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fx / gx   =   (x-3)(x+3)  /  (x-3)(x-4)         factored

     so we can see that x cannot equal 3 or 4   because the denomintaor would be ZERO....undefined/not allowed

Now we can 'cancel out' the x-3 in the numerator and the denominator andit becomes

(x+3)/(x-4)  and    x cannot be 3 or 4

The function is neither even or odd

An even function will be symmetric about the y axis.....this is clearly not the case

It is also not odd.....odd functions are symmetric about the origin....[ it will be symmetric with the line y =x ]

Look at  thee following graph....note that it resembles what  we have.....also note that it is not symmetric about  the line y = x

So....neither....!!!!

oh okay

okay thanks

Sorry....neither even or Odd...   it is not symetrical around origin  either....

Probably odd because it is not symmetrical.

I am a little bit confused?

Note that any portion of  an hour is rounded up.....and we cannot exceed 12 hours

So.....we will only have positive whole numbers in the domain.....and the greatest will be 12

So.....the first answer is  correct

Note that if we rent the boat for 3.5 hours....it puts 4 in the  domain...etc.....

[ This is known as a "step" function ]

Log 4 = .60205    (y1)

Log 6 = .77815    (y2)

Avg rate of change =  slope =   y2-y1  /  x2-x1     = log6-log4   /   6-4  = .088 per unit= .09 (rounded)

Deleted......see cphil's answer below

EVEN FUNCTION : A function with a graph that is symmetric with respect to the y-axis. A function is even if and only if f(–x) = f(x).

Soooooo......    what do you think now?   Does f(x) = f(-x) ?   

Range refers to all the possible y values covered by a function

Note that  the minimum y value is 0  and the maximum is 28....so

Range = [ min, max ]  =  [0, 28 ]

You can only drive a POSITIVE amount of miles, so the DOMAIN would be all positive real numbers.

​ P(even on 2nd run | odd on 1st run) = P(even on 2nd run) ​

The COULD be congruent if the neighbor cuts the 12 ft moulding in to three 4 ft pieces to build his triangle...both would be equalateral triangles.  If the neighbor uses ANY other lengths (say: 5,5,2) , the triangles will not be congruent.

2.2(14.8r - 3.8) =74,9     Divide both sides of the equation by 2.2

14.8r-3.8 = 34.045          Add 3.8 to both sides

14.8r =37.845                 divide by 14.8

r=2.557

Great solution,  heureka......!!!!

Nice.....Melody and heureka....!!!!!

The sides can be the same length, but that would only make the triangles similar not congruent. The angles need to be the same too.

I reallly like that way. It is so short and simple.

Thanks guest.:)

Working to mod 8 ,

\(5^{137}\equiv5.5^{136}\equiv5.25^{68}\equiv5.1^{68}\equiv 5.\)

So the remainder will be 5.

Hello Rom,

i switch it to an alternating series, because the question is: 0.1₂  -  0.01₂ + 0.001₂ - 0.0001₂ + 0.00001₂ +- ...

and the answer is a fraction.

\(\begin{array}{|rcll|} \hline r &=& \dfrac{ {\color{red}-}\dfrac{1}{4} } { \dfrac{1}{2} } = {\color{red}-}\dfrac{1}{2} \\\\ r &=& \dfrac{ \dfrac{1}{8} } { {\color{red}-}\dfrac{1}{4} } = {\color{red}-}\dfrac{1}{2} \\ \ldots \\ \hline \end{array}\)

Limits
I was just reviewing my calc book when I stumbled across this problem.
\(\large { \lim \limits_{x \to \infty} \dfrac{\sqrt{x^4-10x}-\sqrt{x^4-5x^2+7}}{5} }\)

\(\begin{array}{|rcll|} \hline &&\mathbf{ \lim \limits_{x \to \infty} \dfrac{\sqrt{x^4-10x}-\sqrt{x^4-5x^2+7}}{5} } \\\\ &=&\dfrac15 \cdot \lim \limits_{x \to \infty} \left( \sqrt{x^4-10x}-\sqrt{x^4-5x^2+7} \right) \\\\ &=&\dfrac15 \cdot \lim \limits_{x \to \infty} \left( \sqrt{x^4-10x}-\sqrt{x^4-5x^2+7} \right) \dfrac{\left( \sqrt{x^4-10x}+\sqrt{x^4-5x^2+7} \right)}{\left( \sqrt{x^4-10x}+\sqrt{x^4-5x^2+7} \right)} \\\\ &=&\dfrac15 \cdot \lim \limits_{x \to \infty} \left( \dfrac{ x^4-10x-(x^4-5x^2+7)} { \sqrt{x^4-10x}+\sqrt{x^4-5x^2+7} } \right) \\\\ &=&\dfrac15 \cdot \lim \limits_{x \to \infty} \left( \dfrac{ 5x^2-10x-7} { \sqrt{x^4-10x}+\sqrt{x^4-5x^2+7} } \right) \cdot \dfrac{x^2}{x^2} \\\\ &=&\dfrac15 \cdot \lim \limits_{x \to \infty} \left( \dfrac{ \frac{5x^2-10x-7} {x^2} } { \frac{\sqrt{x^4-10x}+\sqrt{x^4-5x^2+7}} {x^2} } \right) \\\\ &=&\dfrac15 \cdot \lim \limits_{x \to \infty} \left( \dfrac{ \frac{5x^2-10x-7} {x^2} } { \sqrt{\frac{x^4-10x} {x^4} }+\sqrt{ \frac{x^4-5x^2+7} {x^4} } } \right) \\\\ &=&\dfrac15 \cdot \lim \limits_{x \to \infty} \left( \dfrac{ 5-\frac{10}{x}-\frac{7}{x^2} } { \sqrt{1-\frac{10}{x^3} } + \sqrt{1-\frac{5}{x^2}+\frac{7}{x^4}} } \right) \\\\ &=&\dfrac15 \cdot \left( \dfrac{ 5-0-0 } { \sqrt{1-0 } + \sqrt{1-0+0} } \right) \\\\ &=&\dfrac15 \cdot \left( \dfrac{ 5 } { 1 + 1 } \right) \\\\ &=&\dfrac15 \cdot \dfrac{ 5 } {2 } \\\\ &\mathbf{=}&\mathbf{\dfrac12} \\ \hline \end{array}\)

Hi Chris, I expect your logic is excellent but I am having trouble following it.

You have just done the same (sort of) as Tertre and me haven't you?

I mean you have developed a pattern and used the pattern (odd and even powers) to answer the question.

I am just a little confused because you have not shown the pattern but I am assuming that is what etc means.