Any number to the power of zero equals 1.
A. To make (7^2)^x = 1
The x has to equal zero
B. (7^0) = 1
So in (7^0)^x = 1
(1)^x , x can be any real number
Both expression have the same denominator: 9x²-1. Thus it must not be 0.
9x²-1=(3x-1)(3x+1)=0, resulting x=+-1/3.
Restrictions: x in R\{-1/3, 1/3}
Adding those expressions:
E=(-x-2)/(9x²-1 ) + (-5x+4)/(9x²-1)=
(-x-2-5x+4)/(9x²-1)=(-6x+2)/(9x²-1)=
(-2)(3x-1)/(9x²-1)=-2/(3x+1)
E=-2/(3x+1)
Answer:
the answer is 2.5
Step-by-step explanation:
add 4+3+2+1=10
add 2+1+8+10+4=25
divide 25÷10= 2.5
Step-by-step explanation:
(a-b)^2=(a^2-2*a*b+b^2)
4*(x^2-2x+1)-9=4x^2-8x+4-9=4x^2-8x-5
if you have to calculate the Δ
there is it: 4x^2-8x-5=0
Δ=b^2-4*a*c=64-4*4*(-5)=64+80=144
-b±Δ/2*a=8±12/8=
=8+12/8=20/8|:4=5/2
=8-12/8=-4/8|:4=-1/2