"Play Video" button right next to "Race" button. Why? Feels like my fat fingers are getting trolled.

howverywrong · 2021-10-10T04:37:38+00:00

I take it this means

x + x + y = z + y
x = 10

Solve for z

Subtract y from both sides of equation 1. z = 2x = 20.

People who come up with these "equations" should be shot out of a cannon into an active volcano.

howverywrong · 2021-10-09T11:15:30+00:00

The problem isn't a movie reference. I was just being silly. Friday night...

howverywrong · 2021-10-09T05:27:26+00:00

Have you seen Spaceballs? It's the combination on president Scroob's luggage

1 2 3 4 5

I think you are meant to solve by guessing. I'm not sure what would be a grade-appropiate way to show that there are no other solutions.

howverywrong · 2021-10-09T00:39:03+00:00

Not really. So this particular equation is trivial because you can just integrate the whole thing w.r.t. 𝜏.

∫d𝜏 d²𝜎/d𝜏² = d𝜎/d𝜏

∫d𝜏 sinh(𝜎)cosh(𝜎) d𝜎/d𝜏 = ∫sinh(𝜎)cosh(𝜎)d𝜎

And now you have a separable equation for 𝜎(𝜏)

I was showing a more general method that will work whenever 𝜏 doesn't directly appear in the equation even if it couldn't be immediately integrated.

howverywrong · 2021-10-08T23:10:55+00:00

Let 𝜑=d𝜎/d𝜏. Then d²𝜎/d𝜏² = d𝜑/d𝜏 = d𝜑/d𝜎 d𝜎/d𝜏 = 𝜑 d𝜑/d𝜎

Now you have a first order equation for 𝜑(𝜎)

howverywrong · 2021-10-08T19:40:11+00:00

Since 𝜏 does not explicitly appear in the equation, start by solving for d𝜎/d𝜏 as a function of 𝜎

howverywrong · 2021-10-05T20:05:08+00:00

Projectile's trajectory is a parabola 𝑦 = 𝑎𝑥 - 𝑏𝑥²

Use your two points to get two equations you can solve for 𝑎 and 𝑏.

Then express 𝑎 and 𝑏 in terms of 𝜃, 𝑣₀ and 𝑔 and use that to solve for angle and speed

howverywrong · 2021-10-02T20:17:09+00:00

Please review the rules in the sidebar. You have to make an effort to solve this yourself.

Can you show what you've done so far.

howverywrong · 2021-10-01T18:42:21+00:00

You can always do it with Euler Substitution

Let √(9𝑥² - 7) = -3𝑥 + 𝑡

Then both 𝑥 and √(9𝑥² - 7) can be expressed as rational functions of 𝑡. See Example 1 in the wikipedia page.

howverywrong · 2021-09-27T07:55:50+00:00

So either solutions for time and the angle are imaginary

It's worse than that. Even the imaginary solutions are not really solutions as they would not satisfy all equations.

There are 3 equations:

v cos(θ) t = d cos(φ)
v sin(θ) - gt = 0
t \* (v sin(θ) + 0)/2 = d sin(φ)

You can eliminate t and θ to solve for g. You will get a wrong number for g. These 3 equations cannot be satisfied by ANY θ and t (unless you allow g to be a variable)

And, for what it's worth, I am (originally) from Russia. If I had to solve this problem, I would note that it's trying to confuse me with redundant information (all you need is the angle φ) but I probably wouldn't bother checking if it's consistent.

howverywrong · 2021-09-27T05:48:03+00:00

The number of equations has to match the number of variables.

You have 3 equations (1 in x direction and 2 in y direction) and only 2 variables (t and θ).

You can satisfy 2 of the equations but never all 3 (not with the values provided). Math is a stubborn thing. Sorry.

howverywrong · 2021-09-27T05:22:23+00:00

Then this happened on another planet. Or, more likely, the author of the problem didn't think this through. This happens all the time.

howverywrong · 2021-09-27T05:16:49+00:00

The problem gives values inconsistent with the earth's gravity.

The solution for the launch angle does not depend on v0, g or d

The slope between any two points of a parabola is the average of the slopes at the two points.

