Recent content by SammyS

Yes. There is an error.

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...

Try it and see.

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 .

And ΔQOA is similar to each of those two.

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...

Yes, it can be factored by grouping. Notice that ##\displaystyle \ 2a^2-1 = a^2 + (a^2-1) =a^2 + (a-1)(a+1) \ .## Group as follows: ##\displaystyle \quad a(a-1)x^2+(2a^2-1)x+a(a+1)=a(a-1)x^2+a^2\,x + (a-1)(a+1)x+a(a+1) ## ##\displaystyle \quad \quad \quad \quad \quad \quad \quad \quad \quad...

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...

@PeroK , Don't you mean ##\displaystyle \frac 1 m ## and ##\displaystyle \frac 1 n ## rather than ##m## and ##n## in this post ?

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...

Not completely. Your solution to part (b) is well organized and we can infer what the variables represent by your labelling on your graph for part (a). It would have been good for you to indicate that the area under the velocity-time graph gives the distance traveled, and that you were using...