All Answers 358150 Answers

I left a few around that you could answer Dragon but you'll have to hunt them down.  

To solve $${\mathtt{84}}{\mathtt{\,-\,}}\left[{\mathtt{8}}{\mathtt{\,-\,}}{\left(-{\mathtt{2}}\right)}^{{\mathtt{3}}}\right]$$, first do the exponents.

$${\left(-{\mathtt{2}}\right)}^{{\mathtt{3}}} = \left(-{\mathtt{8}}\right)$$

Now, subtract $$\left(-{\mathtt{8}}\right)$$ from $${\mathtt{8}}$$.

$$8-(-8)=$$ $${\mathtt{8}}{\mathtt{\,\small\textbf+\,}}\left({\mathtt{8}}\right) = {\mathtt{16}}$$

Finally, subtract $${\mathtt{16}}$$ from $${\mathtt{84}}$$.

$${\mathtt{84}}{\mathtt{\,-\,}}{\mathtt{16}} = {\mathtt{68}}$$

So, $${\mathtt{68}}$$ is our final answer. 

Hope this helps!

You know you get preferential treatment if you are a member, especially when you have such good manners. 

December, for me....

BTW.....I wonder which month all the ugly people were born in ????

CPhill I got all these amazing pics on Google+ , there were many of them I still have some , I'll post those next time and yeah it would look great onFacebook as your avatar ! Melody that cloud is real , I don't remember the places name but its real! And reinout funniest isn't bad either , you make people laugh that's awesome but beautiful is the best tho , btw in which month were u born CPhill and Melody?

Thank you so much :) Just needed the start :)

Oh don't be funny Dragon, don't you know he's a president " a president" obviously he can afford them , but yeah after buying them I guess it would take a lot of time for him to earn what he's spent !

(4-5i)(-8+i)(1-2i)    taking this in parts, we have

(4-5i) (-8 + i)   =    -32 + 40i +4i - 5i² = -32 + 44i + 5   =  -27 + 44i

(-27 + 44i) ( 1 - 2i)  =  -27 + 44i + 54i - 88i²  = 61 + 98i

What's the question here???

horizontal velocity vh = 6*cos(30°) = 6*(√3)/2 = 3√3 m/s

initial vertical velocity vv = 6*sin(30°) = 6/2 = 3m/s

Using s = ut + at²/2 for change in vertical distance in time t, with acceleration a (-9.8m/s²) and initial velocity u (vv = 3m/s) we have

0 = 3*t - 9.8*t²/2 or t = 6/9.8 s (ignoring the t = 0 solution, which just represents staying still!).

The horizontal distance in time t is vh*t or 3√3*6/9.8 m

$${\frac{{\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{\mathtt{6}}}{{\mathtt{9.8}}}} = {\mathtt{3.181\: \!317\: \!809\: \!820\: \!386\: \!9}}$$

distance jumped ≈ 3.18m

HAHAHAHAHAHAHAHAHAHAHAHAHAHA  !!!!!!!!!!!!!!!!!

(Ain't  I crafty ????????)

What a show off!  Anyway,  my gift is to do things the   L  O  N  G     w  a  y    

Yeah...I thought it was a mistake......it was still pretty crafty the way you figured that out!!...I can see it NOW........this was my usual experience in Physics class.....I always understood it better AFTER the tests!! (LOL!!!)

Notice one thing more about this one , if we want to use Melody's algebraic method....taking it from here

5(x-2)(x-4)² < 3(x-2)²(x-4)

We can do this to simplify things.....

5(x-2)(x-4)² - 3(x-2)²(x-4) < 0      factoring, we have

(x-2)(x-4)[5(x-4) - 3(x-2)] < 0

(x-2)(x-4)[5x- 20 -3x +6 ] < 0

(x-2)(x-2)[2x -14] < 0

2(x-2)(x-4)(x-7) < 0

(x-2)(x-4)(x-7) < 0

And notice that we don't have to use the Factor Theorem at all....the "roots" (i.e., critical interval values) are right in front of us !!!

Chris, I did mean 53° not 56°.  Goodness knows where the 56 came from, there isn't a 6 anywhere in the question!  Just put it down to early onset senility!

$${\mathtt{v}} = {\sqrt{{\frac{\left({\frac{{\mathtt{9.8}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{40}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{53}}^\circ\right)}}}\right)}^{{\mathtt{2}}}}{{\mathtt{2}}}}\right)}{\left({\mathtt{40}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{53}}^\circ\right)}{\mathtt{\,\small\textbf+\,}}{\mathtt{15}}\right)}}}} \Rightarrow {\mathtt{v}} = {\mathtt{17.831\: \!156\: \!614\: \!371\: \!128}}$$

v ≈ 17.8m/s

Be careful here.....mutiplying by "a" isn't the same thing a dividing by "a"....

For instance, if we have

3x = 6    

Then I would want to divide both sides by 3 to solve for x.   But, if I multiplied both sides by 3, I would have

9x = 18

And I would only have a multiple of the first equation  ....   and wouldn't be any closer to solving it than I was originally.....!!!

Alan....where did  the "56°" come from???

Did you mean "53°"  instead ???

It depends what x is.

if x=-50

then --50=+50>12

If x=4

then -4<12

Lets look at all the solutions to this

$$\\-x\le 12\\
x \ge -12$$

so any number greater of equal to 12 will make this true.  Any number less than 12 will make it false.  

It is great that you have graphed this Alan - I would not have thought to do that.  

Ignoring air resistance, horizontal velocity is constant at v*cos(56°).  Time is distance over velocity, so time to cross is t = 40/(v*cos(56°).

Change in vertical distance is given by -15 = v*sin(56°)*t - 9.8*t²/2 or 

-15 = v*sin(56°)*40/(v*cos(56°)) - 9.8*(40/(v*cos(56°)))²/2

-15 = 40*tan(56°) - 9.8*(40/(v*cos(56°)))²/2

(9.8*(40/cos(56°))²/2)/(40*tan(56°) + 15) = v²

$${\mathtt{v}} = {\sqrt{{\frac{\left({\frac{{\mathtt{9.8}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{40}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{56}}^\circ\right)}}}\right)}^{{\mathtt{2}}}}{{\mathtt{2}}}}\right)}{\left({\mathtt{40}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{56}}^\circ\right)}{\mathtt{\,\small\textbf+\,}}{\mathtt{15}}\right)}}}} \Rightarrow {\mathtt{v}} = {\mathtt{18.369\: \!394\: \!701\: \!106\: \!074}}$$

or v ≈ 18.4m/s

Yes it was a good way.   I had to go to Wolfram|Alpha to find a calc that would play ball.  

Yes, your answer is exact Melody.  I noted in my answer that mine was approximate; but it was a quick way of doing it!

Alan, you beat me!  Mine is exact I think.  

Not many calculators can deal with such large numbers though.

i have 5 tickets for a lottery that contains a total of 100.000 tickets. 100 tickets are drawn, what are the chances of me winning?

Im splitting them into 100 winning and 99900 losing tickets

1-P(5lose)

1-(99900C5/100000C5)

Very close to 5/1000 = 1/200

That's what I think anyway.    

Perhaps this graph will help

Find the intersection points by setting

1.   7x-3 = 3x-8

2.  -(7x-3) = 3x-8