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hmm what is the equation of f(x)

waiting for moderation of picture, then ill answer

OBT: Get ready!! This is a relatively complicated solution found by "Mathematica 11 Home Edition"

Here are the 3 solutions it found. The process involved is very long and "messy" !!.

x=-9.45649...... or x =-1.01031..........or =10.4668053180....

P.S. If you want the full calculations, I can give them to you.

hmm i dont think there is a solution cause the line just doesnt go through both points at once. sorry :P

o its 6 or -3 nvm

wait it wasnt 6 tho :I

a=6

the fire will never go out

8)

k=-2

you are welcome

8)

well the way it is described, the lines are locked to intersect along the y axis, so they can't intersect at anything with an x value that is not equal to 0, so there is no solution.

This is not a Rubik's cube because those dimensions do not form a cube.

WOW thank you as well!

I can explain how to simplify for you! First, let's solely worry about the square root of a negative number first.

and yet again, the guests beat every real user to answering my question! thank you!

Simplify the following:
4 sqrt(-9) - 2

sqrt(-9) = sqrt(-1) sqrt(9) = i sqrt(9):
4 i sqrt(9) - 2

sqrt(9) = sqrt(3^2) = 3:
4 i×3 - 2

4×3 = 12:
12 i - 2

Factor 2 out of 12 i - 2 giving 2 (6 i - 1):
2 (6 i - 1) =-2 + 12i

how to solve the 7438968940885x68730857893x872956509842976 please

how dare you. JOKES ARE MY THING!!!

+1 your own post all you want, it didnt help me one bit, got no extra points from it XD

At one point in my life (maybe 2 years ago), I watched a tutorial on how to solve the 3x3 Rubik's Cube. I am unsure if I could perform those steps again.  

Unfortunately, I just happen to know that the beginner's method is kinda inefficient because it solves one piece at a time, but that was the only way I could get halfway to solving the cube. 

Sum of the last three terms is       ar^(n - 1) + ar^(n - 2) + ar^(n - 3)  ⇒

ar^(n - 3)  [ r^2 + r  + 1 ]  = ar^(n - 3) [ 1 + r + r^2]

Sum of the first three terms  is     a + ar + ar^2   =  a [ 1 + r + r^2 ]

And the 3rd term is  5  ⇒  ar^2  = 5   ⇒  a = 5 / r^2 

And we're told that  the sum of the lst 3 terms  =  1024 times the sum of the 1st three

So we have that

1024a  [ 1 + r + r^2 ]  = ar^( n-3)  [ 1 + r + r^2]

1024  =  r^(n - 3)

Note that  1024  = 2 ^10....which implies that

2^10  = r^(n - 3)

So    r  = 2      and  n = 13

And the first term is 5 / 2^2  =  5/4

And the third term  is    ar^2  =   (5/4)*2^2  =  (5 /4) * 4  =   5

So.... the last term is

ar ^ (n -1)  = (5/4) (2)^(13 - 1)  =  (5/4) *2^12  =  5120

I assume you have to simplify the expression by factoring. Okay...

${x^2-16\over2x^2-5x-12}={(x+4)(x-4)\over(2x+3)(x-4)}={x+4\over2x+3}$.

To factor the denominator, you don't just find 2 numbers that add up and multiply to whatever if the coefficient of $x^2$ is not 1. Since $2x^2$ factors out to x and 2x, we have to find 2 numbers that, when one is added to twice the other, the result is -5 and when multiplied, the result is -12. The numbers that satisfy this are 3 and -4, so the denominator factors out to $(2x+3)(x-4)$ — the 2x is with the 3 becuase FOIL.

The numerator is easier, since it's just the product of the sum and difference of two numbers: $(x+4)(x-4)$. Cancelling out $(x-4)$, we get the final answer of ${x+4\over2x+3}$. 

"The sum of the last 3 terms of a geometric progression having n terms is 1024 times the sum of the first 3 terms of the progression. If the third term is 5 find the last term. "

.

Yep......bedtime calls...!!!!!

I guess she must have made a mistake, although everything still works together to the same end result...I can see she must have been explaining a "concept" to someone, as she had underlined the "-10t", "+3t" and "-t"....

anycase, I'm putting this to bed now...CPill, you were a great help!!..thanx a lot!!