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Cool, Thanks for that info. How do you know this stuff? You just like to research?

Elijah, this excerpt offers a basic answer to your question.

The Calculator

Some of the history about the calculator comes from Lancelot Link, who watched and documented the development of web2.0calc’s calculator; as he describes it: “...from embryonic abacus to fetal slide-rule, to baby computational engine calculator, and beyond.” “The calculator could offer atomic precision at light-year distances.  ...The baby calculator could literally subtend the angle of an electron at a distance of 10^9 light-years.” If a calculator can to this, then there are always math geeks and wantabes who will test the limits of the calculator.   This is no longer the case. The calculator was neutered years ago to lighten server demand and accommodate more simultaneous users.    The development of this computational engine is based on Javascript also known as “web2” Hence the name, “web2.0calc”.  Prior to this, direct interaction with a complex applet, such as a calculator, required the use of a Java Console to access the side channels of data, else the whole page would have to be regenerated, which included the static data.  

Here’s the source:

80 * 20 = 1600 chips

20 * 5 = 100 students

1600/100 = 16 bags

... how do I know im not talking to a high-school dropout? Or a person who doesnt know squat about math besides what Google feeds him

You don’t know; it’s irrelevant. I only need to know a little more than you for the presented subject and question. Here’s an analogy: Let’s say were walking in the forest, discussing math and chemistry. While we’re walking, a bear comes out of the woods looking for lunch, and charges at us.   We run for our lives. I can’t outrun the bear, but I don’t need to; I only need to run faster than you... The bear gets his lunch (and a severe case of indigestion) and I live another day –and maybe long enough to complete high school.

Vinculum is a high school student. He’s not a dropout, but he’s not completed the requirements for a diploma; yet, he answered your chemistry question.  He knows more than you about this subject, and that all that is required.  I will point out that while his method is correct his final answer is WRONG. He made a math mistake. I recognized it instantly, but I didn’t comment on it because I was curious if you were going to step through the process to learn how to do these problems, or if you were just going to plop the answer on your homework and call it good. If you had stepped through it, you would have caught this mistake.   

How dare I ask a Chemistry Problem when theres literally people asking for simple division problems, fractions, addition. Those people dont get condemed. hmmm... Seems that you just targeted me.

Note that the right hand side is just a constant. 

Take log on both sides:

\(\log_9 (9^y) = \log_9 \left(\dfrac{3^5 \cdot 9 \cdot 6}{27 \cdot 3}\right)\\ y = \log_9 \left(\dfrac{3^5 \cdot 9 \cdot 6}{27 \cdot 3}\right)\)

I will leave the simplification to you.

Multiply by (x^2 - 1) on both sides gives:

\(A(x + 1) + B(x - 1) = x + 8\\ (A + B)x + (A - B) = x + 8\)

Note that this holds true for any real number x.

Therefore, we can compare the coefficients of x on both sides.

\(\begin{cases}A + B &=& 1\\A - B &=& 8\end{cases}\)

Can you take it from here?

Consider the coefficient of x^2 on both sides.

\(5(10) + (-6)(t) + 7(4) = 60\\ -6t + 78 = 60\\ t = 3\)

Yeah, Vinculum's answer is correct.

It's 3, trust me

-Vinculum

That means \(2\lfloor r \rfloor + 2r = 31 + 2\lfloor 4r \rfloor\).

Now, the right-hand side of the equation is an integer and \(2\lfloor r \rfloor\) is an integer. That means \(2r\) is also an integer.

Case 1: Let \(r = x + 0.5\), where x is an integer.

Then 

\(2x + \dfrac12 = 15.5 + 4x + 2\\ 2x = 17 + 4x\\ 2x = -17\\ x = -\dfrac{17}2\)

which is not an integer. Therefore, there are no roots of such form.

Case 2: \(r\) is an integer.

\(2r = 15.5 + 4r\)

\(r = -\dfrac{31}4\)

which is not an integer.

Therefore, there are no such \(r\) satisfying the equation.

By Menelaus' theorem,

\(\dfrac{QO}{ON} \cdot \dfrac{NR}{RP}\cdot \dfrac{PM}{MQ} = 1\\ \dfrac{QO}{ON} \cdot \dfrac12 \cdot \dfrac11 = 1\\ \dfrac{QO}{ON} = 2\\ QO = 2 ON\)

Since QN = 12, QO = 8 and ON = 4.

Since PN = NR and PR = 18, NR = 9.

Then by Pythagorean theorem, 

\(OR = \sqrt{NR^2 + ON^2} = \sqrt{4^2 + 9^2} = \sqrt{97}\)

Here :  https://web2.0calc.com/questions/geometry_95436

Substitute (x, y) = (81, 3 sqrt(2)).

\(3\sqrt 2 = k \cdot 81^{1/4}\\ k = \sqrt 2\)

Then, substitute k = sqrt(2) and x = 4.

\(y = \sqrt 2 \cdot 4^{1/4} = 2\)

So, you just want to solve x^3 + x^2 + x + 1 = 0?

Simplify the quadratic to \(x = 2y^2 -12y + 13 \)

The vertex appears at \(y = -{b \over 2a} = 3\).

Now, plug this in, and solve for x to find the vertex. 

Plug in the point \((0,0)\) into the equation. 

This gives us: \(0a + 0b + c = 0 \), meaning \(c = 0 \). 

Now, plug in the point \((1,2)\) into the equation. This gives us: \(1 =4a + 2b\)

Lastly, plug in the point \((1,-2)\) into the equation. This gives us \(1 = 4a - 2b\)

Solving this system, we find \(b = 0 \) and \(a = {1 \over 4 }\).

Thus, \(a + b + c = \color{brown}\boxed{1 \over 4}\)

https://web2.0calc.com/questions/help-please_52172#r3

It's not correct :(

The roots of 7x^2 + x - 5 = 0 are \(\frac{-1 \pm \sqrt{141}}{14}\)

Plugging these in, you get (a - 4)(b - 4) = 115/7.

-x^2  -6x + 1 + 3x^2 + 4x  - 3  simplifies to

2x^2  - 2x - 2

The coefficient on the x^2  term will  come  from the multiplication of these terms

-5x^2  *  - 2  =    10x^2

-7x *  -2x    =     14x^2

1 * 2x^2     =        2x^2

                          _____

                          26 x^2

I'm willing to believe Wolframalpha, that there are no integer solutions, but your solution is wrong.

Specifically your eq(2) is wrong.

The domain of the function in union of intervals is: \(\left(-\infty \:,\:-8\right)\cup \left(-8,\:\infty \:\right)\)

I would say the easiest way to do it is to just graph and find the distance from one side of the circle to the origin.

The distance from the left end of the circle to the center is 8.

Applying this to the circle area formula:

A=pi*r^{2}

A=pi*8^{2}

You can calculate the rest by yourself by entering pi*8^{2} in a calc.

-Vinculum

"pourquoi sa ne marche pas dès qu'on tape sur le étégale ?"

peut-être que ça ne marche pas, essayez de rafraîchir

-Vinculum