The domain of f/g
consists of numbers x for which g(x) cannot equal 0 that are in the domains of
both f and g.
Let’s take this equation as an example:
If f(x) = 3x - 5 and g(x)
= square root of x-5, what is the domain of (f/g)x.
For x to be in the domain of (f/g)(x), it must be
in the domain of f and in the domain of g since (f/g)(x) = f(x)/g(x). We also
need to ensure that g(x) is not zero since f(x) is divided by g(x). Therefore,
there are 3 conditions.
x must be in the domain of f:
f(x) = 3x -5 are in the domain of x and all real numbers x.
x must be in the domain of g:
g(x) = √(x - 5) so x - 5 ≥ 0 so x ≥ 5.
g(x) can not be 0: g(x)
= √(x - 5) and √(x - 5) = 0 gives x = 5 so x ≠ 5.
Hence to x x ≥ 5 and x ≠ 5
so the domain of (f/g)(x) is all x satisfying x > 5.
Thus, satisfying <span>satisfy all
three conditions, x x ≥ 5 and x ≠ 5 so the domain of (f/g)(x) is all x
satisfying x > 5.</span>
Closest would be 8 43/200
Let x=number 1 let y=number2
there sum is equal to 70
then
number1+number2=70
substituting
x+y=70
y=70-x
let there product will be p
p=xy
substituting
p=x(70-x)
which is choice c
9514 1404 393
Answer:
nπ -π/6 . . . for any integer n
Step-by-step explanation:
tan(x) +√3 = -2tan(x) . . . . . given
3tan(x) = -√3 . . . . . . . . . . . add 2tan(x)-√3
tan(x) = -√3/3 . . . . . . . . . . divide by 3
x = arctan(-√3/3) = -π/6 . . . . use the inverse tangent function to find x
This is the value in the range (-π/2, π/2). The tangent function repeats with period π, so the set of values of x that will satisfy this equation is ...
x = n·π -π/6 . . . . for any integer n
