All Answers 358460 Answers

k^-10  =  1/k^10

So:  9k^-10 / (9k^10)  =  9 / 9k^10k^10)  =  1 / k^20.           (the 9's cancel.) 

Of that $18.00 per hour, how much do you have in take-home pay? 

How many hours do you work per week?

Do you get overtime pay?

How much of that do you spend?

If your take-home pay were $18.00 per hour and you saved all of it, you could have $50,000.00 after 2800 hours  ($50,000.00 ÷ $18.00/hour)  but this is not realistic.

First find the square roots of 144, which are 12 and -12.

So now you know that the square of any integer less than 12 and greater than -12 would be less than 144.

That includes:

0, ±1, ±2, ±3, ±4, ±5, ±6, ±7, ±8, ±9, ±10, ±11

Which would be 23 integers, including zero.

$${f}{\left({\mathtt{x}}\right)} = {{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{7}}$$

Plug in -3 for x:

$${f}{\left(-{\mathtt{3}}\right)} = {\left(-{\mathtt{3}}\right)}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{3}}{\mathtt{\,\times\,}}\left(-{\mathtt{3}}\right){\mathtt{\,-\,}}{\mathtt{7}}$$

Now simplify:

$${f}{\left(-{\mathtt{3}}\right)} = {\mathtt{9}}{\mathtt{\,\small\textbf+\,}}{\mathtt{9}}{\mathtt{\,-\,}}{\mathtt{7}}$$

$${f}{\left(-{\mathtt{3}}\right)} = {\mathtt{11}}$$

Ok thank you alan its very nice =D 

use the inv cos function on your calculator

When working with fractions , I always try to get them to simplest form for multiplication first off 2/4 is 1/2 now we want to get our mixed fraction to a single fraction. 3 1/2 is equal to 7/2. now multiply 7/2 X 7/2  which will be? Remember denominators multiply as well as numerators. I am not going to give you the final answer..:)

Take your last item and get into usable format remember 8 = 8/1 so 2 3/4  change your problem to look like this : 8/1 * 11/4 = X or 8 X 11/4  do mulpication first and divide by denominator 4. Remember denominators are multiplied also.  

 $$\left({\frac{{\mathtt{8}}}{{\mathtt{1}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{11}}}{{\mathtt{4}}}}\right) = {\mathtt{22}}$$ or  $$\left({\frac{{\mathtt{32}}}{{\mathtt{4}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{11}}}{{\mathtt{4}}}}\right) = {\mathtt{22}}$$ in this case you would have $${\frac{{\mathtt{352}}}{{\mathtt{16}}}} = {\mathtt{22}}$$.  I would recommend the first operation for siplicity. But you should understand common denominators and multiplication of denominators, also. Also, always show your work in case you make a mistake your teacher can show you your error(s) 

If you mean 48²:     2304

If you mean √48:    √16 · √3  =  4√3 

Sun 26/10/14

1) Vectors.   Thanks Alan

Yes thanks Alan.  Now it is very clear.  

Does this help?

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@@ End of Day Wrap       Sat 25/10/14    Sydney, Australia        Time 1:10am   (Really Sunday morning)      ♬

Good Evening all,  

Great answers today from Heureka, Geno3141, Rosala, LancelotLink and Chimp Ayumu, Kitty<3, CPhill and Alan.  Thank you.  

I just pulled out a couple of interest posts and here they are:    

1) Calculus - rates - I like these but I got really frustrated because there are only 100cm in a metre.  Doh!  

Thanks Chimp Ayumu, LancelotLink, Chris and Alan (and even I eventually got it right)      

How are we supposed to know if you haven't told us yet?

Good advice Chris.

My question I just asked Alan will most likely be my next Doh! moment.    

You know, sometimes when you are doing something really well it is easier just to roll with it.  

(Like when you roll the body in the carpet to hide the evidence of a terribly dumb m****r)  :)

Alan, I am really confused.  Nothing new there.  I've only looked at the first one.

Does the air speed indicator show the actual speed, or the speed if there was no crosswind?

Could you draw the picture please.  I don't get it.    

Baseball

A. Horizontal component of velocity is vx = v*cos(43°)

    Vertical component of velocity is vy = v*sin(43°)

Horizontal velocity is constant so time to hit ground is t = 188/vx

In this time vertical distance travelled is -0.9m so:

-0.9 = vy*t - (1/2)*g*t² where g = 9.8m/s²

so: -0.9 = vy*188/vx - (1/2)*9.8*(188/vx)²

-0.9 = 188*tan(43°) - (1/2)*9.8*(188/cos(43°))²/v²

v² = (1/2)*9.8*(188/cos(43°))²/(188*tan(43°) + 0.9)  m²/s²

$${\mathtt{v}} = {\sqrt{{\frac{\left({\mathtt{0.5}}{\mathtt{\,\times\,}}{\mathtt{9.8}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{188}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{43}}^\circ\right)}}}\right)}^{{\mathtt{2}}}\right)}{\left({\mathtt{188}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{43}}^\circ\right)}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.9}}\right)}}}} \Rightarrow {\mathtt{v}} = {\mathtt{42.865\: \!673\: \!803\: \!496\: \!16}}$$

v ≈ 43 m/s

yes thats good .. thank you alan 

what about the second one 

Do you mean

$$xlog(x)\times log(x)\\$$

$$\\=x[log(x)]^2$$

That is about it I think.

Note that we're getting 150ml more than we usually do.

So 150/750 = 1/5 = 20% "free"........of course, I always want to ask them.....why don't you just pour the "free" part from 5 bottles into one??.....then, it shouldn't cost me anything...

Plane

A. The angle between south and southwest is 45°. 

The x-component (east) of velocity is -16*cos(45°):  vx = -16/√2 m/s

The y-component (north) of velocity is -(16*sin(45°) + 45):  vy = -(16/√2 + 45) m/s

B. Magnitude:  v_P/E = √(vx² + vy²)

C.  Direction:  θ = tan^-1(vy/vx)  But this will give a result in the first quadrant (i.e. an angle between 0° and 90°).  You will need to convert this to an angle in the third quadrant (i.e. between 180° and 270°).

Anon, I think that you are reading a different question from me LOL

by putting it over 1

ex.

$$12=\frac{12}{1}$$

No LHS not equal RHS

CPhill's answer shows you that there are no real solutions.

The parabola y=3x^2-7x+21   would lie completely above the y axis.

Chris wrote: "I understood your method, Alan......but, am I misreading this?? You say "r" at the "time of interest" = 1m "

Well spotted Chris - you are right:  r = 2/3 m.  The r = 1m is there because I inadvertently set the whole worksheet to round all numbers to the nearest integer, instead of just setting the final result to be rounded to the nearest integer.

The correct value is used in the calculation though.

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