Melody

Lets see

\(\frac{53^{2000}}{53^{1999}}=53^{2000-1999}=53\)

Think about the second one.

The numerator and the denominator are reduced by 2000

    but proportionately the numerator has been reduced by less than the denominator because it was bigger to start with.

   so proportionately... the top was reduced by a smaller factor than the bottom so the entire number got bigger.

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It is like this

\(\frac{10}{8}\;\;compared \;\;to \;\;\frac{9}{7}\\ 1\frac{2}{8}\;\;compared \;\;to \;\;1\frac{2}{7}\\ \)

The second one is bigger.

Thanks itsyaboi,

I'd just like to have a go at it too :)

\({5^{m-2}*2^{m+2}} \over 10^m-10^{m-1}*2\\~\\ =[{5^{m-2}*2^{m+2}} ]\div [10^m-10^{m-1}*2]\\~\\ =[{5^{m-2}*2^{m-2}*2^4} ]\div [10*10^{m-1}-10^{m-1}*2]\\~\\ =[10^{m-2}*2^4 ]\div [10^{m-1}(10-2)]\\~\\ =[10^{m-2}*16 ]\div [10^{m-1}(8)]\\~\\ =10^{m-2-(m-1)}*2 \\~\\ =10^{-1}*2 \\~\\ =\frac{2}{10}\\~\\ =\frac{1}{5}\)

Good spotting CU.

Clelz, No one can read that, or at least I cannot anyway. You need to post your question on the actual forum.   

I'm sure you are correct Rom but I am with itsyaboi.

Talking about needing to go the long way around!

Have you looked at any of these?

Good one Chris  LOL

But they are not frills,  they are important   !!!  :)

The absolute value of a number is the distance it is from 0 onthe number line. 

Distances are never negative. so they can never be less than -5.

So there are no solutions to this .... not in the real number system anyway.

A sine function has the following key features:

Period = π

Amplitude = 2

Midline: y=−2

y-intercept: (0, -2)

The function is a reflection of its parent function over the x-axis.

Use the sine tool to graph the function. The first point must be on the midline and the second point must be a maximum or minimum value on the graph closest to the first point.

the parent funtion is     y=sinx

this has a period of 2pi, and amplitude of 1 and a midline of y=0

\(y=asin[n(x+c)]+m\\ \text{a is the amplitude}\\ \text{m is the midline.}\\ \text{c is the phase shift that you probably don't need to worry about, it is zero}\\ \text{the period is }\frac{2\pi}{n} \)

Since it is reflected over the x axis, the a value will be negative so

Period = π                           n=2

Amplitude = 2                     a= - 2    (becasue of the xaxis reflection)

Midline: y=−2                      m=-2

y-intercept: (0, -2)               hopfully this one will just work.

Can you use that formula I just gave you to work out the equation of your sine curve?

Then graph it in desmos and check that it works :)