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10^10^100=X

X=?

X = (10¹⁰)¹⁰⁰ = 10¹⁰⁰⁰ = 10 * 10⁹⁹⁹

GB and USA:

X = 10 Trecentduotrigintillion

rest of the world

X = 10 Centsexsexagintilliarde

  !

Great to see you back great to hear that but but next morning you need to go back to school again and again

a)  Substitute 9/2 for z in the equation:

2*(9/2)² + k*9/2 - 30 = 6

81/2 + 9k/2 = 36

81 + 9k = 72

9k = -9

k = -1

b) Use this value of k in the original equation:

2z² - z -30 = 6

2z² - z -36 = 0

z² - z/2 -18 = 0

(z + 4)(z - 9/2) = 0

The other solution is therefore z = -4

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From   x=(7y^-1+3)/(5y^-1+7) multiply both sides by (5y^-1+7) to get;

x(5y^-1+7) = 7y^-1+3   Now distribute the x in to the parentheses on the left hand side:

5xy^-1 + 7x = 7y^-1+3      Subtract  5xy^-1  from both sides:

7x = 7y^-1+3 - 5xy^-1    Subtract 3 from both sides:

7x - 3 = 7y^-1 - 5xy^-1   Factor the right hand side:

7x - 3 = (7 -  5x)y^-1   Divide both sides by 7 - 5x:

(7x - 3)/(7 - 5x) = y^-1 

or  y^-1 = (7x - 3)/(7 - 5x)

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(Note that the original function is not defined at x = -7/5, so there is no inverse at this point; and the inverse function isn't defined at x = +7/5).

Find the last two digits of the following sum:

5! + 10! + 15! +... + 100!

20!=2*5*10*other numbers = 100*other numbers      ends in  00

All the factorials bigger than 20! also end in  00  

So add these all up and we get a number that ends in   00

So what is left?

5!+10!+15!  

5! = 2*3*4*5 = 10*12 = 120  the last two digits are 20

10!= 120*6*7*8*9*10 = 6*7*8*9*1200  the last two digits are 00

It follows that the last 2 digits of 15! must also be   00

So the only one that does not end in 00 is   5! which ends in  20

So add them all together and the last two digits must be 20

Just like our guest already said :)

Yup, of course

sum_(n=1)^20 (5 n)! = 93326215454274131171424014083629874498929842578166214433940153388687946517415655599163396465910182277066190137342795918966177385078001619556987788165518636920. This is because of 5! which is =120. The rest of the terms all end in at least 2 zeros.

I've interpreted your curves as y = 58x^2 and y = 138 - x^2  below (though your description isn't totally clear!):

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five times a​ number, subtracted from eighteen​, is triple the number.

let the number be x

Five times the number is 5x

subtract this from 18 and you get              18-5x

This is equal to tripple the number  tripple x is  3x

so

18-5x=3x

add 5x to both sides

18 = 8x

divide both sides by 8

18/8 = x

x =  9/4

x = 2.25

\(59={17}^{x}\)

\({log}_{17}59=x\)

\(x=\frac{log59}{log17}\)

\(x≈1.4391918111143888\)

The last two zero's?

A = π r^2

A is area

π is pi

r is the radius of the circle

Solving an inequality is similar to solving a regular equation; you are isolating the variable, in this case K.

The solutions, if you are wondering, are values for K that satisfies its current restriction (K is greater than or equal to -15). Examples of solutions are \(-14,-13,-12.5,0,\frac{2}{3},1.\overline{78},\pi, \sqrt{21}\). 

I would like to think it is impossible to assign a numerical value for how "helpful" a user is. All the users listed do not answer questions halfheartedly; we are all taking great care in providing answers every day. 

I have updated the graph such that the three vertices of the triangle lie on the circle. 

I have done this to demonstrate what the circumcenter is a tad clearer (if it wasn't already!). As I said before, the circumcenter is the center of a circle where those 3 vertices of a triangle lie on the circle.

The more you know... On the other hand, I'm close to reaching that 1000 ♥ benchmark. Can't wait!

I didn't know that about the circumcenter  

Also...thanks for the detailed graph...it's very helpful !  

Nice job!!!

We  want to use these two identities

cos (x)  = cos (-x)    and      csc (-x)  = -csc (x)

So

cos (-105°)  =  cos (105°)  = -0.26

And

csc ( -105°)   =  -csc (105°) = -1.03  .....so..... - [- csc (105°) ] = csc (105°) =  - [ -1.03]  =  1.03

If f(x) = 4x^2 - x+1, then f(a+1) equals

We are just  substituting (a + 1)  for x.....so we have

4(a + 1)^2 - (a + 1) + 1       simplify

4 ( a^2 + 2a + 1) - a - 1 + 1

4a^2 + 8a + 4  - a - 1 + 1

4a^2 + 7a + 4

15  / 60    = 1 /4

Let x be the percentage.......and we have that......

1 /4   =   x  /100            multiply both sides by 100

100 / 4  =  x   =   25   (percent)

Triangle PRS is equilateral.....so angle PSR  = 60° = angle PSQ 

And PS  = 2   and  QS  = 4

And by the Law of Cosines we have that

PQ  =  sqrt ( 2*2 + 4^2  - 2(4)(2) cos 60 )  =  sqrt ( 20 - 8)  = sqrt (12) = 2sqrt(3)

And by the Law of Sines, we have

sin PSQ / PQ  =  sin PQS / PS

sin 60 / 2sqrt (3)   = sin PQS / 2

(sqrt (3) / 2) / [ 2sqrt (3) ]  =  sin PQS / 2

(1/4)  =  sin PQS / 2

sin PQS  =  1/2  ...    thus PQS  = 30°

So..since PS is the shortest side in triangle PQS, then angle PQS is the smallest angle   = 30°

\(ln(\sqrt e)=\frac{1}{2}\)

ln is just log base "e".

\(log_{e}(\sqrt e)=\frac{1}{2}\)

To convert this expression to an exponent, we take the log base, put it to the power of the otherside of the equation, and then put that equal to what was in the logs brackets.

\(e^{\frac{1}{2}}=\sqrt{e}\)

-16=-26,2t-1/2x9,8t^2

9,8t-32/t=-54,4

\(-16= -26,2t -\frac{1}{2}\times 9,8t^2\\ \frac{1}{2}\times9,8t^2+26,2t-16=0\\ 4,9t^2+26,2t-16=0\)

  a               b            c

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(x = {-26,2 \pm \sqrt{686,44-4\cdot 4,9\cdot (-16)} \over 2\cdot 4,9}\)

\(\color{black}x = \frac{-26,2\pm\sqrt{1000,04}}{9,8}\\ x_1=0,5534\\ x_2=-5,9\)

\(9,8t-\frac{32}{t}=-54,4\)

\(9,8t^2-32=-54,4t\\ 9,8t^2+54,4t-32=0\)

   a              b            c

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(x = {-54,4 \pm \sqrt{2959,36-4\cdot 9,8\cdot (-32)} \over 2\cdot 9,8}\)

\(\color{black}x = \frac{-54,4\pm\sqrt{4213,76}}{19,6}\\ x_1=0,5364\\ x_2=-6,0874\)

  !

Let me try and allow you to visualize this method. I made a square with 2 numbers in it!

We must find 2 numbers that multiply to 23 and add to get 24. What could that be? Well, if you think about it for some time, you'll notice that this is the solution. I have created a box for that, too:

Since the coefficient of the quadratic term is 1, we can factor it immediately as \((p+23)(p+1)\)