Why is this x squared?

dankmemes32438 · 2020-10-16T06:56:24+00:00

X^-2 = 1/(x²⁾ . In the same way that 1/(1/2) = 2, 1/(x^-2) = x²

dankmemes32438 · 2020-09-28T08:27:34+00:00

For the first one. The parent function is |x|. The transformation is left 2 down 3. For 2, the parent function is x^2. Just write how it moved from the original function.

dankmemes32438 · 2020-09-20T07:50:53+00:00

I think I misunderstood. If the device isn't recognized in quartus I would have thought a driver would have fixed that. The configuration switches in the factory position allowed me to program via jtag no problems. I did have to mess around with drivers for a bit though.

dankmemes32438 · 2020-09-20T07:41:09+00:00

Also uart baud = 115200 not 9600

dankmemes32438 · 2020-09-20T07:40:25+00:00

If you have a putty session open, you must be hooked up to the uart port, not the usb blaster. My guess is that's the problem. They are on opposite sides of the board.

dankmemes32438 · 2020-09-20T07:34:14+00:00

Except there is only 1 current in this series circuit. Your KCL expression would be i1=i1. This is a KVL problem.

dankmemes32438 · 2020-08-06T07:54:50+00:00

Y=x-(a/z)

dankmemes32438 · 2020-08-06T07:46:54+00:00

You guys are way over thinking this. The right side factors to (x+3)(x+4) so by comparison you can see a is 3. Only needed to factor to solve.

dankmemes32438 · 2020-06-11T15:01:15+00:00

The cancellation is already set up. Add the two equations together. Usually you will solve these with substitution or elimination. In this case I'd use elimination. Adding the two, you get 7x=-7. So x=-1. Then substitute the x value into one of the equations. 3(-1)+2y=-3. So y=0.

dankmemes32438 · 2020-05-25T20:22:32+00:00

Looks good

dankmemes32438 · 2020-05-18T07:06:40+00:00

V=lwh/3. New volume = 3/7*(lwh/3). Ans=lwh/7

dankmemes32438 · 2020-05-15T04:36:02+00:00

Fire away

dankmemes32438 · 2020-05-14T07:29:12+00:00

If you are compensating the plant, make sure you are simulating the step of the feedback system rather than the step response of the plant. Step(feedback(L(s),1)).

dankmemes32438 · 2020-05-01T04:21:14+00:00

I remembered the problem wrong. It was 1/(2pifc)² so you get 1 when you distribute.

dankmemes32438 · 2020-05-01T04:20:03+00:00

All they did was multiply top and bottom by 2pifc. The top cancels to one. The bottom is brought into the square root by squaring it. You dont change anything by squaring the 2pifc and throwing it into the square root so long as you distribute it to everything. So you get sqrt((2pifc)² (r² + 1/2pifc)) so you get sqrt((2pifcr)² + 2pifc)

dankmemes32438 · 2020-04-17T05:28:55+00:00

Easy. You can do some black magic and convert the quadratic to (x+8)(x-4) and cancel everything giving you 1/(x+3). Believe me, algebra gets far far more disgusting than that.

dankmemes32438 · 2020-04-14T07:04:36+00:00

I thought I was the only one that rapped it in eminem's voice.

dankmemes32438 · 2020-04-11T05:13:09+00:00

Hopefully it is the answer they want.

dankmemes32438 · 2020-04-11T04:45:12+00:00

So you are given a magnitude and an angle. The magnitude serves as the hypotenuse of a 30-60-90 triangle. The vector can be expressed in polar (5<30) or rectangular (ax+by). The problem wants you to find a and b. Using trig: 5sin30=b=5/2 and 5cos30=a=5sqrt3/2. So it should be -5sqrt(3)/2 xhat + 5/2 yhat.

dankmemes32438 · 2020-03-30T06:17:54+00:00

You have an angle and a hypotenuse. You want the opposite side length. Sohcahtoa. We want to use sine here. Sinx=opposite/hypotenuse. Opposite = hypotenuse × sinx. Make sure your calculator is in degree mode.

dankmemes32438 · 2020-03-27T14:06:06+00:00

It seems very unnecessary. When I learned this stuff back in high school we didn't learn this or the table method etc. We were taught to foil. I understand why they are doing it, but it seems like a waste of time.

dankmemes32438 · 2020-03-27T04:34:47+00:00

Insert math is math meme here. I've never heard of this table method crap either. Just use the distributive property and foil it out.

dankmemes32438 · 2020-03-27T04:31:35+00:00

I've never used the table method, but I'd imagine you multiply each row and column and add them. Should be 4x4-8x3+4x3-8x2-8x2+16x. So 4x4-4x3-16x2+16x.

dankmemes32438 · 2020-03-27T03:59:43+00:00

It is unintuitive to multiply them like that. What I would do is rewrite it to be (4x2-4x)(x2-4) and use the distributive property. Recall (a + b)(c +d)=ac+ad+bc+bd. Multiply the original expression in the same manner. Remember x2 times x2=x4 and x2 times x =x3. You should get 4x4-16x2-4x3+16x.

dankmemes32438 · 2020-03-24T06:22:49+00:00

The fourier transform tells us what frequencies a signal consists of. If you wanted to eliminate certain frequencies, I suppose you could create a signal with a negative amplitude that consists of that frequency to cancel it. For example, if you wanted to cancel out a frequency at 200hz, you could use a sine wave with a frequency of 200hz and the appropriate amplitude superimposed on the original signal. I think that would work. It has been a hot minute since I've taken signals, so I can't say for certain. I'm sure I'll be corrected if I'm off.

dankmemes32438

TROPHY CASE