Country Boy

A function, f(x), is continuous at x= a if and only if 1) f(a) exists 2) $\lim_{x\to a} f(x)$ exists 3)$\lim_{x\to a} f(x)= f(a) Since (3) requires both (1) and (2) often we just state (3). Of course, if a function is NOT continuous at x= a, it is discontinuous there, Here, [tex]f(x)= \frac{x^2- 1}{x- 1}[/tex] is not continuous at x=1 becaue $f(1)= \frac{0}{0}$ does not exist. As long as x is NOT 1, [tex]\frac{x^2- 1}{x- 1}= x+ 1[/tex]. g(x)= x+ 1 exists at x= 1, g(1)= 2 and $\lim_{x\to 1} x+ 1= (\lim_{x\to 1} x)+ 1= 1+ 1= 2$ so this new function, g(x)= x+ 1, is continuous at x= 1.

What does "at rest" mean? What does "equilibrium" mean?

Have you ever taken a trigonometry course. If so what can you say about the angles in a right triangle with one leg twice as long as the other? You may need a calculator with trig function and their inverse functions.

If two machines, each having 90% efficiency, work together the result will be (0.9)(0.9)= 0.81 or 81% efficiency. The result will be less efficient, not more.

Okay, you have 0.2 Moles of Potassium and 0.4 WHAT of Sodium Choloride? 0.4 Moles? You say "I really don't know what to do with the compounds present in the solution." Well what are you trying to do? What question are you trying to answer? Your first sentence is a statement of fact about a situation. So what? There are no instructions to do anything and no question is asked. You say it is "the last problem in the activity. What were the other problems? Did they ask you to do something or answer a question? Is it possible that part of this was cut off?

I have no idea what "this object moves in K" could mean.

Of course, Henri Poincare had theories very similar to Einstein's

An acute scalene triangle. Depending on exactly how "isosceles" is defined (it variies) saying a triangle is "non-isosceles" might include equilateral triangles

Change "[tex]f(x)> \epsilon[/tex]" to "[tex]f(x)< -\epsilon[/tex]". (With still [tex]\epsilon> 0[/tex].)

The problem, as you state it, does not ask you to identify minerals, just to describe them.

This is rather an obvious statement. What is your purpose in posting it here?

A rigorous mathematical proof would require using the precise words used in defining the product of two numbers. What definition do you have?

I remember reading "God created the counting numbers- all the rest is the creation of man" but, unfortunately cannot remember where or who to attribute it to. At the time I thought, "Well, yes, it is easy to distiguish between one and two elephants, but what about one or two slime molds? Are the counting numbers really so obvious?"

Strictly speaking it isn't the circuit (note spelling) that is in "series" or "parallel", it is what is in the circuit. First you have to get your electricity from somewhere, either direct current (from a battery) or alternating current (from a wall socket so ultimately from a generator). If it is from a battery then you have one wire from, say, the "negative pole" leading to a light bulb, then another wire back to the "positive pole" of the battery. If you want that battery to supply electricity to two light bulbs, there are two different ways to connect the second bulb. You can take that second wire from the first bulb and instead of running it immediately back to the battery, run it to the second bulb and then a third wire back to the battery. That is in "series". The other thing you could do is take two new wires and, pretty much ignoring the first bulb, run a new wire from the battery to the new bulb then another wire back to the battery. That is in "parallel". Obviously wiring "in parallel" is harder and more expensive than "in series" but it has a major advantage. If you have a large number of light bulbs wired in series and one bulb burns out that interrupts the whole string so all the bulbs go out. If they are wired in parallel only that one bulb will go out and the others will stay on so it is easy to see which bulb needs to be replaced. Some of our older friends may remember Christmas tree lights wired in series and having to replace every bulb one by one to see which was the "bad" one.

This depends strongly on what your "everyday life" is like! If you are an engineer you are likely to use integration regularly. If your "everday life" consists of saying "Do you want fries with that?", not so much!

First, do you understand how "pH" is defined? Everything depends on definitions!

What do you mean by "volcanic landform". Much of the states of Washington and Oregon are covered with a layer of basalt from volcanoes. I certainly would call that a "volcanic landform and it is certainly safe! Even if the hot spots below it should erupt the lava would not be able get past the basalt. (Well, there are a coulple of exceptions to that- such as Mount Saint Helens.)

What does "relative to the other" mean? What "other"? [math]tan(\alpha- \beta)= \frac{tan(\alpha)- tan(\beta)}{1+ tan(\alpha)tan(\beta)}[/math] so [math]tan(2X- \phi_0)= \frac{tan(2X)- tan(\phi_0)}{1+ tan(2X)tan(\phi_0)}[/math]. By the same formula [math]tan(2X)= tan(X- (-X))= \frac{2tan(X)}{1- tan^2(X)}[/math]

Wouldn't it be simpler just to tilt your head?

"Humid air", by the way, will be quite heavy because if the water in it, certainly not lighter than hot air.

I am not sure why you say that collisions "create thrust". Such an engine or even a standard rocket creates thrust even in empty space where there are no air molecules to "collide" with.

Density is "mass divided by volume", d= m/v, so dv= m and v= m/d. That is, volume is mass divided by density. We are told that the mass is 1.899x 10^{27} kg and the density is 1.24 x 10^3 kg per cubic m and we need to divide 1.899/1.24 kg(kg/m^3). Since dividing by a fraction you "invert and multiply" the units are kg(m^3/kg). The "kg" cancel leaving "m^3", cubic meters, as it should be for volume. You also have the numerical division wrong.

"sqrt(1)" does not have any "solutions" because it is not a problem to be solved! The equation $x^2= 1$ has two solutions, x= 1 and x= -1. By definition, $\sqrt{y}$ is the positive number, x, such that $x^2= y$.

In general, [tex]x^2= a^2[/tex] has two solutions, x= a and x=-a. But if [tex]x= \sqrt{a^2}[/tex] then x= |a|. In particular, there is NO x such that [tex]\sqrt{x}= -1[/tex].