Answer:
Step-by-step explanation:
\[2~ sin^2 x+3sin x+1=0\]
\[2sin^2x+2sin x+sin x+1=0\]
2sinx(sin x+1)+1(sin x+1)=0
(sin x+1)(2 sin x+1)=0
either sin x+1=0
sin x=-1=sin 3π/2=sin (2nπ+3π/2)
x=2nπ+3π/2,where n is an integer.
or 2sin x+1=0
sin x=-1/2=-sin π/6=sin (π+π/6),sin (2π-π/6)=sin (2nπ+7π/6),sin (2nπ+11π/6)
x=2nπ+7π/6,2nπ+11π/6,
where n is an integer.
The slope is 3/4
you just count how many it goes up before it goes over to a whole unit
QUESTION:
The code for a lock consists of 5 digits (0-9). The last number cannot be 0 or 1. How many different codes are possible.
ANSWER:
Since in this particular scenario, the order of the numbers matter, we can use the Permutation Formula:–
- P(n,r) = n!/(n−r)! where n is the number of numbers in the set and r is the subset.
Since there are 10 digits to choose from, we can assume that n = 10.
Similarly, since there are 5 numbers that need to be chosen out of the ten, we can assume that r = 5.
Now, plug these values into the formula and solve:
= 10!(10−5)!
= 10!5!
= 10⋅9⋅8⋅7⋅6
= 30240.
G(x)/f(x) will be simplified to (x+3)(x-3)/2-x^1/2,
which will give you [0,4) ∪(4, ∞).
Choice B
