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thanks Happy 

$$\\10^2=x^2+(x+2)^2\\
100=x^2+x^2+4x+4\\
0=2x^2+4x-96\\
0=x^2+2x-48\\
0=(x-6)(x+8)\\
x=6\quador \quad x=-8$$

@@ End of Day Wrap       Tues 2/12/14       Sydney,   Australia        Time 11:05 pm        ♬

Good day ladies and gents, boys and girls,

Lots of fabulous answers from geno3141, JewishNerd, Cue123smile, Mathematician, Sarahho2, Sorasyn, j*z123, heureka, happy7, Alan, BabaLlama and MamaLlama.  Thanks everyone.

Here are some interest posts:

1) Rosala's Laughs     Thanks Rosala

Wed 3/12/14

1) Great fun thread  and Brittany introduces herself properly :)  Thanks Happy7 and BrittanyJ

$$\\-11(3/2)(0.12(-12))\\
=-11(3/2)(-1.44)\\
=+11(3/2)(1.44)$$

$${\mathtt{11}}{\mathtt{\,\times\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right){\mathtt{\,\times\,}}{\mathtt{1.44}} = {\frac{{\mathtt{594}}}{{\mathtt{25}}}} = {\mathtt{23.76}}$$

Thanks Alan  

$$-(8/2(-8))=-(4*-8)=--32=+32$$

x=25-7y

(25-7y)²+y²=25  okay

$$\underbrace{(25-7y)^2}_{=25^2-2*25*7*y+49y^2}+y^2=25 \\\\
25^2-2*25*7*y+49y^2+y^2=25 \\\\
50y^2-50*7*y + 625 - 25 = 0 \\ \\
50y^2-50*7*y + 600= 0 \quad | \quad : 50\\ \\
y^2-7*y + 12= 0 \\ \\
y_{1,2}= \frac{7\pm\sqrt{49-4*12} }{2} \\ \\
y_{1,2}= \frac{7\pm\sqrt{49-48} }{2} \\ \\
y_{1,2}= \frac{7\pm 1}{2} \\ \\
y_1 = \frac{7 +1}{2} = \frac{8}{2} = 4 \qquad x_1 = 25 - 7*4 = -3\\ \\
y_2 = \frac{7 -1}{2} = \frac{6}{2} = 3 \qquad x_2 = 25 - 7*3 = 4\\ \\$$

$$\\\dfrac{-300k^4q^{16}}{6a^5k^5q^2}\\\\\\
=\dfrac{-50q^{14}}{a^5k}\\\\$$

No it isn't

You have left out a "y" that multiplies the 350.

(25 - 7y)² +y² = 25

25² - 2*25*7*y + 7²y² + y² = 25

625 - 350y + 50y² = 25

50y² - 350y + 600 = 0

y² - 7y + 12 = 0

(y - 3)(y - 4) = 0

y  = 3 or y = 4

so x = 25 - 7*3 or x = 4  when y = 3

and x = 25 -7*4 or x = -3  when y = 4 (as Heureka illustrates).

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3x^2 - x - 4,

Alan can probably do this in his head but I cannot

$$3x^2 -1 x - 4$$

I look for 2 numbers that multiply to 3*-4=-12

since they mult to a negative - one must be pos and the other neg.

they ALSO must add to -1    SO the bigger one is neg and they are only one apart.  

That is easy,  -4 and +3

Now make -1x=3x-4x

$$3x^2 -1 x - 4$$

$$3x^2+ 3x-4x - 4$$

NOW factor this in pairs

=3x(x+1) -4(x+1)

=(3x-4) (x+1)

Where is the question - it is not here and I looked under the bed.  

That is a much better idea - Thanks Alan :)

I wish Icould give you double points for that.    

x²+y²=25 x+7y=25   x=? y=?

$$\\\boxed{x^2+y^2=25} \small{\text{ is a circle }}\\
\boxed{x+7y=25} \small{\text{ is a line}}\\$$

$$\small{\text{The line cut the circle in 2 Points.}} \\
\small{\text{Point 1 is ( x=4, y=3) }}\\
\small{\text{Point 2 is ( x=-3, y=4) }}\\$$

2- -3

think of negative numbers as cold numbers and positive numbers as hot numbers

If you take cold out of a room then the room gets hotter.  So  - - = +

Yes, I like it anon :)

Melody, your 2nd answer is correct!

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yes but thats what i wrote and i dont know how to solve it

x=25-7y

(25-7y)²+y²=25

625-350+49y²+y²=25

50y²=-250

y²=-5

Thanks Alan,

I thought I was just taking the upward component.  I cannot do that I am supposing. :)

Is my second answer correct?

Alternatively you could go the other way!

sin(4t) = 2sin(2t)cos(2t)  so  sin(2t)cos(2t) = sin(4t)/2

.

$$\\(Sin2t)(cos2t)\\
=2sintcost(cos^2t-sin^2t)\\
=2sintcost(1-2sin^2t)\\$$

This is not going anyway - what form do you want it in?

Thanks Anon, but you found the percentage wearing sandles 56%

So the percent NOT wearing sandles must be 100-56 = 44%

The first equation can be written as x² + y² = 5²

There is a well-known Pythagorean triple: 3² + 4² = 5²

so it is possible that x and y are 3 and 4.

If x = 3 and y = 4 then the second equation, x + 7y = 25 is not true.  However, if x = 4 and y = 3 then it is true that x + 7y = 25.  Therefore 

x = 4

y = 3

.

If you didn't spot the Pythagorean triple, then, more generally, you would write the 2nd equation as x = 25 - 7y, substitute this into the first equation and solve the resulting quadratic equation in y.

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