Sir-Emo-Chappington

(2x³) * (8xy²) * (3y)

= (2*8*3)*(x³⁺¹)*(y²⁺¹)

= 48 * x⁴ * y³

= 48x⁴y³

The question is a bit confusing as you didn't specify when something ends, so the entirety of "2-1>1/x-1" could be the power you raise "x" to.

For it to even be possible for x to be 0 (as you claim that is the correct answer) without causing a math error, I have to assume you meant:

¹/_((x²)-1) > ¹/_(x-1)

However when x is 0 in this equation, all I get is "-1 > -1" which is a false statement.

I have to assume you have typo'd the question or answer somewhere as a result.

The wording is a bit confusing.

If I understand correctly, you mean to say the train takes 1 hour to travel 120 kilometres. It starts at X speed and slows down by 4_kmh^-1.

However, do you mean the speed reduction happened before it left the first station, or are we suggesting it gradually slowed down by this amount uniformly across the 120_km?

Average speed:

^{120 km}/_{1 hour} = 120_kmh^-1

If we suggest it slowed down uniformly across the 120_km, then 120_kmh^-1 = ^{X + (X-4)} / ₂

120_kmh^-1 = X-2.

X = 118_kmh^-1

If we suggest is slowed down before it even started the 120kmh^-1 stretch, then 120_kmh^-1 = X-4

124_kmh^-1 = X

1 mile = 1.60934 kilometres

14,000 / 1.60934 = 8700_{Miles per hour}

with your "Banana" and "Ends" example, that's more stating:

1 x = 2 y

where x = 2y

So yes this works out fine.

However, using 0/0 you can actually 'prove' that 1 banana = 2 bananas and then draw that to an infinite scale.

1*0 = 0

2*0 = 0

1*0 = 2*0

(1*0) / 0 = (2*0) / 0

1 = 2

$$\left({{\mathtt{3}}}^{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right)}\right){\mathtt{\,\times\,}}\left({{\mathtt{5}}}^{\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}\right) = {{\mathtt{15}}}^{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}\right)}$$

We got a lot of messyness in the powers here. We can split them into more individual numbers to seperate off the constants and variables.

$${{\mathtt{3}}}^{{\mathtt{x}}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{{\mathtt{5}}}^{{\mathtt{x}}}{\mathtt{\,\times\,}}{{\mathtt{5}}}^{-{\mathtt{1}}} = {{\left({{\mathtt{15}}}^{{\mathtt{2}}}\right)}}^{{\mathtt{x}}}$$

$${\mathtt{9}}{\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{5}}}}\right){\mathtt{\,\times\,}}{{\mathtt{5}}}^{{\mathtt{x}}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{{\mathtt{x}}} = {{\mathtt{225}}}^{{\mathtt{x}}}$$

$$\left({\frac{{\mathtt{9}}}{{\mathtt{5}}}}\right){\mathtt{\,\times\,}}{{\mathtt{5}}}^{{\mathtt{x}}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{{\mathtt{x}}} = {{\mathtt{225}}}^{{\mathtt{x}}}$$

Now we can put everything into logarithm form. Remember:

Log(a^b) = b * log(a)

Log(a*b) = log(a) + log(b)

$${log}_{10}\left({\frac{{\mathtt{9}}}{{\mathtt{5}}}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{x}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{5}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{x}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{3}}\right) = {\mathtt{x}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{225}}\right)$$

Let's move over all the 'x' multiples over, so we can handle them together.

$${log}_{10}\left({\frac{{\mathtt{9}}}{{\mathtt{5}}}}\right) = {\mathtt{x}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{225}}\right){\mathtt{\,-\,}}{\mathtt{x}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{5}}\right){\mathtt{\,-\,}}{\mathtt{x}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{3}}\right)$$

Factorise it a little, so we can now split the "x" from the rest of the formula

$${log}_{10}\left({\frac{{\mathtt{9}}}{{\mathtt{5}}}}\right) = {\mathtt{x}}{\mathtt{\,\times\,}}\left({log}_{10}\left({\mathtt{225}}\right){\mathtt{\,-\,}}{log}_{10}\left({\mathtt{5}}\right){\mathtt{\,-\,}}{log}_{10}\left({\mathtt{3}}\right)\right)$$

Simplify the logarithms and re-arrange the forumula

$${log}_{10}\left({\frac{{\mathtt{9}}}{{\mathtt{5}}}}\right) = {\mathtt{x}}{\mathtt{\,\times\,}}{log}_{10}\left({\frac{\left({\frac{{\mathtt{225}}}{{\mathtt{5}}}}\right)}{{\mathtt{3}}}}\right)$$

$${log}_{10}\left({\frac{{\mathtt{9}}}{{\mathtt{5}}}}\right) = {\mathtt{x}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{15}}\right)$$

$${\frac{{log}_{10}\left({\frac{{\mathtt{9}}}{{\mathtt{5}}}}\right)}{{log}_{10}\left({\mathtt{15}}\right)}} = {\mathtt{x}}$$

$${\mathtt{x}} = {\mathtt{0.217\: \!051\: \!613\: \!246\: \!638\: \!7}}$$

- (5x+17a-6) - (-42a-5x)

First we can expand out the equations, since it's all addition and subtraction

-5x - 17a + 6 + 42a + 5x

Now seperate them according the the variables

[5x - 5x] + [42a - 17a] + 6

Simplify it:

