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Neither can we. Even Albert Einstein, Archimedes, Galileo, Homer, Achilles, Heracles, Odysseus, Perseus, Polyphemus, Calypso, Percy Jackson, Harry Potter, Dan Cahill, Seth Sorenson,and Michael Jackson had they're limits.😀 

Forgive me if I'm not happy. It is 9:25 I figured someone would answer it... I guess not. Now I look even more like an idiot in front of my professor.😵👎👎👎👎👎👎👎👎 

In proper English! I may like math but that doesn't mean I'm a mathematician.

sqrt(50)-2sqrt(72)+3(sqrt(18)) = ?

\(\begin{array}{rcll} \sqrt{50}-2\sqrt{72}+3\sqrt{18} &=& \sqrt{50}-2\sqrt{4\cdot18}+3\sqrt{18}\\ &=& \sqrt{50}-2\cdot2\sqrt{18}+3\sqrt{18}\\ &=& \sqrt{50}-4\sqrt{18}+3\sqrt{18}\\ &=& \sqrt{50}-\sqrt{18}\\ &=& \sqrt{2\cdot5^2}-\sqrt{2\cdot3^2}\\ &=& 5\sqrt{2}-3\sqrt{2}\\ &=& 2\sqrt{2}\\ \end{array}\)

Sorry n one is helping ya I would but I didn't understand that

If the remainder is a decimal there are a few options of what to do with it.

      1) If the decimal has zero at the end you automatically remove them.

      2) You can round to the nearest whole number.

      3) You can round to the nearest tenth,hundredth, thousandth,etc.

      4) You can also convert to a fraction, or mixed number.

      5) Also if your dividend and divisor are decimals you can move the decimals place to the left until it dividend becomes a whole number. 

                              - Sylvia

26C5/67C5 =

\(\begin{array}{rcll} \frac{ 26C5 } {67C5} &=& \frac{ \binom{26}{5} }{ \binom{67}{5} } \\\\ &=& \frac{ \frac{26}{5} \cdot \frac{25}{4}\cdot \frac{24}{3}\cdot \frac{23}{2}\cdot \frac{22}{1} }{ \frac{67}{5} \cdot \frac{66}{4}\cdot \frac{65}{3}\cdot \frac{64}{2}\cdot \frac{63}{1} } \\\\ &=& \dfrac{ 26 \cdot 25 \cdot 24 \cdot 23 \cdot 22 } { 67 \cdot 66 \cdot 65 \cdot 64 \cdot 63 } \\\\ &=& \dfrac{7893600 } { 1158917760 } \\\\ &=& \dfrac{ 115 } { 16884 } \\\\ &=& 0.00681118218 \end{array}\)

This polynomial only has two real roots ........see the graph, here......https://www.desmos.com/calculator/8yzxid3bii

They occur at about (0.684, 0)  and at about (1.378, 0)

is it f(x)=3(5/3)^-x a exponential growth function?

\(\begin{array}{rcll} f(x) &=& 3 \cdot \left( \frac53 \right)^{-x} \\ f(x) &=& 3 \cdot \frac{1} { \left( \frac53 \right)^x } \\ f(x) &=& 3 \cdot \frac{1} { \frac{5^x}{3^x} } \\ f(x) &=& 3 \cdot \frac{3^x} { 5^x } \\ f(x) &=& 3 \cdot \left( \frac35 \right)^x \\ \end{array}\)

No it is a decaying function, because: \(\frac35 < 1\)

\( y=a\cdot b^x\\ \text{Example: } y = 3 \cdot \left( \frac35 \right)^x \)

when a > 0 and the b is between 0 and 1, the graph will be decreasing (decaying).

In the unit circle we find by using the periphery angle:

\(\tan{( \frac{\theta}{2} ) } = \frac{ \sin{( \theta ) }}{ 1+\cos{(\theta ) }} \)

49.

\(\begin{array}{rcll} \tan{( \frac{\theta}{2} ) } &=& \frac{ \sin{( \theta ) }}{ 1+\cos{(\theta ) }} \qquad \sin{( \theta ) } = \frac35 \qquad \cos{( \theta ) } = \frac45 \\ \tan{( \frac{\theta}{2} ) } &=& \frac{ \frac35 }{ 1+\frac45 } \\ \tan{( \frac{\theta}{2} ) } &=& \frac35 \cdot \frac{ 1 }{ 1+\frac45 } \\ \tan{( \frac{\theta}{2} ) } &=& \frac35 \cdot \frac{ 1 }{ \frac{5+4}{5} } \\ \tan{( \frac{\theta}{2} ) } &=& \frac35 \cdot \frac{ 5 }{ 9 } \\ \tan{( \frac{\theta}{2} ) } &=& \frac39 \\ \tan{( \frac{\theta}{2} ) } &=& \frac13 \\ \end{array}\)

Press "2nd" key first, then will see "MOD".

Just multiply them together and the product is your LCM.

You were going pretty good, Complex, until after this step

6m - 18 =  3m - 9      add18 to both sides........subtract 3m from both sides

3m = 9                      divide both sides by 3

m = 3

Check this solution in the original problem to see that it "works"

The answer is

\(10000000000000000000000000000000000000000\)

-Complex

Ok lets see...

\(3[2m-6]=3m-9\)

\([6m-18]=3m-9\)

\(3m-6=-9\)

\(3m = -3\)

\(m=-1\)

Hoped that helped...

-Complex

2(x+3)-y=7  →   2x + 6 - y   = 7   →  y  = 2x - 1      (1)

7x-3(y-1)=9  →  7x - 3y + 3  =  9  →  7x - 3y  = 6    (2)

Substituting (1)  into (2), we have

7x - 3[2x - 1]  = 6       simplify

7x - 6x +  3  = 6

x + 3  = 6      subtract 3 from both sides

x = 3            and     y = 2x -1   =   2(3) - 1   =  5

So   (x,y)  = (3, 5)

Here's a graph of the intersection of (1)  and (2)........https://www.desmos.com/calculator/mcvtrkwpyj

Much better.

Typo...

Going to fix it!!

Don't worry, Complex.....we can't either......!!!!!!!

Hello!

Ok lets get down to this.

I don't think you can do anything with remainders on decimals but you can use it on fractions and then convert them.

For example,

\(\frac{8}{5}\)

The answer would be...

1 remainder 3

So, you can put that into a fraction...

\(7\frac{1}{5}\)

Then convert it to a decimal...

\(\frac{36}{5} = 7.2\)

-Complex

I'm probaly forgeting a key princapal or a shorter way.