TheXSquaredFactor

Here is an extremrly basic mockup of a number table from 0-99 for you to reference as I make the adding process easier.

I notice a few patterns here. The tens digit remains the same. Let's attempt to add the numbers 0-9.

\(\underbrace{0+\underbrace{1+\underbrace{2+\underbrace{3+\underbrace{4+5}+6}+7}+8}+9}\\ \)

The numbers enclosed by the underbraces sum to nine (9+0=9, 1+8=9, 2+7=9, 3+6=9. 4+5=9). There are 5 lots of this occurring with the numbers one to nine, so the sum of the tens digit in the first columns equals \(9*5\text{ or }45\). This phenomenon occurs ten times or once per column, so let's multiply this by 10. \(45*10=450\). 

Now, let's do the ones column. The ones column is different because each column increments the amount by 1. The sum of the ones column can be represented by the following sequence. \(0*10+1*10+2*10+...+8*10+9*10\). We can use algebra to group the common factor, so it simplifies to \(10(1+2+3+...+8+9)\). Of course, we already know that the sum of the numbers 1-9 sums to 45 by our first calculation. 10 times that amount yields \(450\). 

Of course, we have only dealt with the sum of the digits from 0-99. However, if you think about it, the sum of 100-109 should be the same as adding the ones digit and the hundreds digits. Of course, this is a special case since the tens digits are all 0. We already know that \(1+2+3+...+8+9=45\). We know that the hundreds digit of the numbers 100-109 sums to \(1*10=10\). However, we have forgotten 110, which has a value of 2. \(45+10+2=57\) for the sum of the digits from 100-110. Now, let's add everything together.

\(450+450+57=957\) or the sum of the digits from 0-110

\(\triangle MNP\sim\triangle MAB\) because a segment located in the interior of the triangle is parallel to a side. You could also prove similarity by Angle-Angle Similarity Theorem. 

Using the above similarity statement, one can create a proportion because each side is proportional. The one I will use is \(\frac{MA}{MB}=\frac{MN}{MP}\). This comes from the similarity statement. 

Although we do not know the length of \(\overline{MA}\) directly, we can find it by subtracting the length of \(\overline{AN}\) from the length of \(\overline{MN}\). When we plug in these numbers, we can then solve for the unknown side length.

Hello again! I see that you are having difficulty finishing the complete simplification of \(\frac{9x+2}{3x^2-2x-8}+\frac{7}{3x^2+x-4}\). 

I am glad to see that you have factored both quadratics correctly. Rewriting the denominators into its factored form results in \(\frac{9x+2}{(3x+4)(x-2)}+\frac{7}{(3x+4)(x-1)}\). Just like the one I solved over here at https://web2.0calc.com/questions/subtracting-rational-expressions#r1, your goal at the moment is to create a common denominator. Unlike the one I linked to above, finding it is a little bit more difficult. In that example, I noted how you want the LCM (least common multiple); however, you really only need a common multiple. It does not have to be the least, but the computation is easier if you do. 

So, what is a common factor of \((3x+4)(x-2)\) and \((3x+4)(x-1)\)? The common factor of 3x+4 is relevant, but it alone will not be enough to figure out a common factor. To figure out a common multiple, just multiply both denominators together. This will always give you a common multiple. It may not be the least one, however. Multiplying both the denominators together yields \((3x+4)^2(x-2)(x-1)\). There is no need to expand yet. Notice, however, that both denominators have a factor of \(3x+4\), so we can divide that from the multiple we have. This gives us the least common multiple of \((x-2)(x-1)\).

Now, let's convert each fraction individually. I'll start with \(\frac{9x+2}{(3x+4)(x-2)}\).

Now, let's convert the next fraction \(\frac{7}{(3x+4)(x-1)}\):

You might notice that there is an issue, though, because the expression \(\frac{(9x+2)(x-2)(x-1)}{(3x+4)(x-2)^2(x-1)}+\frac{7(x-2)(x-1)}{(3x+4)(x-2)(x-1)^2} \Rightarrow \frac{(9x+2)(x-1)}{(3x+4)(x-2)(x-1)}+\frac{7(x-2)}{(3x+4)(x-2)(x-1)}\). You'll see that I factored out some unnecessary factors to create that common denominator. Now, let's add those fractions together!

