To find out if a number is prime you have to find it's factors. if the factors are only 1 and that number it's prime, if the number has other factors then it's composite. hope i helped! have a great day.
Answer:
1
Step-by-step explanation:
16 - 13 = 3
-4 - -7 = 3
3/3 = 1
Answer:
Step-by-step explanation:
Step-by-step explanation:
Equation of straight line is y=mx+c
choose any two points on straight line
for me I choose:(-3,11) and (3,-1)
use these two points to find gradient,m.
m= (-1-11)/(3-(-3))
m= -2
now, y=-2x+c
choose any point on the straight line
I choose point (3,-1)
sub the point into the equation to find c
-1=-2(3)+c
c=5
equation: y=-2x+5
1. so it has to have every x is one y and every y has one x
graph them
we see that the only ones that are one to one are the first one, the 2nd one and the 4th one
2.
solve for x and replace with f⁻¹(x) and y with x
minus 8 and cube both sides
(y-8)³=x-2
add 2
(y-8)³+2=x
replace
f⁻¹(x)=(x-8)³+2
first one
3.
this is a one to one function because theer is no vertical line that could intersect the graph more than once
ah, I see, you're program has a different one to one definition, no horizontal or vertical line can cross, lemme edit te first question again
answer is 2nd option
4. a neat trick is this:
the domain of f(x) is the range of f⁻¹(x)
the range of f(x) is the domain of f⁻¹(x)
the domain of f(x)=1/(3x+2)
hmm, can't divide by 0 so set denomenator to 0
3x+2=0
3x=-2
x=-2/3
domain is all real numbers except -2/3
(-infinity,-2/3)U(-2/3,infinity)
that's the range of f⁻¹(x)
range
ok, hmm, range is from positive initny to 0 then from 0 to positive inifnty, not including 0
so the domain of f⁻¹(x) is (-∞,0)U(0,∞)
range of f⁻¹(x) is (-∞,-2/3)U(-2/3,∞)
first option