Recent content by krebs

Here's the same as above, except the graph is moved up 1 unit on the y axis. Now if we rotate the right side around the point (0,0), it doesn't match up with the left side, so it's not odd anymore.http://tube.geogebra.org/m/1964117

What do you not understand? Here's another example of what an "odd" function will look like. If you rotate the right side 180 degrees around the origin, you get the left side of the curve. That is why it is called "odd". rotating it 180 degrees is the same as flipping it on the x and y axis...

Even function: mirrored on y-axis http://tube.geogebra.org/m/1963927 Odd function: mirrored on the origin http://tube.geogebra.org/m/X9hdpoXV Otherwise it is neither.

Not to nitpick technicality, but Gregory, do keep in the back of your mind that you are calculating velocity, KE and PE relative to the ground.

If you use latex or superscripts, you're more likely to get help. $$4a^2-4ab+b^2-c^2$$ $$(4a^2-4ab+b^2)-c^2$$ $$(2a-b)^2-c^2$$ $$(2a-b-c)(2a-b+c)$$--------------------------------------------- $$a^2+b^2-c^2-2ab$$ Why are you selecting three random terms? you have ## a\space b\space c\space...

Oh, sorry. Use the energy formulas then: $$PE = gh$$ $$KE = \frac{1}{2}v^2$$ There is one point where the ball is not moving. That's when you can use the potential energy formula...

We know that ##V = V_0 - 9.8 t## What is the antiderivative of this formula? Do it on paper and keep track of your units.

Have you taken calculus?

Yes, there are 4 junctions. You will end up seeing that several of the unknowns are equal to each other. Also the ratio of ##\frac{l_2}{l_1} = \frac{\text{Resistance of the left side of the circuit}}{\text{Total resistance of the circuit}}##

Yes.

Correct. Then you need Kirchhoff. Start with your 40 mA total, and reconstruct your circuit piece by piece to calculate the current and voltage through each resistor.

Yep, exactly. Use the same formulas you used to condense the right side to condense the entire circuit. ##\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_n...}## is one of them

Two resistors are in series with each other, and in parallel with another...

I think you might be caught up on the wrong things. We are working backwards along the following proof. Normally I wouldn't post an entire proof, but you've already solved the problem. I want to make sure you have a complete understanding of how that solution worked. There exists a natural...

Sorry, you were wrong in #13 and I didn't catch it. It's a minor error, but ##\frac{1}{t^2 + 7} \leq \frac{1}{t^2}## ##t^2 \leq t^2+7## ## 0 \leq 7## thus, the second inequality is always true.