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How about  

1 and 400

2 and 400

4 and 400

5 and 400

...

I get 45 in total; assuming (1, 400) is different from (400, 1) etc.

and where's the shaded region? :)

Yes, I guess they could be meant to be vectors in 3D Euclidean space. 

For two such vectors, u and v, the dot product is |u|.|v|.cos(theta), where theta is the angle between them.

You can, you even should, so long as you are 'fair' about it.

You should be at least as generous to other people's answers as you are to your own.

And

You should only give yourself a point if your answer actually deserves it.

------------------

Some people make a habit of adding a string of mind-numbing posts and then give themselves multiple points for each post.

Often these same people never give any one else a point.  

This type of behaviour is not acceptable and it does not go unnoticed for long.

NOTE:

I am not accusing LuckyDuck of ever behaving in this way.  

Actually guests logic is simpler than mine and it is also correct.

Nicely done guest!

Use the result at https://web2.0calc.com/questions/determine-binomials-that-represent-dimensions#r1 and replace x by 10 to find the length and width.

The perimeter is then twice the sum of the length and width.

I agree with 6/23

I did it with a probability tree.

Say you chose a   red wood triangle

there are 2*3*4 = 24 blocks, you chose one and there are 23 blocks left

12 of those are NOT wood

Of the 12 that are not wood, 2/3 of them  are not red, that is 8 are not red or made of wood

Of these 8, 3/4 of them are not a triangle so that is 8*3/4 = 6 are not made of wood, or red or a triangle

So that is   6/23 will not share any of the same traits.

If you draw a probability tree it might be easier for you to see.

i think this is supposed to be the dot product

This factors as (6x - 5)(x + 1)  so the binomials are 6x - 5  and  x + 1

"... so you cant multiply two 3x1 vectors..."

This is true in mathematics.  Unfortunately, some pieces of software allow this.  For example, both SMath Studio and Mathcad produce the following:

I use Mathcad a lot, but this is a feature I'm not keen on!

to multiply two matrices the number of columns in the leftmost one has to be equal to the number of rows in the rightmost one, so you cant multiply two 3x1 vectors. if you have two square matrices their multiplication usually DOESNT commute

You didn't actually have to delete it. You only had to edit it and give proper credit to the person or site that it came from.   

Mind you, those questions and answers are copyrighted I think. 

I am not sure what that means exactly, I think you can give a link to it but you probably are not supposed to copy it over.

Thanks LuckyDucky   

Nevermind I ended up figuring it out.

I used 6^2+5^2-2*6*5*cos(65) = 35.64, (sqrt = 5.97) to find the length of the black line,

then 2.5^2+4^2-5.97^2/2*2.5*4 = -0.67, cos^-1(-0.67) = 132.04 to get the angle at b (top triangle)

and finally 0.5*2.5*4*sin(132.04) = 3.71, 3.71 + 13.59 = 17.3m^2

Area = 17.3m^2

A scalar will commute with a vector, which means, for example: 

"Let f be a differentiable function with f(-3)=5 and f'(-3)=3 Let the function g(x)=(x-1)f(x) Write the equation of the line tangent to the graph of g at the point where x=-3"

Answered here: https://web2.0calc.com/questions/reposted-because-the-answer-i-got-last-time-was-incorrect#r9

Let  the slope of the first line  =  m

And let the slope of the second line  =  -1/m

And  the sum of their slopes  = 10   which implies that

m - 1/m  = 10       multiply through by m

m^2 - 1  = 10m

m^2  - 10m - 1  =  0

m^2 - 10m  =  1

m^2 - 10m + 25  =  1 + 25

(m - 5)^2  =  26

m - 5  =  √26

m =  √26 + 5

-1/m  =   -1/ [ √26 + 5]  =  - [√26 - 5]   =  5 - √26 

For the first line we  have that

y = (√26 + 5) ( x -7) + 5 

When x  = 0   the y intercept  is

y = (√26 + 5) (-7)  + 5     =   - 30 - 7√26

And for the second line we have that 

y =  (5 - √ 26) ( x - 7) + 5

When x  = 0   the y intercept   is

y = (5 - √26) ( -7) + 5  =   -30 + 7√26

Here's  a graph to show this  : https://www.desmos.com/calculator/g3ayap2tgm

And the sum of the y intercepts  =  -60

x = r cosθ   =  3√2 cos ( -pi/6)  =  3√2 ( √3/2)  = 3√6/2 ≈ 3.67

y = r sin θ  =  3√2 sin (-pi/6)  = 3√2  (-1/2)  = -3/√2  ≈  -2.12

(x + 2)^2  + y^2  = 4                 x  = r cos θ        y  = rsin θ

( rcosθ + 2 )^2  + (rsin θ)^2   = 4

r^2cos^2θ + 4rcosθ+ 4  + r^2sin^2θ  = 4

r^2  ( sin^2θ  + cos^2θ) + 4r cosθ  =  0                sin^2θ + cos^2θ  = 1

r^2  + 4rcosθ  =  0

r ( r + 4cosθ)  = 0

r - rsin θ   = 4

r - r (y/r)  = 4

r - y  =  4

r  = 4 + y

√[x^2 + y^2]  = 4+ y      square both sides

x^2 + y^2  = 16 + 8y + y^2

x^2  = 16 + 8y     rearrange as

8y  = x^2 - 16       divide both sides by 8

y  = (1/8)x^2  - 2

a=1; b=1;c=lcm(a, b); d=gcd(a, b);if(c==400, goto5, goto6);print"LCM of",a,"and", b,"=",c; a++;if(a<200, goto2, 0);a=1;b++;if(b<200, goto2, discard=0;

LCM of 25 and 16 = 400
LCM of 50 and 16 = 400
LCM of 100 and 16 = 400
LCM of 16 and 25 = 400
LCM of 80 and 25 = 400
LCM of 16 and 50 = 400
LCM of 80 and 50 = 400
LCM of 25 and 80 = 400
LCM of 50 and 80 = 400
LCM of 100 and 80 = 400
LCM of 16 and 100 = 400
LCM of 80 and 100 = 400

Thank you so much!!

We can form a chord connecting the two points where the circle and the square intersect

Call  these points B  and  C

Now triangle BOC is equilateral becuse  OB = OC  = BC  =  1

This means that angle  AOB  = 60°

And we can find the shaded area  between the chord and the edge of the circle thusly :

Area of sector  BOC  - area of the equilateral triangle  =

 pi (1)^2 (60/360) -  (1/2)*(1)^2 * √ 3 / 2  =

[ pi/6  -  √ 3/4  ]

Now  the  height  of  the equlteral  triangle can be found  as

area  = (1/2)base * height

√ 3/4= (1/2)(1) * height

2√ 3/4  = height  =  √ 3/2

If we connect AO  we can call the intersection of this segment and the side of the square Q

And  note that  QO  = 1/2

And  the  remaining rectagular shaded area  has a width of 1

And its  height  =  height of the equlateral triangle  -  OQ  =  √ 3/2 - 1/2  = [ √ 3 - 1] / 2

So  the  remaining rectagular shaded area  = (1) [ √ 3 - 1 ] / 2    =  [ √ 3 - 1 ] / 2

So.....the total shaded area  =  [ pi/6 - √ 3/4] + [√ 3 - 1] / 2  =

 pi/6 + √ 3/2 - √ 3/4 - 1/2  =

[ pi/6 + √ 3/4 - 1/2] units^2  ≈  .46 units^2 

haha i'm probably going to sound really dumb -- but why can't you like your own posts?