Area of circle = $$\textcolor[rgb]{0.5,0,0}{\Pi r^2}$$

where $$\Pi$$ is that number that you start off hating, but later come to love ♥

☞ You do realize that 2 m/s^2 is an acceleration.

First, convert 54 km/hr to m/s

Can we assume the train starts out stationary? (no pun intended)

Then use this equation: v = u + at

Good luck!

4&2/3 is the correct answer

$${\frac{{\mathtt{4.3}}}{{\mathtt{1.609}}}} = {\frac{{\mathtt{4\,300}}}{{\mathtt{1\,609}}}} = {\mathtt{2.672\: \!467\: \!371\: \!037\: \!911\: \!7}}$$

I would muliply each set by 1000(3 zeros behind the 1 in 100 and 3 numbers behind the decimal point in 1.609) and then go from there. Or, you can set up the problem and make the 4.3 a 43 and 1.609 to a 16.09 and then work it out that way.. hope this helps :)

21 is the jouk

19 is the true anser

 10+9=19

9 + 10 = 19

Hope this helps.

~dogeminer

Sorasyn, these are great examples :)

With you first example

$$\left({\frac{{\mathtt{4}}}{{\mathtt{5}}}}\right){\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}}{{\mathtt{7}}}}\right) = {\mathtt{6}}$$

It is easier if you multiply both sides  by 35 right from the beginning.

$${\mathtt{35}}{\mathtt{\,\times\,}}\left(\left({\frac{{\mathtt{4}}}{{\mathtt{5}}}}\right){\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}}{{\mathtt{7}}}}\right)\right) = {\mathtt{35}}{\mathtt{\,\times\,}}{\mathtt{6}}$$

$$\left({\mathtt{7}}{\mathtt{\,\times\,}}{\mathtt{4}}\right){\mathtt{\,\small\textbf+\,}}\left({\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}\right) = {\mathtt{210}}$$

$${\mathtt{28}}{\mathtt{\,\small\textbf+\,}}{\mathtt{20}}{\mathtt{\,\times\,}}{\mathtt{x}} = {\mathtt{210}}$$

$$$Use the formula$\\\\
\frac{y_2-y_1}{x_2-x_1}$$

x1=12.7, y1=5.4

x2=-10.2, y2=2.5

It is the locus of all points equidistant from a single point. (the centre)

See cue123smile, you really do have CDD. A very virulent form of it, too. Melody caught it from you.

Here is Melody, who only yesterday could calculate a fair fare to pay the troll’s toll by solving high-level circumnavigation equations for warped triangular troll-toll bridges; now, reduced to counting pennies in a kindergarten class for the slow and dim-witted.

This is your fault!

Get some treament for your disease, and while you are at it, wipe that 123smile off your face!

Thanks you anon 

ok so 

96km/hour

$${\frac{{\mathtt{96}}{\mathtt{\,\times\,}}{\mathtt{1\,000}}}{\left({\mathtt{60}}{\mathtt{\,\times\,}}{\mathtt{60}}\right)}} = {\frac{{\mathtt{80}}}{{\mathtt{3}}}} = {\mathtt{26.666\: \!666\: \!666\: \!666\: \!666\: \!7}}$$

$$\begin{array}{rll}
KEnergy&=&gravitational\; Potential \;energy\\
0.5mv^2&=&mgh\\
h&=&0.5v^2/g\\
h&=&0.5*(26.\dot6)^2/9.8 \;m\\
h&=&0.5*711.111/9.8\;m\\
h&\approx&36.28m
\end{array}$$

Is this correct?  It does sound better.  

If this one is correct then what was wrong with my logic the first time?

Thank you! I get it now!

Say the card is good for a total of N songs

By the end of Tuesday's shopping there are N - 3 -5 songs left on the card.

On Wednesday, she downloads half of these, so we can say the Wednesday

shopping used up (N - 8)/2 more songs

So by the end of Wednesday's shopping, she has downloaded how many songs all told?

write your answer involving N and some numbers ....

that's a start ..... 

nm got the answer

4.3

÷

1.609

=

your answer would be $265.5 you divide 295 dollars by 10 then subtract your answer you got from 295- 

It's correct your answer would be 0.90 repeted:) hope this helps

W*f is that"........ How do I answer that

I'm not sure what I did to get Courier text and all those blank lines.

☹

It's correct, it happend to me then I got it

6 and 9 are okay now :)

Lol I would

part (b)

It doesn't gain as much gravitational PE as it would if losses were absent

so friction losses = mgh_{ideal -}  mgh_actual

oh baballama!