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20^20 prime factorized is 2^40 x 5^20. Thus, using the prime factorization trick to find the # of factors, 20^20 has 861 factors. 

10^18 prime factorszed is 2^18 x 5^18. Thus, using the prime factorization trick to find the # of factors, 10^18 has 361 factors. 

Since all factors greater than 10^18 in 20^20 are a multiple of 10^18, thus there are 861 - 361 + 1 = 501 factors that are a multiple of 10^18 when choosing one from 20^20. 

(The reason I had a "+ 1" is because one factor of 10^18 is 10^18, which is also a multiple of 10^18. Thus there are 360 factors that do not work in 20^20.)

That means, the probability of choosing a divisor from 20^20 and having it be a multiple of 10^18 is 501/861.

Simplifying, we get the probability as 167/287.

You said it XxmathguyxX.

Remember Guest, the formula for the area of a trapezoid is the combined bases x the height x 0.5.

Great step-by-step process and well detailed explanations.

Great job BuilderBoi! Keep working hard!

ok the area of a trapezoid = \(\frac{(base(1)+base(2))\times(height)}{2}\)

so \(\frac{(12+15)\times 6}{2}\)=\(27\times3=81\)

so 81

We have ourselves here a trapezoid. The area of a trapezoid is h(base1 + base2) / 2.

Plugging in the values we see the area is 6(12 + 15)/2 = 3(12 + 15) = 3(27) = 81.

Thus, the area of the trapezoid is 81 units squared. 

I think the way to go for this problem is to substitute in a few values in terms of other variables. 

I multiplied equations 1 and 2 to get \({x^2y^5z\over{xz}}=720\).

Thus, \(xy^5 = 720\).

Not quite sure what to do from here, maybe you can give it a go Guest! (try substituting in variables in terms of other variables, and try plugging it into the expression)

Good luck :D!

we were calculating different things i was calculating the arc ab and cphill was calculating the line ab

The believe that the only value not in the domain is 3, which makes the denominator 

of f(x) = 0.

I believe that the denominator of g(x) will never become 0.

ABCD is a rectangle, so that means AB is one of its sides. AB has length 4. That means the other side of the rectangle has length 32/4 = 8 units. That means the x coordinate of point C and point D has to be 8 units away from the origin. 

Note: In the problem it says x > 0 so x cannot be -8.

Thus, x = 8.

omg thank you so much!!

Oh yes! Thank you for your input everyone! But I caught a quick error! The solution ITSELF is a + bi, the solutions AREN'T a, and b. Thus, we might have to use the quadratic formula to find the solutions.

Applying the quadratic formula \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\). We get -5 +- sqrt(25 - 72) / 2. 

Thus the two solutions are \(-{5\over2}+{\sqrt{-47}\over2}\) and \(-{5\over2} - {\sqrt{-47}\over2}\). 

In this case, it doesn't matter whether b is negative or positive since we are squaring it eventually. a is 5/2. Thus, a^2 = 25/4 and b^2 = -47 or -188/4. 

That means, \(a^2 + b^2 = -{163\over4}\), which approximates to -40.75.

Don't cheat on your homework!

12% is found by multiplying by 0.12, so:  (x - 0.12x)·1.06

where  x = original price

           0.12 represents 12% of x

           x - 0.12x represents the cost after 12% has been removed.

           1.06 is used to find the final price (the "1" is used to give the original price,

                       the extra ".06" is used to give the tax, so "1.06" represents the

                       original price plus the applied tax.

The expansion of  (x + a)³  is  x³ + 3x²a + 3xa² + a³

So the expansion of x³ is 1.

This is an AO PS homework problem.  If you need help, then post on the message board.

Distribute the equation, and you get \(20+20i-25i-25i^2\)

After simplifying, you get: \(-25i^2-5i+20\)

Substitute -1 for \(i^2\): \((-25)(-1)-5i+20\)

Simplify again, and we find that the answer is: \(\color{brown}\boxed{-5i+45}\)

ax + by  =  e   --->   · d   --->     adx + bdy  =  de

cx + dy  =  f    --->   · -b   --->   -bcx - bdy  =  -bf

                                       --->     (ad - bc)x  =  (de - bf)

                                       --->                  x  =  (de - bf) / (ad - bc)

ax + by  =  e   --->   ·-c   --->     -acx -  bcy  =  -ce

cx + dy  =  f    --->   · a   --->      acx + ady  =   af

                                       --->     (ad - bc)y  =  (af - ce)

                                       --->                  x  =  (af - ce) / (ad - bc)

which match the solutions that you get when you use Cramer's formula.

If you don't know Vieta's formula:

1)  Use the quadratic formula to find the two solutions.

2)  Square each solution and add them together.

It won't be as quick, but you won't need to memorize another formula.

Q·E  +  (11 - 5i)·(3 - 2i)  =  (11)(3) + (11)(-2i) + (-5i)(3) + (-5i)(-2i)

                                      =  33 - 22i - 15i +10i²

                                      =  33 - 22i - 15i - 10

                                      =  23 - 37i

(QE)·D  =  (23 - 37i)·(2i)  =  (23)(2i) + (-37i)(2i)  =  46i + 74

-11?

Using Vieta's, a+b= -5 and ab = 18. a^2 + b^2 = (a+b)^2 - 2ab.                Reasoning: (a+b)^2 = a^2 + 2ab + b^2, so subtracting 2ab will get a^2 + b^2, our requested sum. 

So our final answer is 25 - 36 = -11.

Instead of doing the entire problem by getting each root exactly, why not try and use Vieta's formula to get ab and a + b instead? 

ab = 18 and a + b = -5. Thus, a^2 + b^2 = (a + b)^2 - 2ab. Then a^2 + b^2 = 25 - 36.

Thus, a^2 + b^2 = -11. 

Plan:  to find the equation of the line that passes through the points (3,9) and (x,7).

This line will be perpendicular to the line that passes through the points (4,2) and (3,9).

First, find the slope of the line that passes through (4,2) and (3,9).

     slope  =  (9 - 2) / (3 - 4)  =  7 / -1  =  -7

Second, from this, find the slope of the line that passes through (3,9) and (x,7).

The slope of this line will be the negative reciprocal of the previous slope:  m  =  1/7.

Third, use the point-slope form to find the equation of the line that passes through

(3,9) and (x,7) using the point (3,9):  y - 9  =  (1/7)(x - 3)

Fourth, replace the y-value of this equation with the y-value of the point that we want to find.

   7 - 9  =  (1/7)(x - 3)

Last, solve this formula for x ...

Instead of listing, why not write an equation? If we multiply all 4 equations together, we get:

\(a^2b^2c^2d^2=36^2\)

Thus abcd = 36. 

If you need any help, just think about trying to substitute variables in terms of other variables such as a = 6/c, and try doing that and plugging it back into the main equation abcd = 36. 

Hopefully this method (the normal method) will help you find ALL sets faster.

The best way is to use this method in the future, or else for more difficult problems it may be almost impossible to brainstorm every single set. Good luck...

:D