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Thanks Lance, you are right of course :))

$$\\\sqrt{48/16}=\sqrt{3}$$

Miss Melody should this be 3t(t-2)

Then t=0 and t=2

$${{\mathtt{2}}}^{{\mathtt{9}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}\left({{\mathtt{9}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{9}}}^{{\mathtt{2}}}\right){\mathtt{\,\small\textbf+\,}}{{\mathtt{9}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{9}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{9}}{\mathtt{\,-\,}}{\mathtt{2}} = {\mathtt{987}}$$       

Thanks Rosala, you have answered my question - the cars are just on dirt. 

Photos tell people things that you do not think about because you just take it for granted :)

I think this is interesting because I would not see a road like that in Australia.

We have dirt roads or we have tared roads, we do not have roads that are half and half :)

 Hi Mellie :)

$$\\In $\triangle ABC$, $AB = 5$ cm, $BC = 10$ cm, and the altitude drawn to $\overline{AB}$ is 8 cm. $\\$What is the number of centimeters in the length of the altitude to $\overline{BC}$?$$

Let h=altitude to BC

When I tried to draw the pic I realized that the altitude to AB had to lie outside the triangle.

That is, angle CAB is an obtuse angle.

Area of triangle ABC = 1/2 * 5*8 = 20cm^2       (error fixed - thanks Chris) 

Area of triangle also = 1/2 * BC * altitude to BC = 1/2 * 10* h = 5h cm^2

so      5h=20

h=4

The altitude to BC is 4 cm

Do you have a picture Mellie ?

Perhaps you need to write it in plain English too.  

Looks like tin has got another person in his side...TJ...*rosala and dragonlance turn into their red and yellow ranger suits and attack with their sword and blasters on tin and TJ...looks like a bad fight is gonna take place*

The object is at rest when velocity = 0

$$\\0=3t^2-6t\\\\
0=3t(t-6)$$

the object is at rest initially and at rest again after exactly 6 units of time :)

$${\frac{\left({\mathtt{7}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{6}}}}\right)}{\left({\frac{{\mathtt{5}}}{{\mathtt{9}}}}\right)}} = {\frac{{\mathtt{129}}}{{\mathtt{10}}}} = {\mathtt{12.9}}$$

well i have no idea..i think this road was made last year or so...the cars are just parked beside the road on the simple ground.....i dont really bother for detail.....:/

Answered here

If you mean (3x41x18) + (3x41x91), then it means 2,214 + 11,193=13,407

If by 2,6 you mean 2.6(in English-speaking countries), then:

2.6 x 1/2 =2.6/2=1.3

sqrt(81)/1/(sqrt(81)=81

$$\\|x-3|-|x+1|/2*|x+1|=1\\\\
|x-3|-\frac{|x+1|}{2}*|x+1|=1\\\\
|x-3|-\frac{|x+1|*|x+1|}{2}=1\\\\
|x-3|-\frac{(x+1)^2}{2}=1\\\\
|x-3|=1+\frac{(x+1)^2}{2}\\\\
|x-3|=\frac{x^2+2x+3}{2}\\\\
x-3=\frac{x^2+2x+3}{2}\qquad or \qquad x-3=-\frac{x^2+2x+3}{2} \\\\
2x-6=x^2+2x+3\qquad or \qquad 2x-6=-x^2-2x-3 \\\\
0=x^2+9\qquad or \qquad 0=-x^2-4x+3 \\\\
x^2=-9\qquad or \qquad x=\frac{4\pm\sqrt{16+12}}{-2} \\\\$$

$$\\x^2=-9\qquad or \qquad x=\frac{4\pm\sqrt{4*7}}{-2} \\\\
x^2=-9\qquad or \qquad x=\frac{4\pm2\sqrt{7}}{-2} \\\\
x^2=-9\qquad or \qquad x=\frac{2\pm\sqrt{7}}{-1} \\\\
x^2=-9\qquad or \qquad x=-2\pm \sqrt{7 }\\\\
x=\pm\sqrt{- 9}\quad or \quad x=-2+\sqrt{7 } \quad or \quad x=-2-\sqrt{7 }\\\\
$So it appears that the only real roots are $\\\\
x=-2+\sqrt{7 } \quad or \quad x=-2-\sqrt{7 }\\\\$$