You wrote the above in response to @pasmith's Post #4 .
Maybe adding a few clarifying words will help you understand.
Keeping ##a## fixed, ##3a^2 + 2b## is strictly increasing in the direction of increasing ##b##, and keeping ##b## fixed, ##3b^2 + 2a## is strictly increasing in the direction...
Right.
##0.8## radians ##\approx 45.8^\circ## . Of course, the given value for the solution for ##x## has only one significant figure. It should be something closer to ##0.7767## radians.
It does not really matter that @Hill disagrees.
The given problem, as you eventually stated in post #9, is quite clear. Your recent replies to @FactChecker reinforce that the given problem refers to 2D grid on a planar surface.
Judging by your method of solution, you have the constraint that the hypotenuse has a fixed length.
For ##c=5##, your solution gives ##a=\dfrac {5}{\sqrt 2}\approx 3.5355\,.\ ## So that ##b=\dfrac {5}{\sqrt 2}\ ## as well. This gives ##a+b=5\,\sqrt{2\,}\approx 7.071\,.\ ##
This is actually the...
I suspect that the problem asks for the circumference and area of the region you found the circumference of.
The area of this region is even easier to determine .
You asked how to go from
##\displaystyle \quad (x+1)^2 a^2+(1-x)(x+1)a-x##
to
##\displaystyle \quad ((x+1)a+1)((x+1)a-x)##
You have mentioned using grouping to factor in the OP. That's what @Hill uses in Post #16.
Adding some detail to doing the grouping:
First you need to split the "middle...
This can also be factored as a quadratic in ##a## .
Using your ##\displaystyle \text{Autodesk Sketchbook}^{\circledR} ## image we get
##\quad a(a-1)x^2+(2a^2-1)x+a(a+1)##
##\displaystyle \quad =(x^2+2x+1)a^2-(x^2-1)a-x##
##\displaystyle \quad =(x+1)^2 a^2+(1-x)(x+1)a-x##
##\displaystyle...
Suppose, as you propose, that for the battery on the left we have 2A flow out and 1A flow in.
After 1 second of time in this situation, this battery would have sent out 1 Coulomb more charge than what it had received. Conservation of charge should tell you that this battery would then have...
I agree with you.
The Rayleigh Criterion says that when two distant lights sources (wavelength, ##\lambda\,##) are viewed through a circular aperture of diameter, ##D##, they may be resolved provided that there angular separation, ##\theta## (in radians) is such that
##\displaystyle \quad...
I have previously replied to this post, but at that time I did not realize that this was the correct for ##v(t)## on the interval, ##\displaystyle \ 2\le t\le12\,,\ ## due to the "y-intercept" being ##-3.6## while that of the velocity graph was zero.
I suppose that rather than using the...