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You can go to "insert" and click on Formula's or "symbols".

GuestSep 12, 2013

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192000.
There is a calculator at this website and you could use it, Alice.

GuestSep 12, 2013

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It's done the same way as if positive. 2 is 8 quarters, plus 1 quarter; make it negative: -9/4 (You were probably taught to multiply the denominator, 4, by the whole number, 2, to get 8, then add the numerator. That's the algorithmic shortcut, but the first way helps to understand what is actually happening.)

GuestSep 12, 2013

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IF you want 100 to fifth power then I suspect most mathematicians would just write it like this.

100^5.

The 5 denotes the power and would written as a a superscript. I have tried replicating that here but the text buttons above are not working for me.

However 100 is itself 10^2. So it would be better to write it as ((10^2)^5) This is where to evaluate it you now multiply the indices and so your answer is 10^10. If you want to expand that then 10^10 is one followed by 10 zeroes. That number has a name and is a billion. Hope this helps.

GuestSep 12, 2013

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10,000,000,000
Ten Billion.

GuestSep 12, 2013

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n(8^10) = n(8.10) = n(80)

GuestSep 12, 2013

Sep 11, 2013

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2/3. If you type it into the Web 2.0 calc and press enter, it will do it for you.

GuestSep 11, 2013

+166

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The answer to that question is 9. (DUH!!!!!!!!!!!!!!!!!!!!!!!!!) how could u not know that?

gamrgrrl156Sep 11, 2013

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I think you have a typo and should read:

f(3x) = 3f(x).

GuestSep 11, 2013

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Not sure who came up with the answer 'n' but it isn't right.

I think you need to factorise (simple):

9n - 9 = 9(n-1).

I am guessing this is part of an equation: 9(n-1) = 0? Which implies n = 1

GuestSep 11, 2013

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Ok - I assume you know what a prime number is? So all numbers can be expressed as product (ie. multiplying together) of the appropriate prime numbers. You always start with the lowest prime no first:

So in the case of 72 = 2 x 36 = 2 x 2 x 18 = 2 x 2 x 2 x 9 = 2 x 2 x 2 x 3 x 3

and 96 = 2 x 48 = 2 x 2 x 24 = 2 x 2 x 2 x 12 = 2 x 2 x 2 x 2 x 6 = 2 x 2 x 2 x 2 x 2 x 3

GuestSep 11, 2013

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So you need to put this expression in the form of:

(ax + p) (bx + c) where a, b, p and q are positive or negative numbers. such that when you expand the brackets (ie. both terms in the first brackets are multiplied with both terms in the second brackets) you get back your original expression.

So the clue is that we have to multiply ax and bx together to get 4x^2. So this could mean a = 1 and b = 4 or a = 2 and b = 2 or a = 4 and b = 1. Similarly, we much end up with a -9, this means p = 1 and q = -9 or p = -1 and q = 9 etc.

So we need a bit practice but if you play the options you will find that the factors are: (x - 1) (4x + 9). Why? Well, let's expand out the brackets:

x . 4x = 4x^2
x . +9 = +9x
- 1 . 4x = -4x
-1 . +9 = -9

Add them up: 4x^2 + 9x - 4x - 9 = 4x^2 +5x - 9

GuestSep 11, 2013

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Kalaiah19:
:| what is 9n - 9?

I think it is just 'n'

GuestSep 11, 2013

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The answer should be 1!
(5+5)(5+5) ÷ (5+5)(5+5)
(10)(10) ÷ (10)(10)
100 ÷ 100
1

The reason is when ever a number is divided by itself it equals one.
The only easiest way to get the answer to equal 100 is if the division sign was an equal sign. Maybe the problem was copied wrong?

(5+5)(5+5) = (5+5)(5+5)
(10)(10) = (10)(10)
100 = 100
100

GuestSep 11, 2013

+259

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However, there's another way to solve this equation using 2 Bezout theorems about roots of polynomial. According to the 1st one, the polynomial p(x) can be exactly divided by the binomial (x - a) if x = a and the quotient is p(a), and according to the 2nd one if the polynomial with the whole coefficients have roots that are the whole numbers then these roots are divisors of the polynom's constant term.

In this case we should apply the theorem 2 first and look for divisors of 6. They are -1,1;-2,2;-3,3;-6;6. Then we insert into the equation them to make sure whether one of them fits it. The suitable integer is 3. As the coeffs of the expression are the whole numbers the polynomial is being divided by the binomial x - 3 exactly, so its 1rst root is x = 3. To find the other roots we should divide the polynomial x^3-x^2-4x-6 by x - 3. Performing long division we get the quotient x^2 + 2x + 2, so the full arrangement will look as (x - 3)(x^2 + 2x + 2) = 0 The first factor has the root already had been calculated (x = 3). The second equation has no roots among the real numbers but has two roots among the complex numbers: x2 = -1+ i, x3 = -1 - i

solve(x^3-x^2-(4x)-6 = 0,x)

scrutinizerSep 11, 2013

+3146

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[input]abs(9)[/input]

adminSep 11, 2013

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GuestSep 11, 2013

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24x - 54 = x
24x = x + 54
23x = 54
x = 54/23
x = 2.3478

GuestSep 11, 2013

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Division by zero is by means not possible.

n/0 = 0 would result into n = 0*0 = 0^2. If you set n = 0: 0 = 0^2 sqrt(0) = 0. But 0/0 = (0^1)/(0^1) = 0^(1-1) = 0^0 = 1 != 0.
Same for all other integers. 5/0 = 0 5 != 0^2.

GuestSep 11, 2013

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KING OF THE CURRYMUNCHERs

GuestSep 11, 2013

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Mike:
First you should add the fractions together :

(7*9)/(8*9) + (8*8)/(9*8) -> 63/73 + 64/72 = (63+64)/72 = 127/72

factors of 72 : 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72

factors of 127 : 1, 127 (prime number)

So that is as simple as it gets.

STFU, GTFO, SMD, GG, WP, CUNTBAGGER, SWAGLORD

GuestSep 11, 2013

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Shut up Mike and let other like minded people have a go... You're hogging up all the nerd space and there aint no space left for your fat a*s
Jks, ur awesome.
$10 bucks ur asian

GuestSep 11, 2013

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doctor math:
6 and 3/4

SORRY MISREAD i mean 6 and 1/4

GuestSep 11, 2013

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GuestSep 11, 2013

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GuestSep 11, 2013

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