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Is this question 20 - 3.58? or 20 divided by 3.58?

If it's 20 - 3.58: 16.42

If it's 20 divided by 3.58: 

5.58659217877

i cant make an account right now but i can when i get my phone tonight thanks 

okay thank you do you think you could help with my question

well first of if your new here you should make a new account so that you can have more interactive times on this site and be able to use the message center and DM other users

well its just a calc with texting and the forum is where you have get help with sme math and give it too

You think so...I thought his whole little story was hilarious

What?

If to be is to not be then saying that you are "dandy" or just fine means that in actuality you aren't. But if you aren't actually fine, then that means you are fine right??? it is a never ending cycle in which one is neither ok but very ok at the same time. Meaning this is an unsolvable question

I'm just dandy

Pretty good, you?  

There are 8 possible outcomes .......   tth hth etc     Only ONE does not have a t in it   hhh    the other 7 outcomes all have at LEAST one t

so

7/8

In both diagrams, the tangent line is given. By definition of derivative, the derivatives of the functions at the point of tangency is just the slope of tangent line.

Therefore, by calculating(or approximating? :D) the slope of the tangents, we get \(p'(2) = 2\) and \(q'(2) = \dfrac{-1}{3}\).

And then, we see the graph of p(x) passes through (2,3), and the graph of q(x) passes through (2, -1). So p(2) = 3 and q(2) = -1.

By product rule of differentiation:

\(\quad r'(2)\\ = p(2) q'(2) + q(2) p'(2)\\ = 3\cdot \dfrac{-1}{3} + (-1)\cdot 2\\ = -1 - 2\\ = -3\)

There we go :)

\(\quad\displaystyle \int \sqrt x \left(1-x^2\right)\;\mathbb{d}x\\ \text{Substitute }u^2 = x\text{.}\\ =\displaystyle \int u \left(1-u^4\right) (2u) \;\mathbb{d}u\\ =2 \displaystyle \int \left(u^2 - u^6\right) \;\mathbb{d}u\\ = \dfrac{2u^3}{3} - \dfrac{2u^7}{7} + \mathbf{C}\\ = \dfrac{2x\sqrt x}{3}-\dfrac{2x^3\sqrt x}{7} + \mathbf C\)

Thank you ;)

Thanks Rom, I just want to see if i can do this myself :)

Pat wants to select 8 pieces of fruit to bring in the car for the people he's driving to Montana with. He randomly chooses each piece of fruit to be an orange, an apple, or a banana. What is the probability that either exactly 3 of the pieces of fruit are oranges or exactly 6 of the pieces of fruit are apples?

I am going to assume that it is ok to have exactly 3 oranges AND exactly 6 apples 

     .... Nope that is impossible since there are only 8 peices of fruit.

8 peices of fruit

I am going to have  the apples first, then the bananas and last the oranges

I'l use stars and bars too.

8 stars   2 bars

total number of possibilities is   10C2 = 45

number of combinations if there are exactly 3 oranges    OOO|    1 2 3 4 5 6        6C1= 6

number of combinations if there are exactly 6 apples        1 2 3 | a a a a a a a     3C1= 3

So I think that the answer is    

P(exactly 3 oranges or exactly 6 apples) = 9/45  = 1/5

I got the same as you ROM. 

That doubles the chance that it is correct   

Nice answer Max :)

\(5^{a_{n+1}-a_n} - 1 = \dfrac{1}{n+\frac{2}{3}}, n\geq 1\)

\(k > 1, k\in \mathbb Z^+\) and \(a_k\) is an integer.

Let us do something to the recursive definition of the sequence.

\(\dfrac{5^{a_{n+1}}-5^{a_n}}{5^{a_n}} = \dfrac{1}{n+\frac{2}{3}}\\ 5^{a_{n+1}} = \left(\dfrac{3}{3n+2}+1\right) \cdot 5^{a_n}\\ 5^{a_{n+1}} = \left(\dfrac{3n+5}{3n+2}\right) \cdot 5^{a_n}\\ a_{n+1} = a_n + \log_5\left(\dfrac{3(n+1)+2}{3n+2}\right)\)

You see somehow it is telescoping :P

Now list some values of a_n for small n.

\(a_1 = 1\\ a_2 = 1 + \log_5\left(\dfrac{8}{5}\right) = \log_5(8)\\ a_3 = 1 + \log_5\left(\dfrac{8}{5}\right) + \log_5\left(\dfrac{11}{8}\right)=1+\log_5\left(\dfrac{11}{5}\right)=\log_5(11)\\\)

You observe that: \(a_n = \log_5\left(3n+2\right)\). 

Now we got the explicit formula for a_n. We are halfway done :) (Nice :D)

If a_k is an integer, then \({3k+2}\) must be a power of 5.

Now the question becomes a number theory question: What is the least value of k such that k > 1 and 3k + 2 is a power of 5?

Let \(5^x = 3k+2\).

\(5^x \equiv 2 \pmod 3\)

\(2^x \equiv 2 \pmod 3\)

This congruence holds true when x is an odd integer.

So the least value of k can be obtained by the equation \(5^3 = 3k+2\). (Why not 5¹?)

\(3k + 2 = 125\\ k = \dfrac{123}{3} = 41\)

Voilà!

Let \(f(n) = \log_{10}\left(S(n)\right)\), where S(n) is the sum of elements of the nth row in the Pascal's triangle.

When n = 0, f(n) = log₁₀(1) = 0

So \(\dfrac{f(n)}{\log_{10}2}\) = 0 when n = 0.

When n = 1, f(n) = log₁₀(2)

So \(\dfrac{f(n)}{\log_{10}2}\) = 1 when n = 1.

When n = 2, f(n) = log₁₀(4) = 2log₁₀(2).

So \(\dfrac{f(n)}{\log_{10}2}\) = 2, when n = 2.

We observe the pattern that \(\dfrac{f(n)}{\log_{10}2} = n \;\forall n \in \mathbb N\).

That's OK. Thanks for your attempt to help. 

Not necessarily...If Guest meant the fraction 1.483 with a repetent above the 3, the decimal is completely different. I believe that Guest was thinking about this decimal since it is a lot harder and a more difficult math problem. 

                                         _

This is your decimal: 1.483, which is basically 1.483333333...

We can do this: we want to take away decimal part, including the repeating part.

X = 1.483 repeating 3

      10X = 14.833 repeating 3

  -       X = 1.483 repeating 3

_______________________

        9X = 13.35

X = \(\dfrac{13.35}{9} \Rightarrow \dfrac{1335}{900} \rightarrow \dfrac{89}{60} \Rightarrow \boxed{1\dfrac{29}{60}}\). 

1 29/60 or 89/60 is the closest fraction to 1.483 repeating 3. 

Hope this helps,

- PM

The answers are approximate, so if you check the values, your results might be a few cents more expensive than the original cost. 

By the way, is this a real math question? I feel like the similarity in the topic of the math problem (ribeyes) and your username is not a coincident.