To simplify the function, we need to know some basic identities involving exponents.
1. b^(ax)=(b^x)^a=(b^a)^x
2. b^(x/d) = (b^x)^(1/d) = ((b^(1/d)^x)
Now simplify f(x), where
f(x)=(1/3)*(81)^(3*x/4)
=(1/3)(3^4)^(3*x/4) [ 81=3^4 ]
=(1/3)(3^(4*3*x/4) [ rule 1 above ]
=(1/3) (3^(3*x)
=(1/3)(3^(3x)) [ or (1/3)(27^x), by rule 1 ]
(A) Initial value is the value of the function when x=0, i.e.
initial value
= f(0)
=(1/3)(3^(3x))
=(1/3)(3^(3*0))
=(1/3)(3^0)
=(1/3)(1)
=1/3
(B) the simplified base base is 3 (or 27 if the other form is used)
(C) The domain for an exponential function is all real values ( - ∞ , + ∞ ).
(D) The range of an exponential function with a positive coefficient and without vertical shift is ( 0, + ∞ ).
Answer:
5.8 m
Step-by-step explanation:
Using the Pythagorean Theorem (a² + b² = c²) you get the equation a² + 1.5² = 6² (or 1.5² + b² = 6²). Then you solve from there and rounding to the first decimal place you get 5.8
35 has 4 divisors, hence two factor pairs: 1*35 and 5*7. Each corresponds to a set of perfect squares that differ by 35
One pair is ((35±1)/2)^2 = {17^2, 18^2} = {289, 324}
The other is ((7±5)/2)^2 = {1^2, 6^2} = {1, 36}
You FOIL
9x^2-15x+21x-35
Combine like terms
9x^2 + 6x - 35
