left(4x-15\right):\quad -11nI+4nxI\\In\cdot \:4+In\left(4x-15\right)\\=4nI+nI\left(4x-15\right)\\\\\:In\left(4x-15\right):\quad 4nxI-15nI\\\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b-c\right)=ab-ac\\a=In,\:b=4x,\:c=15\\=In\cdot \:4x-In\cdot \:15\\=4nxI-15nI\\=In\cdot \:4+4nxI-15nI\\\mathrm{Simplify}\:In\cdot \:4+4nxI-15nI:\quad -11nI+4nxI\\In\cdot \:4+4nxI-15nI\\\mathrm{Group\:like\:terms}\\=4nI-15nI+4nxI\\\mathrm{Add\:similar\:elements:}\:4nI-15nI=-11nI\\=-11nI+4nxI\\

\mathrm{Expand\:}In\left(5x+19\right):\quad 5nxI+19nI\\In\left(5x+19\right)\\=nI\left(5x+19\right)\\\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b+c\right)=ab+ac\\a=In,\:b=5x,\:c=19\\=In\cdot \:5x+In\cdot \:19\\=5nxI+19nI\\\\-11nI+4nxI=5nxI+19nI\\\\\mathrm{Add\:}11nI\mathrm{\:to\:both\:sides}\\-11nI+4nxI+11nI=5nxI+19nI+11nI\\Simplify\\4nxI=5nxI+30nI\\\mathrm{Subtract\:}5nxI\mathrm{\:from\:both\:sides}\\4nxI-5nxI=5nxI+30nI-5nxI\\\mathrm{Simplify}\\-nxI=30nI\\

\mathrm{Divide\:both\:sides\:by\:}-nI;\quad \:I\ne \:0\\\frac{-nxI}{-nI}=\frac{30nI}{-nI};\quad \:I\ne \:0\\\mathrm{Simplify}\\\frac{-nxI}{-nI}=\frac{30nI}{-nI}\\\mathrm{Simplify\:}\frac{-nxI}{-nI}:\quad x\\\frac{-nxI}{-nI}\\\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{-b}=\frac{a}{b}\\=\frac{nxI}{nI}\\\mathrm{Cancel\:the\:common\:factor:}\:n\\=\frac{xI}{I}\\\mathrm{Cancel\:the\:common\:factor:}\:I\\=x\\\mathrm{Simplify\:}\frac{30nI}{-nI}:\quad -30\\\mathrm{Apply\:the\:fraction\:rule}:

\quad \frac{a}{-b}=-\frac{a}{b}\\\mathrm{Cancel\:the\:common\:factor:}\:n\\=\frac{30I}{I}\\\mathrm{Cancel\:the\:common\:factor:}\:I\\=-30\\x=-30;\quad \:I\ne \:0

3 0
3 years ago
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