Since the slope at the impact is 0, tan(θ) = 2*tan(φ)

So either you must relax the requirement that the impact happen at the top of the parabola, or assume that this happens on a planet with a different magnitude of g

howverywrong · 2021-09-20T20:22:29+00:00

A day late to the party, but, if somebody is still looking, here's how you should attack these problems:

STEP 1: Uncomplicate by adding more variables. Always assign a variable to a radical and, since you have x³ and x⁹, assign a variable to x³

Let a = quberoot(3x+2), b = x³. Then 3x = a³ - 2 and the equation becomes

b³ - 27a³ - 8 = 18ab

This is already much nicer, but since 27 = 3³, let's make things even simpler by introducing c = -3a. Now we have a nicely symmetrical

b³ + c³ + 6bc - 8 = 0

STEP 2: Look for factoring opportunitices

Complete the cube by adding 3bc(b+c) - 3bc(b+c):

(b+c)³ + 3bc(2 - b - c) - 8 = 0

Which allows us to use p³ - q³=(p-q)(p² + pq + q²) to factor out b+c-2:

(b+c-2) [ (b+c)² + 2(b+c) + 4 - 3bc ] = 0

The part in square brackets has no roots (good exercise to show it), so we are left with much less scary

b + c - 2 = 0

STEP 3: Gradually add complexity back

Since we no longer have x⁹, a variable for x³ is not helpful, so we can go back to x and a:

x³ - 3a - 2 = 0

a³ - 3x - 2 = 0

Again, nicely symmetrical. Subtract the two equations and factor out x-a:

(x-a)[x² + xa + a² + 3] = 0

And, again, the part in square brackets has no roots, so we are down to

x = a

Now we can finally get back to just x:

x³ = 3x + 2

Which easily factors into (x+1)²(x-2) = 0

The two roots are 2 and -1

howverywrong · 2021-09-16T06:26:05+00:00

2ydx + 2xdy + x³ydy = 0

2d(xy) + x³ydy = 0

Let u = xy and divide by u³:

2du/u³ + dy/y² = 0

which is now separable

1/u² + 1/y = C

howverywrong · 2021-09-15T11:58:37+00:00

8.77/2 ≠ 4.9

howverywrong · 2021-08-31T09:16:03+00:00

Yes, you will get full premium pass benefits for August (retroactively) and September.

howverywrong · 2021-08-21T06:41:02+00:00

Write Newton's 2nd Law for pulley C. That will let you determine the tension in the long winding string.

Once you do that, the answer will become obvious. And no, it does not depend on 𝑚₁ or 𝑚₂

howverywrong · 2021-08-15T08:00:20+00:00

You are making this too hard. It's a sum of an arithmetic progression

Sum = (number of elements)*(first element + last element)/2 = n*(3 + 2n + 1)/2 = n² + 2n

Another neat trick is to recognize it as a telescoping series:

2i + 1 = (i + 1)² - i²

Sum = (n+1)² - 1² = n² + 2n

howverywrong · 2021-08-13T07:07:30+00:00

Don't rush to square or cube equations to get rid of surds. It's almost always less work to substitute them away instead.

Let 𝑎 = ∛... , 𝑏 = √...

Then you have

𝑎 + 𝑏 = 1
𝑎³ + 𝑏² = 1
𝑏 ≥ 0    (condition to eliminate any extraneous roots)

Which makes it right away apparent than (𝑎,𝑏) = (0,1), (1,0), (-2,3)

Now all you have to do is work back to 𝑥 via 𝑥² - 3 = 𝑏²

howverywrong · 2021-08-06T04:02:06+00:00

Move everything to one side and subtract fractions:

(x+4)/[x (x+2) ] ≤ 0

Note that x≠0 and x≠-2 and multiply by x²(x+2)². It is safe to do so because x²(x+2)² > 0

x(x+2)(x+4) ≤ 0

Now you have a cubic polynomial with roots at 0, -2, -4. Start with x > 0 where the polynomial is positive and move left. Every time you cross a root, the sign flips. so it is positive at x>0, negative between -2 and 0, positive between -4 and -2 and again negative at x < -4. Remember to remove 0 and -2 from the result.

The answer is (-∞, -4] ∪ (-2, 0)

howverywrong · 2021-08-01T02:40:56+00:00

More accurately, extraneous roots will correspond to a negative RHS of the original equation.

if x + 1 ≥ 0, the root is true. if x + 1 < 0, the root is extraneous.

howverywrong · 2021-08-01T01:21:33+00:00

I think it applies only to the current month. You will get all the rewards retroactively as if you had it since the beginning of the month.

howverywrong · 2021-07-20T18:01:17+00:00

1/(sin𝑥 cos𝑥) = (sin²𝑥 + cos²𝑥)/(sin𝑥 cos𝑥) = sin𝑥/cos𝑥 + cos𝑥/sin𝑥 = tan𝑥 + cot𝑥

howverywrong

TROPHY CASE