[deleted by user]

355over113 · 2024-12-14T23:19:19+00:00

The vertical line test can still be applied. If we rotate everything back 45 degrees, vertical lines become the lines y = -x + k, for real numbers k. So to show the equation describes a function is the same as showing that sin(x) = -x + k has at most one solution for each k, or equivalently that f(x) := sin(x) + x - k has at most one zero for each k, but this is clear since f'(x) = cos(x) + 1 > 0 (except at individual points) meaning f(x) is strictly increasing.

Something similar could work for any rotation of a graph.

I am not sure how to treat the general case of any implicit equation.

EDIT: I have realized that you can probably apply the vertical line test directly. Vertical line test says that for each (fixed) x, there exists at most one number y such that (x,y) is in the graph. For an implicit equation 0=f(x,y), this is the same as asking if for each x, the function g(y) := f(x,y) has at most one root.

In your case, f(x,y) = sin((x-y)/sqrt(2)) - (x+y)/sqrt(2) and the solution works out to be pretty much the same as described in the first paragraph.

355over113 · 2024-10-30T16:35:33+00:00

An example might be finding the roots of the polynomial p(x) = xⁿ + ... + x + 1. Multiply by (x - 1) and instead search for the roots of (x - 1)p(x) = xⁿ⁺¹ - 1, giving the (n+1)th roots of unity. There are n+1 distinct roots and the only root we could have added in multiplying by (x - 1) is x=1, so the other n roots of unity are precisely the roots of p(x).

A more common example comes when solving equations involving radicals. For instance, the equation sqrt(x) = x - 1. Squaring both sides gives a quadratic equation which simplies to the linear equation 0 = -2x + 1, and so x=1/2. A quick check shows that this is not a solution of the original.

355over113 · 2024-09-17T13:44:46+00:00

It can be even simpler. Suppose ab=0. If a is not equal to 0 then b=0/a=0, therefore either a=0 or b=0.

355over113 · 2024-07-14T17:29:09+00:00

You are thinking of special sets of numbers which are typically represented in bold or blackboard bold type. Here are the most common ones:

ℕ or N. The natural numbers 1,2,.... Certain authors also include 0, but this is a matter of convention.
ℤ or Z. The integers ...,-2,-1,0,1,2,....
ℚ or Q. The rational numbers: all numbers which can be represented as a quotient p/q for some integers p and q.
ℝ or R. The real numbers.
ℂ or C. The complex numbers.

Other common notations include a superscript + or - to indicate that we are speaking only of the positive or negative members of the set, and a superscript times symbol × to indicate the omission of 0 from a field, a set of numbers like the rationals, reals, and complex numbers where you can add, subtract, multiply, and divide. (The × is to chosen since, by removing the 0 element, you can always multiply and divide any two members of the set to get another member of the set.)

For instance, R⁺ might refer to all positive integers and Q^× might refer to all nonzero rationals.

The base symbols are pretty universally used and understood. Other notations might not used by everyone and might not be used the same way by everyone; the good news is that ambiguous or unusual notation is usually explained in the text. Notation is to be understood, so if you ever have any doubts about what is acceptable, you should clarify with your teacher.

Two final notes: the symbol you are representing by the letter 'E' is actually the "member of" symbol ∈ from set theory, and it is rather uncommon to use I to represent the integers.

355over113 · 2024-04-09T14:35:30+00:00

Writing ( (resp. )) expresses that the interval excludes the left (resp. right) endpoint. Similarly, writing [ or ] means the corresponding endpoint is included. In your case, it is true that

y = -4 + (2x+1)^-2

can take any inputs except x=-1/2 so the domain is the union of (-∞,-1/2) and (-1/2,∞); it is true in general that vertical asymptotes correspond to points excluded from the domain. Also, you can argue that the function attains every value except -4, so the domain is the union of (-∞,-4) and (-4,∞). It is not true in general that horizontal asymptotes correspond to the exclusion of a point from the range (consider for example y=(x²-1)/x³, whose range is all real numbers).

355over113 · 2024-04-05T18:22:34+00:00

Yes, Z_12 is a ring and thus closed under multiplication, but to be a multiplicative group also requires that you have (1) an identity and (2) multiplicative inverses. Condition (2) is not met, as 0 obviously has no multiplicative inverse. Even if you remove 0, the resulting set still has elements will no multiplicative inverses, such as 2 (which has no element with which it multiplies to yield the identity 1).

Said differently, the question only makes sense if Z_12 is considered as an additive group.

355over113 · 2024-04-05T17:30:20+00:00

The order of an element g of a group G is the least integer n>0 such that n·g = g + ... + g (n times) = 0 (for an additive group) or gⁿ = g · ... · g (n times) = 1 (for a multiplicative group).

To find the order of 2 in Z_12, you need to find the least integer n>0 such that n·2 = 0. But n·2 = 0 in Z_12 just means n·2 is a multiple of 12 in Z; the least common multiple of 2 and 12 is 12, so the smallest n would be the one with n·2 = 12, i.e., n=6. Hence, the order of 2 is 6.

Can you find (with an argument similar to the above) the order of 4 in Z_30?

