Answer:
how are you defining tens and ones' places for words?
Step-by-step explanation:
je parle francais
Answer:
Domain: {-2, -1, 0, 1, 2, 3}
Range: {-5, -3, -1, 1, 3, 5}
Step-by-step explanation:
The domain is the set containing all the x-coordinates, and the range is the set containing all the y-coordinates.
Domain: {-2, -1, 0, 1, 2, 3}
Range: {-5, -3, -1, 1, 3, 5}
So, f[x] = 1/4x^2 - 1/2Ln(x)
<span>thus f'[x] = 1/4*2x - 1/2*(1/x) = x/2 - 1/2x </span>
<span>thus f'[x]^2 = (x^2)/4 - 2*(x/2)*(1/2x) + 1/(4x^2) = (x^2)/4 - 1/2 + 1/(4x^2) </span>
<span>thus f'[x]^2 + 1 = (x^2)/4 + 1/2 + 1/(4x^2) = (x/2 + 1/2x)^2 </span>
<span>thus Sqrt[...] = (x/2 + 1/2x) </span>
Answer:
(x+8) ^2=64
Step-by-step explanation:
that will give you your two solutions: x= -16 and x=0. I hope this helps!!!
