√129 is less than 
.
Let us first solve for √129.
➝
➝
Now,
➝
➝
Clearly,
➵
➵
Hence, √129 is less than 
.
Answer:
Step-by-step explanation:
Since we believe it
The number sequence is that those which can be described as 
and that is not ending and recurring. Figures that are unreasonable are also the ones, that are not and therefore are non-ending, non-recurring.
The number sequence is 
Rational amount since it is classified as .
is really a rational number.
is a real function, since it does not end but repeats it.
The rational number 
Unreasonable number 
.
F = 2 solutions
G = 2 solutions
H = infinitely
J = 2
Hello there!
To find the increasing intervals for this graph just based on the equation, we should find the turning points first.
Take the derivative of f(x)...
f(x)=-x²+3x+8
f'(x)=-2x+3
Set f'(x) equal to 0...
0=-2x+3
-3=-2x
3/2=x
This means that the x-value of our turning point is 3/2. Now we need to analyze the equation to figure out the end behavior of this graph as x approaches infinity and negative infinity.
Since the leading coefficient is -1, as x approaches ∞, f(x) approaches -∞ Because the exponent of the leading term is even, the end behavior of f(x) as x approaches -∞ is also -∞.
This means that the interval by which this parabola is increasing is...
(-∞,3/2)
PLEASE DON'T include 3/2 on the increasing interval because it's a turning point. The slope of the tangent line to the turning point is 0 so the graph isn't increasing OR decreasing at this point.
I really hope this helps!
Best wishes :)