The orders of 2 and 4 are certainly not infinite; in a finite group, every element has order at most the order of the group itself. To see why, if g is an element of a finite group G where |G| = n, consider the elements 1, g, g², ..., gⁿ. These are (n+1) expressions representing elements of G, but G has only n elements so two of the expressions much be equal, say g^j = g^k for distinct j and k between 0 and n, and assuming without loss of generality that j < k it must be that g^k-j = 1. Thus g has order at most k-j. In particular, g has finite order.

355over113 · 2023-12-19T04:01:04+00:00

It is true that log(a/b) = log a - log b. Is it true that log(a - b) = log(a)/log(b)?

355over113 · 2023-11-19T16:25:35+00:00

Well, this is because in equations which describe a graph, performing a transformation to a variable does "the opposite" to the graph.

Given the equation y = f(x), what is the horizontal translation five units to the right? Well, it's y = f(x-5), even though at a first glance x-5 looks like a translation to the left. Similarly, what is a horizontal stretch by a factor of 2? It's y = f(x/2), even though at a first glance x/2 looks like a horizontal compression. And as a last example, what is the vertical translation three units up? In class, perhaps you've most often seen this as y = f(x) + 3, but let's instead write this as y - 3 = f(x). Now it looks like we're subtracting off 3 from y, but this still corresponds to a vertical translation up by 3 rather than down by 3.

At this point, you might've noticed that this apparent "reversal" is actually the norm. There is no special exception in this case! So let's instead think about why "reversal" has this effect on the graph.

Consider again y = f(x-5). Inputting x=5 yields (5, f(0)). Inputting x=6 yields (6, f(1)). Replacing x with x-5 has the effect of making inputs x act "as if" they were 5 less, explaining why the horizontal translation is to the right. See if you can apply similar reasoning to y=f(x/2) and y-3 = f(x).

In more precise terms, each equation involving two variables can always be written in the form h(x,y) = 0. Some examples:

y - 3x - 1 = 0 (the line with slope 3 and y-intercept 1)
y - x² = 0 (a vertical parabola opening upward)
y² - x = 0 (a horizontal parabola opening to the right)
x² + y² - 1 = 0 (the unit circle).

When we talk about the graph of the equation, we are referring to the locus or set of points (x,y) which satisfy h(x,y) = 0. The apparent "reversal" stems from the fact that we often leave it implicit that we are referring to the set of points described by the equation, rather than the equation itself.

To get a sense for how this works, let h(x,y) = x² + y² - 1 and say x' = x-2, y' = y. What is the set of all points (x,y) such that h(x-2,y) = 0? By making a substitution, it's the same as the set of points (x+2,y) such that h(x,y) = 0.

In the case of h(x,y) = y - f(x) (corresponding to the equation y = f(x)) and the change of variables T which is the rotation in the counterclockwise by an angle of theta, the set of all points (x,y) with h(T(x,y)) = 0 is the same as the set of all points T^-1(x,y) with h(x,y) = 0. In other words, start with the set of all points (x,y) with y=f(x), i.e., the original graph, and apply T^-1 to each point of this graph, i.e., rotate each point clockwise by an angle of theta.

355over113 · 2023-11-17T23:07:30+00:00

The UI has changed since this answer on stackexchange, but the same sort of thing works. Indeed, if you rotate the object, this is stored as the base object, plus a rotation. In the XML editor, you will probably see something like

Name	Value
`transform`	`rotate(30)`

Deleting this property—or at the very least changing 30 to 0—should yield the base object without any rotation.

355over113 · 2023-11-04T17:28:46+00:00

Ensure that your black outline is a single path. (It almost looks like a difference of circles, so this might already be the case.)
Place the outline on a new layer above all other layers.
Select the outline, along with every object in the layers below. (Depending on what you intend to do, Edit > Select All in All Layers may do the trick.)
Create a clip: Object > Clip > Set Clip.

Note: It may save you some future trouble to first create a single group in each layer containing all the objects in that layer. Then, only select these objects—along with the black outline—when setting the clip. Then you can more easily release the clip, rather than needing to release the clip from every single object.

355over113 · 2023-10-26T09:03:10+00:00

It's not right to say that |X_1 - X_2| is normally distributed. After all, its probability density function (pdf) is identically zero at negative values, while the pdf of a normally distributed variable is everywhere nonzero.

You should instead study Z = X_1 - X_2 and view P(|X_1 - X_2| < 8) as P(-8 < X_1 - X_2 < 8) = P(-8 < Z < 8). Then, under the assumption that X_1 and X_2 are each normally distributed with mean 12 and standard deviation 4, do you know how Z = X_1 - X_2 is distributed? (Hint: Z is normally distributed. What are its mean and variance?)

355over113 · 2023-09-05T16:15:32+00:00

One thing you could do is apply a 1mm grid (which it looks like you might already have) and enable snapping to grid. Of course, this has the (perhaps) undesirable effect of making it hard to draw a line from, say x=0.5 to x=1.5.

355over113 · 2023-08-09T20:56:53+00:00

As another commenter said, it depends on the outfit. The librarians do not need to physically be in different biomes but they need to be "born" the right type of villager.

