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y(x) = x(x + 3)   =  x^2 + 3x

(a)  We have this difference quotient 

[ (x + h)^2 + 3(x +h) - (  x^2 + 3x) ]

_________________________               simplify

            h

[  (x^2 + 2xh + h^2 +  3x + 3h)   - (x^2 + 3x) ]

___________________________________

                   h

[ 2xh + h^2 + 3h - ]

________________

           h

h [ 2x + h + 3 ]

____________

        h

2x + h + 3          and when h → 0, we have

2x + 3    ...... this is the gradient at any x  value on the curve

(b)   The gradient  of the curve at (-5, 10) is

2(-5) + 3   =      -7

\(\dfrac{\dfrac a b }{\dfrac c d} = \dfrac a b \dfrac d c\)

\(10~\dfrac 1 2 = \dfrac{21}{2}\\\)

\(\dfrac{\dfrac{21}{2}}{\dfrac 3 4} = \dfrac{21}{2} \dfrac{4}{3} = \dfrac{84}{6} = 14\)

Thanks guest, your praise really pleases me   

more than 3 does not include 3 

so the answer is 1/2 *1/2 = 1/4

I suggest that you repost and present your question properly.  

Guest, your answer makes no sense at all.

Probability of tails: 1/2

Probability of landing on 3, 4, 5, or 6: 4/6

Probability of both: 1/2 * 4/6 = 1/4

Attn:  tankai79

Of couse you do not know what s/he is  talking about .    You are 11

So why did you answer at all?  I think that is a bit rude.

Just so you could give yourself a point ?   That is even ruder.

sorry it should be f(x+1)

Here's my best attempt, ME....

Probability  of rollling

2 on first pair and 2 on second pair   =   (1/36) * (1/36) = (1/36)^2

3 on first pair and 3 on second pair  =  (2/36) * (2/36) = (2/36)^2

4 on first pair and 4 on second pair = (3/36) * (3/36) = (3/36)^2

5  and 5      =    (4/36) * (4/36)    =   (4/36)^2

6 and 6      =     (5/ 36) * ( 5/36)   = (5/35)^2

7 and 7       =    (6/36) * (6/36)

8 and 8  = (5/36)^2

9 and 9  = (4/36)^2

10 and 10 = (3/36)^2

11 and 11 = (2/36)^2

12 and 12 = (1/36)^2

So.....the probability is

( 2 [ 1^2 + 2^2 + 3^2 + 4^2 + 5^2] + 6^2 )  /  6^4     =   

146 / 1296  = 

73 / 648

thank you

Thanks, PM........!!!!!

I'm assuming that your lunch break comes during the work hours

So....you really are paid for 4.5 hrs per day

So....the weekly amount is

4.5  *  5 *  15     =   $337.50

Equation of asymptote  :   y = -2

I believe the "parent" is     f(x) = (1/3)^x

This graph shifts the parent  down by  2 units

So.....the equation  of g(x)   is

g(x)   =  ( 1/3)^x  - 2

I assume this is

3x /   [ 2x^2 - 18 ]        if so, we can wrtite this as

(3x) ( 2x^2 - 18)^(-1)

First derivative

3 ( 2x^2 - 18)^(-1)  - 3x (2x^2 - 18)^(-2) (4x) =

3( 2x^2 - 18 )^(-1) - 12x^2 (2x^2 - 18)^(-2)   =

3 ( 2x^2 - 18)^(-2)    [ (2x^2 - 18 - 4x^2 ] =

3 (2x^2 - 18)^(-2) [ -2x^2 - 18 ] =

3 ( 2x^2 - 18)^(-2) [ -2 (x^2 + 9 ] =

-6 (x^2 + 9) (2x^2 - 18)^(-2)

Second derivative

- 6  [   2x (2x^2 - 18)^(-2)  - 2 (x^2 + 9) (2x^2 - 18)^(-3) (4x) ]   =

-6 [  2x (2x^2 - 18)^(-2)   - 8x (x^2 + 9) (2x^2 - 18)^(-3) ]  =

-6 [ 2x (2x^2 - 18)^(-3)  ] [   (2x^2 - 18) - 4( x^2 + 9) ]   =

-6 [ 2x (2x^2 - 18)^(-3) ]   [ 2x^2 - 18 - 4x^2 - 36 ]   =

-6 [ 2x (2x^2 - 18)^(-3) [ -2x^2 - 54 ]   =

6 [ 2x (2x^2 - 18)^(-3) (2)(x^2 + 27) ]  =

12 [ 2x (2x^2 - 18 )^(-3) (x^2 + 27)   =

24x (x^2 + 27) / [ (2^3 (x^2 - 9)^(-3) ] =

24x  (x^2 + 27) / [ 8 ( x^2 - 9)^(-3) ] \

3x (x^2 + 27) (x^2 - 9 )^(-3)

We know the midpoint formula is \((\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})\) . Plugging our values in, we get \((\frac{4+x_2}{2}, \frac{5+y_2}{2})=(1,-2)\) . Solving for \(x_2,\) we attain \(\frac{4+x_2}{2}=1, 4+x_2=2, x_2=-2\) and \( \frac{5+y_2}{2}=-2, 5+y_2=-4, y_2=-9\) . Thus, the order pair is \(\boxed{(-2,-9)}\) .

Congrats Melody!! This question is on the Internet and the "Best" answer is about a page long by some guy, which eventually arrives at the same answer of 72!!!. Your method is so simple and intuitive that, almost, anybody can understand it.

The probability of Gimli getting more heads than Legolas is: ~ p = 1 - 1004/1005 =~0.000995 !!

Now the answer isn't incorrect.

Sorry, but I don't think the graph shows.

Thanks, I think I made the mistake of saying you could choose 2 of the same number.

What about the heads before the last flip? Gimli could've gotten all tails until the 2009th flip, and then Gimli didn't flip more heads than Legolas (unless he also got all tails).