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You'll note that it takes three sticks to make a triangle.  So your structure with 135 sticks must comprise 45 individual triangles, by "individual" meaning only the ones with a point at the top.  For example, in the illustration, the center one contains 3 individual triangles. 

So, if there are 45 triangles total, that means the bottom row must contain 9 triangles.  If the bottom has 9 triangles, then so do both sides, and therefore there will be 9 sticks along the side. 

_.

The vectors are [3,5], [5,-3], [4,1].

The inequality is 4x + 5y - 8 > 0.

x=39 for real numbers..

     Find the last digit of (2 + sqrt(3))^(17) + (2 - sqrt(3))^(17) = 5,285,770,564                                                             

Here is a hint:

Multiply first then divide then add and subtract. (ORDER MATTER!!) (From left to right)

Another hint:

BODMAS Is the rule you should follow , (We solve brackets (B) then orders (O) which is powers,exponents etc, then Division (D) then multiplication (M) then addition (A) then subtraction (S)) Most importantly, from left to right, if M is first on the left then you should start with M 

M have the same value as D 

A and S have the same value 

In both order matters tho! From left to right again!!

Solution obtained by this website calculator:

90*73/34+73-50 = 216.2352941176470588

Without a calculator: 

\(90*\frac{73}{2*17}+23\)

\(45*\frac{73}{17}+23\)

now you need to multiply 45*73 

\(\frac{3285}{17}+23\)

Long division..

\(193.24+23=216.24\)

egyptian fractions

can all fractions with a numerator of two be written as the sum of two different unitary fractions with mathematical proof

I assume:
\(\dfrac{2}{n}=\dfrac{1}{a}+\dfrac{1}{b} \)

if \(n\) is even then we set \(n=2k\):

\(\mathbf{\dfrac{2}{n}}=\dfrac{2}{2k}=\dfrac{1}{k}\mathbf{=\dfrac{1}{k+1} + \dfrac{1}{k(k+1)} }\)

if \(n\) is odd then we set \(n=2k-1\):

\(\mathbf{\dfrac{2}{n}}=\dfrac{2}{2k-1}\mathbf{=\dfrac{1}{k} + \dfrac{1}{k(2k-1)} }\)

  A = 

{1, 2}

{1 , 2, 3}

{1, 2, 4}

{1, 2, 5}

{1, 2, 3, 4}

{1, 2, 3, 5}

{1, 2, 4, 5}

{1, 2, 3, 4, 5}

I'll leave you to count them!

Good work EpicWater.

It is always good when you can solve it by yourself!   

What's the complexity of the number 15?

15.......{ (1+1)*(1+1)+1 } * (1+1+1)........8

The integer complexity of the number 15 is 8

1) One suggestion would be to make a new post of your own.

Ask if any users can help you with your adblocker problem.

2) Maybe also some you know in person, like a friend or relative, (not on the forum) might know more than you and might be able to help you.

3)  Here is one site that might help.  

OR maybe do your own search, you might find better help.

Like this is for internet explorer, maybe you need instructions for Google Chrome or some other browser.

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EP just made a careless mistake, it is no big deal. 

It does not even detract from his answer since only the method is important.

5x + 8y = -9      subtract   5x  from both sides

8y  =   -5x   - 9                divide both sides  by  8

y = (-5/8)x - (9/8)

The slope of this line   is   -5/8

So.....the  slope of a perpendicular line is     8/5

Since this line passes  through   (10,10)  we have that

y = (8/5) (x - 10) +10     simplify

y = (8/5)x  - 80/5 + 50/5

y= (8/5) x  -  30/5

m = 8/5      b = -30/5

So

m +  b =    -22/ 5

Call D  the distance  that the first car is from point A when the second car passes point A

Then the remaining  time for the  first car  to  travel 75 miles from point A  must be 

(75 - D)  / 60

And since the second car catches the first car 75 km  past point A  and it travels at  75km/h it must take this car 1 hour to  catch the  first car after passing point A

So.....the time it takes the first car to travel (75 - D) km   must be just  1 hour

So

(75 - D) / 60  = 1 

D =  15

So...the first car is 15 km past point A when the second car reaches point A

So.....the time it takes the  first car to travel 15 km =  15 minutes

So......the second car must pass point A 15 minutes after the  first

Vertical line must pass through both points of inersection of circles's circumferences, and the line L must pass through the centers of both circles. 

and you know where can i ask to this users? i don't want a stop to use this calculator, i love this tool sorry for my english :)

Good Job!!! Melody, you are correct. I also got 48. A bit late on your response, but I answered the problem before seeing you post this. 

The guest is correct

See this image

Since  AB  = AC

AD = AD

BD  = CD

Then by SSS   triangle BAD is congruent to triangle CAD

The angle BAD   = angle CAD

So   angle ABC is bisected

The answer is 3-2i

OH!

I see

It's (30 + 19i)/(4 + 9i) 

not 30+   19i/4  +9i

AB and AC must be equal!!  

OK......here's another way to solve this

C(6, n) (6x)ⁿ (  1 / (3x) ) ^{6 - n}        and we can write

C (6, n)  ( 6)ⁿ (x)ⁿ  ( 1 / 3)^6-n ( 1 / x) ^6-n

C (6, n)  (6)ⁿ( 1/3)^6-n  ( x)ⁿ  ( 1/x)^6-n

C(6, n) (6)ⁿ (1/3)^6-n ( x)ⁿ  ( x^-1)^6-n

C(6, n)  (6)ⁿ (1/3)^6-n (x)ⁿ ( x)^{n - 6}

C(6, n) (6)ⁿ(1/3)^6-n ( x) ^{2n - 6}

We need to   have    2n - 6  = 0    to eliminate the variable, x

Then   n  must    = 3

So  we have

C(6, 3) (6)³ (1/3)^6-3   

20 * 216 * (1/3)³

20 * 216 *  (1/27)  =

20 * 8  =

160

Nice, Dragan.....!!!!

You lost me at C(6,3).

If (3,6) is on the graph of y = g(x), then g(3) = 6.

we know that h(x) = (g(x))²

that means h(3) = (g(3))²

so h(3) = (6)²

h(3) = 36

therefore, the point (3, 36) is on the graph of y = h(x)

you can probably sum up the coordinates yourself :)