ilorty

I agree with Melody. Sometimes, people may not know how to answer the problem, and then they skip it and move on to a different problem. (like me, sorry for being picky...)

Also, if you are actually confused about this question, and don't want to confess it, you could post anonymously, (but then what's the purpose of creating an account?)

Nobody is calling you a liar, as that is rude. But, if you really need help with the problem, just say that you need help. Nobody is here to judge you (unless you are rude about your question...but that is not judgement...that's just discipline) It also increases the posibility that your question is answered :)

Concerning the math question that you had...I'm confused...how do you multiply\add points? Anybody able to explain? (maybe even you, TigerNathan, as you solved the problem already) It would be good for my learning experiance.

:)

Have you gotten the answer, or do you need more help?

Before you look at answer: Try to divide using long division... watch this video for more information: https://www.khanacademy.org/math/algebra-home/alg-polynomials/alg-long-division-of-polynomials/v/dividing-polynomials-1

Tried it already? Now, here is how I did it:

 What can you do to 12y^3-6y^2 to make it easier to divide? (*cough* factoring out a 6? *cough*) 

So, after I factored out a 6, I got 6*(2y^3-y^2)/(2y-1). This was easy to divide, and I got 6*y^2, which is 6y^2

OR

If you were smart enough to see if right after factoring, you could multiply the bottom by 6. This is 12y-6. Now, the difference between the top and the bottom? y^2. So, you still end up with 6y^2

:)

Since you already proved that n(2n + 1)(7n + 1) is divisible by 3, now you need to see how it could be divisible by 4. We have already seen that 2n+1 is odd, so n(7n+1) has no be divisible by 4. This happens when n is divisible by 4, or when 7n+1 is divisible by 4. Now, we first need to see when 7n+1 is divisible by 4. As you see, this happens if and only if -n-1 is divisible by 4. (by subtracting 8n) From there, can you generalize an equation that is true for n? (In the problem, you do not have to list out e v e r y   s i n g l e option, as there are infinite possibilities.)

(please ask for more help if needed!)

:)

Here is one of the many ways you can approach this problem:

As you see, there is a pattern: eachtime, you are taking away a digit, starting from 0.

So, we have

6075, 2025, 675, 225, 75, 25 (see the pattern? the last 2 digits are always the same)

Using this pattern, we see that for the last 2 digits to remain the same, it cannot go any further without becoming a fraction. So, there are a total of 6 integers.

We first need to find the prime factorization of 540. This is 2 × 2 × 3 × 5 × 3 × 3, which is 2^2(3^3)(5^1).

I never solved a system of equations by graphing before, but let me try!

What I did (and kind of adding on to geno's answer):

We know that the median is 7. So, we can put 7 down as d.

We also see that for the mode to be 7, it has to occur the most. The most it can occur is 3, because if it occured less, than there would be 2 modes, and if it occured more, than g would not be that large. 

Now, for the mean to be 7, the a+b+c+7+7+7+g has to equal 49.

Subtract 21 from 49, and you get 28. From 28, you should see that to get the largest possible value of g, you need the smallest possible values of a, b, and c. Since the mode has to be 7, you can only have 2 1's. That leaves you over with one 2.

Subtract all of that from 28, and you get 24 for g.

:)

Try dividing the shape up into triangles.

You should end up with 4 triangles: ECD, CFE, BFC, and ABF.

Now, use the trangle area formula and some pythagorean (to find the heights) to get the area of each of those triangles. Then, add it up.

:)