Now, let's consider the restrictions. All the factors of the original problem \((3x+4),(x-2),\text{and}(x-1)\) cannot be divided by zero. Set them equal to zero to figure out the forbidden values. 

\(\frac{3x-4}{x^2-3x+2},x\neq -\frac{4}{3},1,2\)

There is a special formula that will aide us in this problem. It is the following:
 

\(K=\frac{1}{2}mv^2\)

K = the kinetic energy, in joules

m = the mass, in kilograms

v = velocity in meters per second

We have all that information already! Let's use it to find the mass of this cheetah.

You are so close!

Let's start with the quadratic \(3x^2-2x-8\). The factors of 8 are 1,2,4, and 8. One of the factors would have to be negative. Since the coefficient of the quadratic term is 3, a prime, it is easy to see that this quadratic will be factored into \((3x+\text{__})(x+\text{__})\). You saw that 2 and -4 are factors of 8, so you filled in the the blanks as follows. \((3x+2)(x-4)\). This is the right idea. Sometimes these require some trial and error. Keep trying until you get the right binomial. 

Yet again, you have the right idea with the second binomial. The factors of 4 are 1,2, and 4. Keep trying the factors and changing the signs and changing which binomials those factors belong in. Remember that one must be negative! With time, you'll figure it out. 

If you have any further questions, please ask!

\(\frac{x^2+2x^4+3x^6+...+1005x^{2010}}{2x+4x^3+6x^5+...+2010x^{2009}}\)

I see that both the numerator and denominator both have a GCF. The numerator has a GCF of x^2, and the denominator has a GCF of 2x. Let's factor that out.

\(\frac{x^2(1+2x^2+3x^4+...+1005x^{2008})}{2x(1+2x^2+3x^4+...+1005x^{2008})}\)

As you can see here, the sequence in the middle cancels out! That leaves us with a simplified expression.

\(\frac{x^2}{2x}\)

Don't stop yet! This fraction can be simplified even more to \(\frac{x}{2}\) because of the common term of x. Wow! I never would have guessed that the monstrosity would simplify to something nice!

In order to simplify this expression, we must create a common denominator. Otherwise, subtracting complicated fractions is not feasible. Currently, \(\frac{a^2+1}{a^2-1}-\frac{a-1}{a+1}\) does not have this, so we must create one. 

The denominator of the first term, \(a^2-1\) can be factored because it is a difference of two squares. Sometimes, factoring the denominator fully is helpful since it can make identifying the least common multiple all that more painless. In this case, \(a^2-1=(a+1)(a-1)\).

We now have \(\frac{a^2+1}{(a+1)(a-1)}-\frac{a-1}{a+1}\). Well, would you look at that! It is clear here that \((a+1)(a-1)\) is the least common multiple of the denominators since \(a+1\) is a factor of the product of binomials. Let's manipulate \(\frac{a-1}{a+1}\) so that we can create the desired common denominator.

Ok, we have now changed the problem from \(\frac{a^2+1}{a^2-1}-\frac{a-1}{a+1}\) to \(\frac{a^2+1}{(a+1)(a-1)}-\frac{(a-1)(a-1)}{(a+1)(a-1)}\). This transformation is indeed a positive one because we can now combine the fractions. We have now overcome the first stumbling block. 

Precisely!

Let's use the exponent rules to simplify this expression.

\(\left(2x^2\right)^{-4}=2^{-4}\left(x^2\right)^{-4}\) by the power of a product exponent rule.

\(2^{-4}\left(x^2\right)^{-4}=2^{-4}x^{-8}\)   by the power of a power exponent rule.

\(2^{-4}x^{-8}=\frac{1}{2^4x^8}\)               by the negative power exponent rule.

\(\frac{1}{2^4x^8}=\frac{1}{16x^8}\)                     This step does not involve an exponent rule but rather this is just simplification.