25a + 6

$${\frac{{\mathtt{\,-\,}}\left({\mathtt{7}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{85}}}}\right)}{\left({\mathtt{6}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{85}}}}\right)}}$$

Put the co-efficiencts into the surds by squaring them...

$${\frac{{\mathtt{\,-\,}}{\sqrt{{\mathtt{85}}{\mathtt{\,\times\,}}{\mathtt{49}}}}}{{\sqrt{{\mathtt{85}}{\mathtt{\,\times\,}}{\mathtt{36}}}}}}$$

Simplify...

$${\mathtt{\,-\,}}{\frac{{\sqrt{{\mathtt{4\,165}}}}}{{\sqrt{{\mathtt{3\,060}}}}}}$$

Which can be represented as:

$${\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\sqrt{{\mathtt{3\,060}}}}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{4\,165}}}}\right)$$

So we can put the surds together, again by squaring it:

$${\mathtt{\,-\,}}{\sqrt{{\mathtt{4\,165}}{\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{{\sqrt{{\mathtt{3\,060}}}}}^{\,{\mathtt{2}}}}}\right)}}$$

Simplify...

$${\mathtt{\,-\,}}{\sqrt{{\frac{{\mathtt{4\,165}}}{{\mathtt{3\,060}}}}}}$$

Now we can simplfy the fraction:

$${\mathtt{\,-\,}}{\sqrt{{\frac{{\mathtt{833}}}{{\mathtt{612}}}}}}$$

And further...

$${\mathtt{\,-\,}}{\sqrt{{\frac{{\mathtt{49}}}{{\mathtt{36}}}}}}$$

11666666666666666666666

Let's add some commas to make it easier to read... (Or periods in some countries)

11,666,666,666,666,666,666,666

Now, the level of magnitude of the 11 is 10^21, also known as "Sextillion".

So we have:

11 Sextillion, 666 Quintillion, 666 Quadrillion, 666 Trillion, 666 Million, 666 thousand, 6 hundred and 66

or: 

Eleven Sextillion, six hundred and sixty six Quintillion, six hundred and sixty six Quadrillion, six hundred and sixty six Trillion, six hundred and sixty six Million, six hundred and sixty six thousand, six hundred and sixty six.

$${\sqrt{{\mathtt{x}}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}}}$$

Let's combine the surds. To do this, square one of them (so put it in terms of it's square root)

$${\sqrt{{{\sqrt{{\mathtt{x}}}}}^{\,{\mathtt{2}}}{\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}}$$

Now simplify since obviously the square of a square root is...

$${\sqrt{{\mathtt{x}}{\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}}$$

Expand out the equation inside the surd and you get:

$${\sqrt{{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{x}}}}$$

Assuming it is not compound interest (Since if it was, the daily gain would change on a yearly basis):

1 year of interest = 10,000 * 5%

= 10,000 * ¹/₂₀

= ^10,000/₂₀

= 500

1 day of interest = ⁵⁰⁰/₃₆₅

= $1.37

200 * ¹/₄ * ¹/₂ * ¹/₅

= 100 * ¹/₄ * ¹/₅

= 25 * ¹/₅

= 5

Depends how you mean "add".

If you mean something like -7 + 4, then look at which number is larger. The larger number subtracted by the smaller number gives you the resulting value, and if the larger number is the negative one, then the resulting value is negative. so -7 + 4 = - (7  -4) = - (3) = -3

Consider positives as "removing" negative values, and negatives "removing" positive values. It's easier to work this way as long as you identify which value is bigger.

If you mean adding a negative number as though it was a positive number, this is represented as "|x|" meaning you get the value of x in it's positive form (So if x = -8 then |x| = 8, and if x = 4 then |x| = 4).

-3 + -5 + 1

Consider that "+ -" is just another way of saying "minus" or " - " (I actually do this all the time personally. Makes it easier to understand numbers as negatives and not negative as a function).

So we have:

- 3 - 5 + 1

Combine the negatives together:

- 8 + 1

Now combine these two together:

 -7

to type the inverse trigonomic function, enter "a" before it.

This gives "acos", "asin" and "atan" as the invesre functions. (In your case it would be "asin")

Suggest (2x - 3)/(x-2) = 1

2x - 3 = x - 2

2x - x = -2 + 3

x = 1

Suggest (2x - 3)/(x-2) = -1

2x - 3 = -1(x - 2)

2x - 3 = -x + 2

3x = 5

x = 5/3

x = 1

x = 5/3

(6121.23N) / (78.54x10-^6 )

First since we have some scientific notation present, I'll turn this whole thing into scientific notations:

= ([6.12123 * 10³]N) / (7.854 * 10^-5)

Now seperate out the magnitudes and combine them

= [(6.12123 / 7.854) * 10^3--5] N

Complete the fraction

= 0.7793773873185638  * 10⁸ N

And now round it for simplicity and make it comply with Scientific Notation

= 7.794 * 10⁷ N

Alan, that is not fully correct. While atan is the same as "tan^-1()", that "-1" is not actually it's power: it's merely a common notation to mean the inverse function (So if you had "3" instead that would be saying the tangent function 3 times).

Instead, tan to the power of -1 is referred to as "cot()" or "cotangent".

In simpler terms:

Suggest: y = tan(x)

∴ tan^-1(y) = x

∴ y^-1 = cot(x)