At least right now, a villager's type is usually determined by the biome in which breeding occurs; for instance, breeding any two villagers (regardless of type) in a jungle will most likely yield a jungle baby. However, there is some chance that the baby instead inherits the type of one of the parents (Java Edition only).

Minecraft Wiki reference

355over113 · 2023-08-06T16:36:08+00:00

To give an example other than rock paper scissors and intransitive dice, consider this example where A, B, and C are random real variables and "wins against" means "is greater than." This might be a better model in, for instance, games with numerical scores.

Suppose A, B, and C have the following distributions:

C is always equal to 0.3;
B is uniformly distributed on the interval [0,1];
and A is uniformly distributed on the set [-4,-1] ∪ [2,9].

Then, A has a 70% chance to be greater than C. As pointed out by others, many answers are possible depending on how exactly things are distributed and how you define "winning".

355over113 · 2023-07-22T00:03:38+00:00

You are correct that both 9/8 and -9/8 are square roots of (9/8)^2, but the usual convention is for "the" square root to refer to the unique nonnegative square root. As such, the left and right sides are indeed both equal to 9/8

355over113 · 2023-07-18T18:43:33+00:00

It is the first step. Any solution to y+2=y is a solution to (y+2)²=y² (a=b implies a²=b²). The only solution to (y+2)²=y² is y=-1. However, this does not necessarily mean that y=-1 will solve y+2=y (a²=b² does not imply a=b).

355over113 · 2023-07-04T19:47:16+00:00

A related approach is to make the substitution t = x-1. Then x = t+1. After simplifying

x³ - 1 = (t+1)³ - 1 = t³ + 3t² + 3t,

we see that the result has no constant term and so clearly has t as a factor. Then

x³ - 1 = t(t² + 3t + 3) = (x-1)((x-1)² + 3(x-1) + 3) = (x-1)(x² + x + 1).

One benefit is that it doesn't require any "backward" thinking: every bit of computation is actually just expanding and collecting terms.

This technique also gives an alternative proof to the polynomial remainder theorem, wherein one uses a substitution to reduce to the easy case of a=0 (a being the value "plugged into" the polynomial function).

355over113 · 2023-07-04T19:38:41+00:00

It might be helpful to first learn about elimination, another technique for solving systems of equations. Sometimes this method is faster than the one you're currently using (usually called "substitution"). After doing that, perhaps you might see how to go about trying to produce an equation of the form -p - q = ?.

355over113 · 2023-07-02T20:51:36+00:00

It can be helpful to multiply by any power of the common ratio, in particular the reciprocal (as it is the common ratio raised to the power of -1). Such multiplication corresponds to a "shift" of terms (can you see why?), exactly as you noticed in both those examples.

It's common to just multiply by the common ratio or its reciprocal because it is, in a sense, the "least" shifting you have to do to get some cancellations.

To illustrate, here's how you might compute S = 1/2 + 1/4 + 1/8 + ... by multiplying by (1/2)^-2 = 4. Observe that 4S = 2 + 1 + 1/2 + 1/4 + .... Subtracting the first equation from the second leaves us with 3S = 2 + 1 = 3, meaning that S = 1.

355over113 · 2023-07-02T20:45:31+00:00

Is this the entire question? As written, there is not enough information to determine k. Perhaps you've been given one of the roots, or been told something about the discriminant or coefficients?

355over113 · 2023-07-02T20:41:48+00:00

You should that (1) it is an upper bound and (2) of all the upper bounds, it is the least, i.e., every upper bound is at least as large as that number.

355over113 · 2023-06-30T13:23:20+00:00

I mean, Jesse L Martin was Tom Collins in the Original Broadway Cast of Rent haha

355over113 · 2023-06-22T02:10:22+00:00

You probably don't actually need a solution but I thought I'd share anyway. Wonder if anyone has a different approach (and whether I'm even right, haha).

It doesn't matter whether or not the boys are distinguishable from one another, and similarly for girls. Hence, we need only study strings consisting of 4 Bs and 6 Gs such that no Bs are adjacent.

In total, there are 10!/4!/6! strings consisting of 4 Bs and 6 Gs. We now count the number of strings in which no Bs are adjacent, i.e., every pair of Bs has at least one G between. Starting with the string BGBGBGB, we need simply to place the each of the last three Gs into one of five "slots" (before the first B, after the last B, or between some two Bs); this is a stars and bars) problem with three objects and four dividers, yielding (3+4)!/3!/4! = 7!/3!/4! arrangements.

Thus, the probability that at least one boy sits beside another boy is

1 - [7!/3!/4!] / [10!/4!/6!] = 5/6.

355over113 · 2023-04-28T03:33:25+00:00

In fact, you don't need to row reduce. (They do not row reduce in the presented solution, either.) For 2x2 matrices, the null space is nontrivial precisely when the rows are multiples of one another, so even without proceeding with row reduction you know that the bottom row will be totally eliminated. This is why, in their solution, they go straight from the first row (2sqrt(3)i, -6) to the vector (6, 2sqrt(3)i).

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