Okay, so let's break this down.
Note: != means not equal to. This would be an equal sign with a slash between it in written form.
1) We have the equation (x-1)/(x-2) = (x-k)/(x+6)
2) We can identitify the non permissible values, aka the values that make the denominator equal to 0 (we cannot divide by 0, gives us an error). Now, if we set both denominators to equal 0 and solve, we will get the following: x-2 = 0; x != 2 and x+6 = 0; x != -6. This means that x cannot equal -6 or 2
3) Okay, now we know what x cannot equal to. But the question asks what are the non permissible values of k. We can solve for x in this case to create an equation that will allow us to find the non-permissible values of k.
1) We start with the equation (x-1)/(x-2) = (x-k)/(x+6).
2) We can cross multiply which will give us (x+6)(x-1) = (x-2)(x-k)
3) Foil and combine like terms for both sides. x^2 + 5x - 6 = x^2 - xk - 2x + 2k
4) Subtract x^2 from both sides. x^2 + 5x - 6 - x^2 = -xk - 2x + 2k
5) We can then isolate the other x^2 on the left side (+ x^2 - x^2). Then we get 5x - 6 = -xk - 2x + 2k
6) We can add xk to both sides. Then we get 5x - 6 + xk = -2x + 2k
7) Then we can add 2x to both sides. We then get 5x + 2x - 6 + xk = 2k
8) Combine 5x and 2x. We get 7x - 6 + xk = 2k
9) Add 6 to both sides. We get 7x + xk = 2k + 6
10) Factor 7x + xk. We get x(k+7) = 2k + 6. Or you could arrange it like x(7+k) = 2k + 6. Whatever works for you.
11) We can now comfortably solve for x. Divide both sides by (k+7). We get x = 2k+6/k+7.
12) Now we can find the non permissible values of k. Set the denominator to equal to 0. k + 7 = 0 and solve. Subtract 7 from both sides and we get k = -7.
Solution (word it however you want but these would work):
1) k != -7
2) The non permissible values of k for the equation (x-1)/(x-2) = (x-k)(x+6) is k != -7
3) The equation (x-1)/(x-2) = (x-k)(x+6) has no solution for k when k = -7
1) Since we know that f(x+T) = f(x)
2) and f(x) = pi(x)^2 - cos (x), we can plug in f(x+T) for anywhere there is an f(x) for our manipulated side (left).
Since this is a proof, there are going to be two sides.
3) Then we get pi(x+T)^2 - cos(x+T) = pi(x)^2 - cos(x). Our left side going to be manipulated while our right side is going to be our "constant" with the original expression pi(x)^2 - cos(x).
4) We can expand the expression which gives us pi(x^2+2(x)T+T^2) - cos(x+T) = pi(x)^2 - cos(x)
5) Then we can foil the pi, which gives us pi(x)^2+ 2pi(x)T + pi(T)^2 - cos(x+T) = pi(x)^2 - cos(x)
6) We can then subtract pi(x)^2 from both sides which will leave us with 2pi(T)x + pi(T)^2 - cos(x+T) = -cos(x)
I might've switched the T and the x in some places (such as 2pi(x)T) compared from my previous answer (where it was 2pi(T)x), but it doesn't affect the actual proof. We can still test our proof just fine in this case. Where you arrange the x and T's is mostly whatever is easier to understand. It's multiplication, so it's fine to flip them generally. It's like 4x5 is the same as 5x4.
I already explained this earlier but here's a more detailed explanation for the contradiction of the proof.
7) Testing the proof:
Now since we are given the hint that f(x) is a periodic function of period T when T > 0 for all x in the domain of f. We can test T = 0. Plug in 0 where ever there's a T and we get -cos(x) = -cos(x). This isn't what we are looking for because this shows that f(x) is a periodic function when T = 0 and is also a valid proof. Not a contradiction.
So since we need to show that f(x) is not a periodic function, we need to make the proof impossible. Since we tried T = 0, we can also test x = 0. Plug in 0 where ever there's an x, and we get pi(T)^2 - cos(T) = -1. This shows that f(x) is not a periodic function when x = 0 and is not a valid proof. It's a contradiction.
