Math Is Fun Forum

Who's cost is this? Is that the price a store pays for something? If so, then you just have to divide the regular price by 1.7.

To see why, remember that to mark it up 70%, the store multiplied the cost by 1.7. Algebraically,
P = 1.7*C

So, to go back to the cost from the price, just do the opposite: divide by 1.7. Algebraically, you solve for C:
C = P/1.7

I think that, on the second question, he means "log base 3 of 3 + log base 9 of 1".

If so, it's easy. 3¹ = 3, and 9^0 = 1. So, 1 + 0 = 1.

See my post, "What's a dx?". I'm still not 100% clear on that one, even after they explained it to me.

Some more thoughts.

You can calculate the angle that your line makes with the horizontal from the slope. Say D is the distance between point1 and point2. Dy = rise of slope, Dx = run of slope. So,

∠β = arcsin (Dy / D).

The plane of rotation is perpendicular to this angle. So, to get the angle that the plane of rotation makes with the x/y plane, you just add π/2: ∠Z = β + π/2

This lets you find the z component: Lz = L sin ∠Z

For Lx and Ly, we need to drop a perpendicular from the high end of L. The top angle here is (π/2 - ∠Z) = ∠θ. This forms a triangle on the "ground", with hypotenuse given by H = sqrt( L² - Lz² ).

Now, Lx = H sin θ
and Ly = H cos θ

We're done! In summary,

r = 3d distance formula
ω = angular velocity of point (rate of rotation, radians/sec)
n = 2π / ω
L = 2πr sin (π/n)

And the rest, from this post. (Man, I'm lazy. Too lazy to summarize.)

Awesome, that was really quick!

∀ ∠ ∃ θ

I'm going to type out loud here.

You want the point to travel in a circle, so my intuition tells me to use the circular functions.

Define the rate of rotation as radians per unit time. Let ω = rate of rotation.
Let n = number of ω's per circle = 2pi/ω

r = radius of circle = found by distance formula (sorry I'm too lazy to do this for you )

R1 = radius from center to original point
R2 = radius to translated point

Length L that point travels per unit time: L = 2*r*pi * sin (pi/n)

R1, L, and R2 form a triangle.

L is a vector quantity, so you can find its components using vector arithmetic. To do this, I'll first define my axes. I'll put x and y on the horizontal plane, and z on the vertical (altitude). I'll also call the components of the vector Lx, Ly, and Lz.

Imagine viewing the system from directly overhead, so the z component is entirely suppressed. Now, vectors can be freely repositioned as long as their magnitude and direction are not altered. So, we'll put the corner of  R1 and L at the origin.

Drop a perpendicular to the x-axis, and you have a right triangle, the legs of which are Lx and Ly. We know all of the angles of the triangle (R1 L R2) -- one is ω, and the other two are (π/2 - ω). All we need is the angle that L makes with the x (or y) axis, and the problem is solved.

But, I can't think of a way to find it right now, and I need to go to bed. Later maybe, or someone else can find it. Or you

Air resistance is a tough nut to crack, because it can behave in complex ways. My physics textbook gives only two:

R = bv

where R is the resistive force, a vector quantity directed opposite the direction of motion; v is the velocity of the object; and b is a constant (you know, one of those "constants" that varies in every situation) that represents the properties of the medium and shape/size of the object.

R = (1/2) DpAv²

where D is the "drag coefficient", p (should be a rho) is the density of air, A is the cross-sectional area of the object, and v is again the velocity. This formula is for objects moving at high speeds, such as planes, cars, and meteors.

Hmm...guess not. But, I think it would be much easier, or at least faster (for me, anyway) to use these than to copy and paste from the top.

Do we have [sup]sup[/sup] and [sub]sub[/sub] tags?

Here are some thoughts to get you started. I don't have time right now for a full and proper solution, but this is the direction I think you'll need to go to get one. And since time is not on your side, I might not get to the solution before you need it...so here we are.

The points A, B, and C form one of the four triangles of the tetrahedron. (Kinda cool that adding one point creates three triangles...) Imagine that triangle with the sphere touching one face. The diameter of the sphere is 2 (note that ?5 is about 2.24). Now, in order for all three points to be able to connect to point D (on the opposite side of the sphere) with a straight line, there is some lower limit on how close the points can be to S.

To simplify the problem, I would let (SA = SB = SC) = minimum possible distance from S. That also makes ABC an equilateral triangle and ABCD a regular tetrahedron. (I'm not 100% sure that's true, but that's what my intuition says.) It also makes the triangles SAD, SAB, SAC, SBC, etc. not only similar but identical.

Now, consider the triangle SAD (don't let the name fool you, it's happy to be alive). Since SA = SD, you can calculate how long SA must be in order for the line AD to pass outside of the sphere. Just drop a perpendicular from S to the point where AD touches the wall of the sphere. This will give you right triangles with the length of one side and all of the angles known, which is enough information to find the lengths of all the other side.

Armed with this information, you should be able to solve the problem.

(And check to make sure my unproven intuitions are valid--I typed this rather quickly.)

That problem statement is greek to me. Is that how they gave it to you? I'm sorry.

Because it might not be the batteries, but the buttons.

To motivate you not to overdraw.

Laziness. Paint is easier to check.

Vapor pressure and quick exhaustion of oxygen.

Habit.

Must be Mayan.

Reflexes. You can't see a bullet (well, maybe Superman could).

A linguist with a sense of irony.

They didn't.

The color isn't involved with bubbles.

I won't do it and you can't make me.

This one doesn't need A's and C's. You can just break it up, like so:

      2x²                   x                       1
-------------  -  --------------  +  ---------------
(x+1)(x-1)²     (x+1)(x-1)²       (x+1)(x-1)²

Expand the denominator:

          2x²                        x                         1
------------------  -  -------------------  +  ---------------
x³ - 3x² - x + 1      x³ - 3x² - x + 1       (x+1)(x-1)²

                         2x²
------------------------------------------
2x²[ (x/2) - (3/2) - (1/2x) + (1/2x²) ]

               x
--------------------------
x[ x² - 3x + (1/x) - 1 ]

                      1                                       1                            1
------------------------------------  -  ----------------------  +  --------------- 
(x/2) - (3/2) - (1/2x) + (1/2x²)      x² - 3x + (1/x) - 1        (x+1)(x-1)²

Ugly but true.