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Well heck,   that makes sense.....hadn't thought of it that way....THANX, Melody !  

WOW, that is more like it - I might just sneak one before Geno gets here..... ok?........

OK.....here you go, Geno........

Come on Chris, you can do better than that !

You could at least offer Geno some champagne or some chocolates or something :))

What Melody said........!!!!

Thanks ElectricPavlov Here is something for you also to learn from :)

Value of k for which x+k is a factor of x^3+kx^2-2x+k+4 is

Just use the factor theorem

If x+k is a factor then

f(-k)=0

\((-k)^3+k(-k)^2-2(-k)+k+4=0\\ -k^3+k^3+2k+k+4=0\\ 3k+4=0\\ k=\frac{-4}{3}\\ \)

------------

Why does this work... It is very simple.

I'll show an example

\(f(x)=x^2+5x+6\\ f(x)=(x+1)(x+6)\\ f(-1)=(-1+1)(-1+6)=0\\ f(-6)=(-6+1)(-6+6)=0\\ SO\\ \text{If x+a is a factor of of a polynomial then f(-a) must =0 } \)

55/75 x 100% = 73.3 % Won

55/75  also equals  11/15 games won

Let x be the number

the smaller number is then x-10

Now  12 (x-10) - x  = 45

          12x - 120 - x = 45

           11x = 165

              x = 15         then the smaller number is   x-10 = 5        then the sum is   20

For an arithmetic sequence   Sn = n/2 (a1+an)

208 = 3/2(10+a3)

a3=128.66

10              a2                  128.66                (128.66+10)/2 = D  = 59.33

10        10+59.33         10 + 59.33+ 59.33

10         69.33                128.66

an then becomes       an= 10 + 59.33(n-1)          or     an = 59.33n - 49.33

Here is the graph:

There is no way to answer this unless you give us the PRICES of Kona and Plain.....

I will assume you MEANT  Force = 0.02 Newtons  and distance = 300 m   (not m/s)

Work = Force x Distance

W =  .02 (300) = 6 joules

Good one guest!  You made me laugh :))

Hi 315,

I don't know of any civil engineers here but we have many professional guests.

Why don't you just post your question ?

can someone help me, I need to find the non simplified difference quotieint at 0 of the function

f(x) = -|x|

Maybe I do not understand the question but there is no gradient at x=0.

The curve is not smooth at this point. 

The curve is not differentiable at this point.

y=-2x^2-1   we can write this as

 y= -2(x − 0) ^2 - 1        and in the form   y= a(x − h) ^2   +  k.....     the vertex is at (h, k)  =  (0, -1)    and this parabola is vertical......so he axis of symmetry is   x  = 0

3(2-k)= -3k+2   simplify

6 - 3k  = -3k + 2     add 3k to both sides

6 = 2    and this is impossible.....therefore, no solution

I checked the forum’s professional registry; none of the engineers on here are civil.

Most of them are grumpy and have bad attitudes.

Ok I think i just made too many thoughts about the question. The math way seems logical, but just thinking about the question itself kinda confused me.

Here's an addendum to this :

Let a  = m      and  [ 2b + 3c]  =  n

So we are looking for the term in (m + n)^6  where  a = m has a power of 1....and this is given by

C(6,1) mn^5  =  6an^5

So....we have

6a [ 2b + 3c]^5

And now we are looking for the term in (2b + 3c)^5  where  [2b]  has a power of 2....and this is given by:

C(5, 2) [2b]^2 [3b]^3  =   10 * 4b^2 * 27c^3  =  1080 b^2 c^3

Putting this all together, the term we are looking for is given by

6a [ 1080 b^2 c^3]  =   6480ab^2c^3 .......   and 6480  is the coefficient

4x - 3 + 2x = 8x -3 -x   simplify

6x  - 3   =   7x   - 3       add 3 to both sides

6x  = 8x         subract  8x from both sides

- 2x  = 0      divide both sides b  - 2

x =  0

Car                                   Truck

|--------|--------|--------x--------|

x is where they meet...car has travelled 3/4 and truck has travelled 1/4 (in the opposite direction) (because truck travels at 1/3 speed of car)

time units for car to get to x = 3         for truck to get to x is also 3

time units for car to back up to start to let truck pass = 3 x 3 = 9

time units for car to continue forward to other side = 4

Total time for both vehicles to be clear would then be   9+4= 13 time units from point x start

Time units for truck to get to x = 3

Time units for truck to back up to its origin = 3x3 = 9

Time units for truck to then cross thru detour and clear the ither end = 3x4 = 12

Total time if for both vehicles to clear if TRUCK is the one to back up = 9+12 = 21 time units

SO it is faster if the CAR backs up to the start to let the truck pass and then the car continues on its way across.