Solution: The function is not periodic when x = 0. Unless when T = 0, the function is periodic.
I actually forgot to show the expanding and foiling part on my actual assignment since I did it on a separate piece of paper. So it looked like I skipped two steps lol.
I actually figured it out. My phrasing is probably a bit off but it makes sense (to me at least)
We end up with 2pi(x)T + pi(T)^2 - cos(x+T) = -cos(x) after some simplifying, and distributing/expanding.
If we let x = 0
Then we end up with pi(T)^2 - cos(T) = -1. This makes our proof impossible. It's a contradiction because when we let x = 0, the function is not periodic.
UNLESS, if we let T = 0, then our proof is valid. We end up with -cos(x) = -cos(x) which then makes our function periodic. However, this isn't what we are looking for, as we aren't trying to find what makes the function periodic.
I think you might be correct but I'm not too sure. If we wanted to make the proof impossible, we need an expression such as a positive equals a negative and/or non matching sides. So I do believe that you can't go further for your answer besides simplying both sides if applicable.
The answer key given for this particular question is a dead end for me (can't see the equations), so I have to rely on others so I can understand this. There can be different solutions to this, so any general proof would work otherwise.
Thanks for the help though! I will wait until some other users answer this so I can get more feedback and compare.
6:7 = 6/7 = 0.857
4:5 = 4/5 = 0.8
Solution: 6:7 > 4:5 (6:7 is greater than 4:5).
So to make this simple, ratios are basically fractions. When you want to compare ratios, you generally convert them into fraction form. If you are allowed, simplify the fraction into decimals to get the exact/approximate values. After that, you are able to compare easily.
I'm not sure what grade you are in but I'm sure that since this is basic math, you aren't able to use a calculator. It's obvious that using a calculator is a no brainer for this, and is obviously a boring method. It won't help you learn what you need to learn at this stage. But, there's two ways you can compare fractions easily without using a calculator and it'll get your brain going.
1) Long division (the division algorithm of two integers):
- Put the number of denominator on the outside and the number of the numerator on the inside. So for 6:7, the 7 goes on the outside and the 6 goes on the inside. For 4:5, the 5 goes on the outside and the 4 goes on the inside.
- Then you solve the long division for both ratios. Make sure you know your basic multiplication, put your zeros and decimals in the right spot, and bring down numbers as necessary to subtract.
- Lastly, compare the two. We normally use the symbol ">" for greater than and the symbol "<" for less than.
2) Pie chart/rectangle square method (only works for small numbers):
- Make sure your ratios are converted to fraction form. In this case, 6:7 = 6/7 and 4:5 = 4/5.
- Draw a circle or rectangle for each ratio.
- Then for each, draw triangles/squares that coordinate to the amount in the denominator (biggest number usually). So that is 7 and 5 respectively.
- For ratio 6:7, you want to shade 6 out of the 7 triangles/squares. For the ratio 4:5, you want to shade 4 out of the 5 triangles/squares.
- Now compare and figure which one is greater than the other. They're pretty close tbh, but 6/7 is the larger one in this case.
- You can then write 6:7 > 4:5 as your answer.
Note: The pie chart/rectangle square method isn't the most accurate for comparing since you can be easily fooled, but it's good for small numbers. This method isn't practical but it is a semi-fast way to compare. However, the long division method is much prefered because it's the most accurate and approximative you can get. The long division method will get your algorithmic thinking and mental mathematics skills on point when you need it later on. The long division method also requires cleanliness during the algorithm because you can easily make mistakes if you aren't careful. Otherwise, it's a good challenge.
Ratios will come up in basic probability as well. If you are able to master the tougher aspects of this and have some general algorithmic thinking, you'll blast through the probability unit. Probability is quite annoying and tedious if you don't have some common sense and know your basic mathematics skills that you are taught at your specific grade level; such as counting, guessing, ratios, fractions, mean/median/range, etc.
Mathematics is tough for a lot of people, and even I struggle immensely on subjects that I should grasp at my age.
Let me know if I made any mistakes.