Find the equation of a line parallel to y=3x−6that passes through the point (1,-5).
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Answer:
Seven-thirds
Step-by-step explanation:
The give equation is .
The leading coefficient is 9
The constant term is 7
According to the rational root theorem, the ratio of factors the constant terms to that of the coefficient of the leading term are all possible rational roots of the given polynomial.
Base on this theorem, or seven thirds is a potential root because the numeration is a factor of 7 and the denominator is a factor of 9.
Hello,
y=2^(-x)
y=2^(2x)+3
==>2^(2x)+3=1/2^x
==>2^(3x)+3*2^x-1=0 (1)
Let's assume u=2^x
(1)==>u^3+3*u-1=0
which as 3 roots
u=0.322185354626 or
u = -0.161092677313 + i1.754380959784 or
u = -0.161092677313 - i1.754380959784.
Let's take the real solution
0.322185354626=2^x
==>x=ln(0.322185354626) / ln(2)
x=-1,6340371790199...
an other way is
f(x)=2^(3x)+3*2^x-1
f(-2)=1/64+3/4-1=-15/64 <0
f(-1)=1/8+1-1=1/8>0
==> there is a solution betheen -2<x<-1
y=2^(-x)
y=2^(2x)+3
==>2^(2x)+3=1/2^x
==>2^(3x)+3*2^x-1=0 (1)
Let's assume u=2^x
(1)==>u^3+3*u-1=0
which as 3 roots
u=0.322185354626 or
u = -0.161092677313 + i1.754380959784 or
u = -0.161092677313 - i1.754380959784.
Let's take the real solution
0.322185354626=2^x
==>x=ln(0.322185354626) / ln(2)
x=-1,6340371790199...
an other way is
f(x)=2^(3x)+3*2^x-1
f(-2)=1/64+3/4-1=-15/64 <0
f(-1)=1/8+1-1=1/8>0
==> there is a solution betheen -2<x<-1