mobro

Note: They aren't necessarily farming upvotes, because there is also a downvote feature for not-so-great posts.

Also, Web 2.0 Calc is for posting problems, regardless of where it comes from. The point here is to learn, not to cheat. How can you assume that they're cheating ? How do you know that they haven't gotten it wrong, and just want an explanation in general?

You can't just assume things like this, it isn't fair to the other users.

Also, while wrong answers are pesky, they are encouraged in many ways! Then, when the actual solution is posted, the wrong answerer and the person who asked the question get to understand the concept !

Wait .. did you just copy exactly what CPhill did?

To begin, note that this is working out as a sort of a pattern, where \((\frac{1}{5})^0 = 1, (\frac{1}{5})^1 = \frac15\), and then it continues on to where the exponential power keeps increasing by 1. Let's express this in sigma notation:

\(\sum _ {n=0} ^ {\infty} (\frac15)^n\), because there is no ending to the original sum, so we assume that it goes on forever to infinity. Now, we evaluate the sum, using the formula for geometeric series: \(a(\frac{1-r^n}{1-r})\), where we know a is the first term, which is 1. R is the common ratio, which we see that we are raising 1/5 to some power, so the common ratio is 1/5. Then, we have n, which is the number of terms. This is infinity, because the index of the summation ends at infinity. So, writing this out gives:

\(a(\frac{1-r^n}{1-r}) \implies 1(\frac{1-(\frac15)^{\infty}}{1-\frac15})\).

When raising something to the infinite power, we conclude that it is 0, so for the numerator, we have 1 - 0, which is 1, and for the denominator, we have 1 - 1/5, which is 4/5. 1 divided by 4/5 is 1 times 5/4, so the answer is \(\boxed{\frac54}\).

Please correct me if I made a mistake, I am always willing to learn! 

Any thoughts?

I'm not Chris, but I can help you with this 

First, we need to find the square roots of the 2 squares, because the squares of them will give us 32 and 63. So, the square root of 32 is \(\sqrt{32}\), which simplifies to: \(\sqrt{16} \cdot \sqrt2 \implies 4\sqrt2\). 

We can use this same thing to find the square root of 63. 

\(\sqrt{63} \implies \sqrt9 \cdot \sqrt7 \implies 3\sqrt7\)

So when finding the square roots, our fraction becomes \(\frac{\sqrt{32}}{\sqrt{63}}\), which using our steps makes the fraction \(\frac{4\sqrt2}{3\sqrt7}\)

To rationalize this, we need to multiply \(\sqrt7\) on both sides, so this gives us: 

\(\frac{4\sqrt{2} \cdot \sqrt7}{3\sqrt7 \cdot \sqrt7} \implies \frac{4\sqrt{14}}{21}\)

That makes a=4, b=14, and c=21. Adding 4+14+21 give us \(\boxed{39}\)

I believe it was because somebody reported your question. So it was probably deleted. When I clicked the link on my answer to your question, it took me to an invalid address.

Can you clarify on your question? Do you mean solving each value seprately? If so, can you ask it in a seperate question so no one gets confused? I'll try to answer it.

So we know that xy=20, xz=35, and yz=14. If we multiply them together to form an equation, we get:

\(xy \cdot xz \cdot yz = 20 \cdot 35 \cdot 14\)

Finding the product, we have: 

\(x^2y^2z^2 = 9800\)

This is the same as 

\((xyz)^2 = 9800\)

which finding the square root gives us 

\(xyz=\sqrt{9800}\)

Simplifying, we find: 

\(\sqrt{9800} \implies \sqrt{4900 \cdot 2} \implies \sqrt{4900} \cdot \sqrt2 \implies \boxed{70\sqrt2}\)

That's a quicker way, but I think it's better if they understand how to do it.

Let's start by making a fraction. 

I'll start by giving a tip for these problems. Whenever you do these problems, the fraction should be written as \(\frac{\text{number}}{\text{total}}\)

Here, we have a known number, 14, which is the number of strikes. Therefore, 14 should be the numerator. If she had 70 attempts, then that means she tried 70 times to bowl. So, 70 would be our denominator, our total. Now we can solve:

\(\frac{14}{70} \implies \frac{14 \div 7}{70 \div 7} \implies \frac{2}{10} \implies 0.2\)

The fraction equals 0.2, but that's not a percentage. To find percentage from a decimal, we multiply by 100, so we get:

\(0.2 \cdot 100 \implies \boxed{20\%}\)

(sorry if i'm not allowed to answer, i'm new.)