A)
The discriminant (radicand) is √(b^2-4ac), let us call this "d" for the discriminant.
If:
d<0, there are no real solutions (though there are two imaginary ones)
d=0, there is one real solution
d>0, there are two real solutions.
In this case, d=12^2-4(4)9
d=144-144
d=0
So there is one real solution.
B)
9x^2-30x+25=0
9x^2-15x-15x+25=0
3x(3x-5)-5(3x-5)=0
(3x-5)(3x-5)=0
(3x-5)^2=0
x=5/3
x=1 2/3
Answer:
no solution
Step-by-step explanation:
12x − 18y = 27
4x − 6y = 10
Multiply the second equation by -3 so we can eliminate x
-3(4x − 6y) = 10*-3
-12x +18y = -30
Add the first equation to this new equation
12x − 18y = 27
-12x +18y = -30
------------------------
0 = -3
This is not a true equation, so there is no solution
Answer:log₂(
)
Explanation:Before we begin, remember the following:logₐ(x) - logₐ(y) = logₐ(
)
alog(x) = log(xᵃ)
Now, for the given we have:3 log₂(x) - (log₂3 - log₂(x+4))
log₂(x²) - log₂(
)
log₂(
) = log₂(
)
Hope this helps :)
Answer:
I think A or C I'm not sure tho