Vinculum

Hey BuilderBoi, 

I have a question...

How is it 3?

Using the floor rule \(\mathrm{If}\:\lfloor \:n\:\rfloor =x\:\mathrm{then}\:x\le \:n \)

We get \(5\le \sqrt[3]{n}<6\)

That is \(125\le \:n<216\)

-Vinculum

Hello, 

Take a look at the graph below:

The points where they intersect are (1, -4);(2, -5)

-Vinculum

Hello, 

This question was answered by CPhill,

Please take a look:

 https://web2.0calc.com/questions/algebra_13724

If you have a question, reply back.

-Vinculum

Hello, 

So to convert from percents to decimals you move the decimal right 2 spaces. 

Example: 45.0% = 0.45

Or you can divide the percent by 100.

Example: We have 67.0%, 

Divide by 100

67.0/100 = 0.67

-Vinculum

\(2x-5\le \:-x+12-3x+19\)

\(2x-5\le \:-x-3x+12+19\)

\(2x-5\le \:-4x+12+19\)

\(2x-5\le \:-4x+31\)

\(2x\le \:-4x+36\)

\(6x\le \:36\)

\(x\le \:6\)

Interval notation: \((-\infty \:,\:6]\)

-Vinculum

The question is not complete

\(-4\left(x+4\right)>x+7+5\left(x+2\right)\)

\(-4x-16>6x+17\)

\(-4x>6x+33\)

\(-10x>33\)

\(10x<-33\)

\(x<-\frac{33}{10}\)

\(Interval\:Notation: \:\left(-\infty \:,\:-\frac{33}{10}\right)\)

-Vinculum

Hi Guest, 

So...

\(\frac{y}{3}+1=\frac{\left(y+3\right)}{y}\)

\(y^2+3y=3\left(y+3\right)\)

\(y=3,\:y=-3\)

The number of values of y: \(2\)

The above expression can be simplified to:

 \(\sqrt{\cfrac{1+\sqrt{\cfrac{1-\sqrt{\cfrac{1-\sqrt{\cfrac{1+\cfrac{\sqrt{3}}{2}}{2}}}{2}}}{2}}}{2}} = \sqrt{ \frac{1 + \cos\left(\frac{13\pi}{48} \right)}{2}} \)

The above value is the cosine of \(13\pi/96\)

So the answer to the question is: \(13\pi/96\)

-Vinculum

\(8+x<4-3x-7<5x-4+11\)

\(8+x<4-3x-7\quad \mathrm{and}\quad \:4-3x-7<5x-4+11\)

\(x<-\frac{11}{4}\quad \mathrm{and}\quad \:x>-\frac{5}{4}\)

Since they are overlapping intervals:

No Solution

-Vinculum

Lancelot Link???

Isn't that the thing with the monkey ????

LMAO, As an Asian I agree!!!

But We do have friends other than asian people lol