There are two cases to consider.
A) The removed square is in an odd-numbered column (and row). In this case, the board is divided by that column and row into parts with an even number of columns, which can always be tiled by dominos, and the column the square is in, which has an even number of remaining squares that can also be tiled by dominos.
B) The removed square is in an even-numbered column (and row). In this case, the top row to the left of that column (including that column) can be tiled by dominos, as can the bottom row to the right of that column (including that column). The remaining untiled sections of the board have even numbers of rows, so can be tiled by dominos.
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Perhaps the shorter answer is that in an odd-sized board, the corner squares are the ones that there is one of in excess. Cutting out one that is of that color leaves an even number of squares, and equal numbers of each color. Such a board seems like it <em>ought</em> to be able to be tiled by dominos, but the above shows there is actually an algorithm for doing so.
Answer:
Step-by-step explanation:
So there are two types of variables we need to focus on, the first is the dependent variable, the second is the independent variable ( which is the one in question). You asked about it in relation to mathematics, although even across the board, in science they both really mean the same thing. Here's an example, say you are taking a test, and the rubric for grading the test is 5 points (p) for every question (q) answered correctly. So given that information you can conclude that (p) is equal to 5 x the amount of questions you got right (p = 5q). The amount of points in this equation relies on how many questions you get right, you can't logically say that question (q) is equal to 5 points (p), because the questions don't depend on the points, the points depend on the amount of questions you get right. So that means since the points (p) depend on the # of questions you answered correctly, it is the dependent variable, there for by process of elimination you're independent variable is (q) since they don't rely on the points, the points rely on the questions. It all boils down to how well you can recognize this, and how complicated your question is, but hopefully this provided some insight.
<h3>Answers are:
sine, tangent, cosecant, cotangent</h3>
Explanation:
On the unit circle we have some point (x,y) such that x = cos(theta) and y = sin(theta). The sine corresponds to the y coordinate of the point on the circle. Quadrant IV is below the x axis which explains why sine is negative here, since y < 0 here.
Since sine is negative, so is cosecant as this is the reciprocal of sine
csc = 1/sin
In quadrant IV, cosine is positive as x > 0 here. So the ratio tan = sin/cos is going to be negative. We have a negative over a positive when we divide.
Because tangent is negative, so is cotangent.
The only positive functions in Q4 are cosine and secant, which is because sec = 1/cos.
Answer: 1-17/30, or 47/30
Step-by-step explanation:
Find the common denominator. In this case, it’s 30. Multiply the numerators to fit the new denominator. (10•3=30, so 9•3=27, and 2•10=20.)
Your new fractions are 27/30 and 20/30. Now add the numerators.
27+20=47.
Your new fraction is 47/30.
Now check if you can simplify. Since 47 is a prime number, this fraction is in its simplest form. However, it is an improper fraction, so you can simplify it to 1-17/30.
