If I've read that correctly, you are asking if cube_root of all of x over 27 can be written in the form kx to the power p
ie ³√(x/27) = kx^p
Yes, with k = 1/3, p = 1/3
Cube root can be expressed as something to the power 1/3 since (x^a)^b = x^(ab) and cube roots are such that (³√x)³ = x and 1/3 × 3 = 1.
³√(x/27) = (x/27)^(1/3) = (x^(1/3))/(27^(1/3)) = (x^(1/3))/3 = (1/3)x^(1/3)
→ k = 1/3, p = 1/3.
w=2*pi*sqr(x/t) First divide each side by 2*pi w/(2pi)=sqr(x/t) Next square both sides (w/2pi)^2=x/t Next multiply both sides by t t(w/2pi)^2=x
w=3pi*sqr(x/t) First divide each side by 3pi w/3pi=sqr(x/t) Next square both sides. (w/3pi)^2=x/t Next multiply both sides by t t(w/3pi)^2=x
I think you have it backwards. The two integers with the square root of 61 BETWEEN them are 7 and 8.
Which two numbers doesStartRoot 128 EndRoot lie between on a number line? 11.0 and 11.1 11.2 and 11.3 11.3 and 11.4 11.4 and 11.5
The domain of f is x is R (if imaginary roots are permitted, and there is nothing in the question to suggest otherwise). The domain of g is R excluding x = 5 So the domain of f + g is R excluding x = 5 and the domain of f/g is R excluding x = 0