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converted 565 km^2 *

Oh jea, your right. I only converted 5.6 km^2 while it is 56 km^2. Mhmm that makes my calculation even more irritating having only 5.4Gw/h with one high or 10.8 Gw/h with two highes.

2+2=4

I'm not sure what you mean by dollars, but I'll treat that as nothing. 

x^2+4x+5 --> h would be the x value of the  vertex 

To find the vertex we need to you the formula: -b/2a 

This formula is for the axis of symmetry, but the axis of symmetry is also the x value of the vertex. 

Lets state our letters: 

a=1

b=4

c=5 

Plug and chug into the formula

-4/2(1) 

-4/2=

-2. So h =-2 

Now lets find a and k 

a is our strech factor which is in front of our x^2. 

So a is =1 

Now to find our k value which is the y value of the vertex, we plug in our x value for everytime we see x

(-2)^2+4(2)+5 

4+8+5

=17 so the k=17

Now lets plug into our vertex formula

a=1, h=-2 and k=17

y=a(x-h)^2+k

NOTE for our x value of our vertex which is h we flip signs when we put it into the formula

So: y=(x+2)^2+17 

Hope that helps 

It actually is possible.

134

To climb mountains!!!.

129

doggo splish splash

a meme lol

As far as I know, it has only 3 bases.Base 10, base e(natural log) and base 2. Look for the "log" key. Just underneath it there is a small arrow. Click on that arrow and you will see the other two, ln, base 2.

a dog

PMT=PV. R.{[1 + R]^N/ [1 + R]^N - 1}

170.22 =8,200 x R {[1+R]^60 / [1+R]^60 - 1}

0.020758536= R*{[1+R]^60 / [1+R]^60 - 1}, solve for R

There is no DIRECT solution for R (interest rate), but through iteration and interpolation, the computer gives a rate of 9% compounded monthly.

An animal with 4 legs 

what is a dog

Your formula is : E = 0.5 * A * p * h^2 *g. All the numbers you have plugged in appear to be OK except one. A, the size of tidal barrage is 56.5Km^2 is way too big. It could be 5.65 Km^2 =5.65e6. So that you total will be:1,817,989,200,000J. Now you have 2 high tides per day.So, 1,817,989,200,000J x 2 =

3,635,978,400,000J. Therefore, the mean power generation potential = Energy generation potential / time in 1 day =3,635,978,400,000J /86,400s =42,083,083 / 10^6 =42 MW - per day.

Sorry, that is the best I can do. The only Math error in your calculations is 56.5Km^2 =5.65e+7 NOT 8.

correction!!!

120=2*2*2*3*5

Yes....I saw after I posted that the 64 was 8^2 and the other 8 was an exponent.....

Perhaps Wikipedia has the wrong answer !!   Perhaps look somewhere else...they may have made a typo ......

Im still editing too much, welp cant edit my first post thou since its a reply.

I didnt multpiplied it by two, but i mentioned it in my main post  in the last sentence. But still with multiplying it doesnt get close to the 550 Gw/h stated on wiki. Also where did u get the 64 and 8^2 in ur first reply? Shouldnt there be only one squared tidal high? 

Jup, its 8 meters. I squared 8, so it gave me the 64 in my calculation. I also transfered every unit into meter. I used the sample calculation as a refference. I didnt  divided it by  86400 seconds, as i didnt wanted mw/day, but gw/h to get a better refference.

I got the equation from many reliable resources, also the one you linked :)

E=1/2 Apgh^2    from

http://hydropower-tidalpower.blogspot.com/2009/07/energy-calculations_07.html

Which has a sample calc.....which will step you through your calculations to see where you might have erred

  your '8' for tidal height....is that in meters? (should be...not in feet)

Did you divide by the seconds in a day (86400)   /  Did you multiply by 2 (tides per day) as in the example????

Let me know !     G'Day !!

I THINK this means 'solve for 's' '

v^2 - u^2 = 2 as

(v^2-u^2)/(2a) = s

.5  x  64  x  1025   x   5.65 x10^8    x  8^2   x   9.8 = 116232704 x 10^8      

                      kg/m^3      m^2                m^2         m/s^2   =   kg  m^2/s^2    (correct units?)

where did 64  come from?   8 should be squared according to your equation......

Can you answer these questions for me.....   and where did you GET this eqution?

I'll see if I can help further....

Nice job completing the puzzle Hecticar !     The PROBLEM wasn't that difficult, but the erroneous presentation (which you figured out) was !        Keep up the good work !!