kumold

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21 Answers

+42

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there should be conditions:
r+4>0
r>0
r+1>0 --->r>0

kumoldJun 8, 2012

+42

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it means that 10 ^{1+1.36437635384187i}=-10
here i = (-1)^1/2

kumoldJun 8, 2012

+42

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kumoldJun 8, 2012

+42

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whole part of 3.14 is 3

kumoldJun 8, 2012

+42

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Depends on what kind of calculator you have.

kumoldJun 8, 2012

+42

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kumoldJun 8, 2012

+42

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Vertical asymptote is x=3
and Oblique asymptote is y=1-x, which is found like this:
y=(-x ²+4x+3)/(x-3)=-x+1+6/(x-3)
since 6/(x-3) <> 0, y=-x+1+6/(x-3) <>-x+1

kumoldJun 7, 2012

+42

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for decomposition to fractions, firstly we need to find roots of equation 6x ²+7x-3=0
Discriminant of equation
D=49+72=121=11 ²
roots:
x _1/2=(-7±11)/12= -3/2 and 1/3
so, second-degree polynomial 6x ²+7x-3 can be written as 6(x+3/2)(x-1/3) or (2x+3)(3x-1)
from there fraction 22/(6x^2+7x-3)=22/(2x+3)(3x-1) can be written as
22/(2x+3)(3x-1)=a/(2x+3)+b/(3x-1)
now let's find a and b
a/(2x+3)+b/(3x-1)=(a(3x-1)+b(2x+3))/(2x+3)(3x-1)=(3ax-a+2bx+3b)/(2x+3)(3x-1)=(x(3a+2b)+3b-a)/(2x+3)(3x-1)
here we get set of equations:
3a+2b=0
3b-a=22

3a+2b=0
a=3b-22

3(3b-22)+2b=0
a=3b-22

9b-66+2b=0
a=3b-22

11b=66
a=3b-22

b=6
a=3*6-22=-4

so, fraction decomposition of 22/(6x^2+7x-3) will be 6/(3x-1)-4/(2x+3)

kumoldJun 6, 2012

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16 ^(x-4) = 9 ^(-9x)
log ₁₀ 16 ^(x-4) = log ₁₀9 ^(-9x)
(x-4) log ₁₀ 16=-9x log ₁₀ 9
x(log ₁₀ 16 + 9 log ₁₀ 9)=4 log ₁₀ 16
xlog ₁₀ (16*9 ⁹)=4 log ₁₀ 16
x=4 log ₁₀ 16/log ₁₀ (16*9 ⁹)

kumoldJun 6, 2012

+42

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what do you mean by "make fractions"?

kumoldJun 6, 2012