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This one is easier than it looks

Any "x" that makes a denominator 0  will not be in the domain

So

x^2 - 7  = 0

x^2  = 7        take both square roots

x = ±√7   but.....notice that the sum of these = 0 

x^3 - 8  = 0      the only real number that makes this = 0  is 2

x^4 - 9  = 0

x^4  = 9       take both 4th roots

x =  ± ⁴√9     and, like before, the sum of these two roots  = 0

So.....the sum of all the numbers not in the domain  is just 2

From looking at this, we can see that

CD  =  14 - (2 + 5)

CD  =  7

And...

AC  =  2 + 5  =  7

BD  =  5 + 7  =  12

So, the ratio of AC to BD   =   \(\frac{AC}{BD}\)   =   \(\frac{7}{12}\)

THANKYOU!!!

Very impressive, hectictar.....!!!!

Forget that college stuff....you're already too smart for your own good....LOL!!!!!

This is an old one known as the "locker problem".....let's take a more simple example that will lead us to where we want to go

Let's take an example with just 10 students...o = open, c = closed

Student 1     o o o o o o o o o o 

Student 2     o c o c o c  o c o c 

Student 3     o c c c o o  o c c c

Student 4     o c c o o o o o c c

Student 5     o c c o c o o o c o  

Student 6     o c c o c c o o c o 

Student 7     o c c o c c c o c o     

Student 8     o c c o c c c c  c o 

Student 9      o c c o c c c c o o 

Student 10    o c c o c c c c o c 

Do you notice which lockers remain open??  ...  1, 4 and 9....

In other words......the ones whose numbers are perfect squares

So....the lockers that remain open are just the ones that are perfect squares from 1 - 1000

And these are  {1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961}

You might be curious as to why this happens....

Notice that  non-sqares aways have an even number of divisors [ including themselves]

So that each "non-square" locker is visited an even number of times

For example....locker 6   is visited by the first student who opens it, by the second  student who closes it, by the third student who opens it and then by the sixth student who closes it and it is never touched again

But consider locker 9...it's opened by the first student, closed by the third and opened by the ninth and never touched again....!!!!...in other words, it is visited an odd number of times....the same number as the number of its proper and improper divisors....!!!! 

Let's find the two points on the parabola that touch the line y = 5 .

y  =  x² + 3x + 2             We want to find the  x  values when  y  is equal to  5  .

5  =  x² + 3x + 2             Subtract  5  from both sides of this equation.

0  =  x² + 3x - 3              Use the quadratic formula to find  x  .

x  =  \({-3 \pm \sqrt{21} \over 2}\)

There are 15 odds from 1-30 inclusive

And there are 15 evens from 1-30 inclusive

It might be easiest to choose the last number, first

We have 5 odds that are multiples of 3 from 1-30 inclusive

Then we have 15 ways to choose the second even number 

Finally, we have 14 ways to choose the first odd [ we have already chosen one of them as the last number ]

And we have 5 evens that are multiples of 3 from 1-30 inclusive

Then we have 15 ways to choose the second odd number 

Finally, we have 14 ways to choose the first even [ we have already chosen one of them as the last number ]

So we have these two possibilities

( 14 odds ) ( 15 evens) (5 odds that are  multiples of 3) =  750 combinations

(15 odds) (14 evens)  ( 5 evens that are multiples of 3) = 750 combinations

So there are 750 + 750  = 1500  possible combinations

The numerator will be defined for all x

If the discriminant  of   x^2 - x + c  is < 0, there will be no real x's that make the denominator 0

So...we have that

1 - 4c < 0

1 < 4c

1/4 < c

So....c > 1/4  and the smallest positive integer  where c > 1/4  is when c  =1

There are 4 of them:

This isn't as difficult as it seems

8^(-9)  = 8^(-6) + ?       can be written as

1 / 8^9  =   1/8^6  + ?      subtract   1/8^6  from both sides

1 / 8^9  -  1 / 8^6  =  ?      getting a common demoninator on the right we have

[ 1 -  8^3 ] / 8^9  =  ? 

-511/ 134,217,728 = ?   

The discriminant must be greater than zero, so:

15^2 - 4*8*c > 0

225 - 32c > 0

This will be the case for c = 1, 2, 3, 4, 5, 6 and 7

The product is 7! → 5040

Here's another approach

If a line intersects a parabola at only one point, it is tangent to the parabola at that point.

So....the slopes  are equal at that point

The slope of the parabola at any point is   2x + 2

The slope of the line is a constant, 6

Equate the slopes

2x + 2  = 6

2x  = 4

x = 2

So.....this is the value where the slopes are the same.....subbing this into  both functions gives us

(2)^2 + 2(2) + 7  = 6(2) + b

15  =  12 + b

b = 3

It's obvious that 0 and -1 are two of the numbers

Let's simplify this

           1                               1                     1                    1                x + 1

---------------------    =   ----------------  =     ----------  =    -----------  =    -----------   

1  +      1                    1  +   1                 1  +  x           2x + 1            2x + 1

         -----------                  ---------                 ------        --------    

          1  +   1                    x + 1                   x + 1        x + 1

                  ----                  ------

                   x                      x

And  -1//2 makes the denominator of this fraction 0

So..... the sum of the three numbers not in the domain is :

0 + (- 1) + ( - 1/2 )   =  -3 / 2

\(\sum_{k=1}^{50}k^k=\)

8948298532620477012317180002541949344142954252890052803...

627765484931151082399454417125

(The dots just mean continued on the next line.)

Thanks, Alan....this was a good one to learn from....I've added it to my "Watchlist"..!!!!

240 (1/4)  =  240 / 4  =  60

300 / 50000  =   3 / 500  = 6/1000 =  .006  = .6%  are defective

If f(0)=0, f(1)=2, f(2)=5 and f(x)=ax^2+bx+c, what is the value of f(3)

If f(0) = 0, then c  = 0

So we have these two equations  for f(1)  and f(2)

a(1)^2  + b(1)  = 2   →   a + b   = 2       (1)

a(2)^2  + b(2)  = 5   →  4a + 2b = 5      (2)

Multiply (1) by - 2   →  -2a - 2b = -4        add this to (2)

2a  = 1     →   a = 1/2

And subbing this into (1) to find b

1/2 + b  = 2      →   b = 3/2

So our function is

f(x)  =  (1/2)x^2  + (3/2)x

So   f(3)  =   (1/2) (3)^2  + (3/2) (3)  =   9/2  + 9/2  =  18/2  = 9

We have \((1+kX)^n\rightarrow 1+nkX+\frac{n(n-1)}{2}(kX)^2+...\)

So \(nk=-4\\\frac{n(n-1)}{2}k^2=\frac{15}{2} \)

You now have two equations for two unknowns. Can you take it from here?

The straight line must be tangent to the parabola.

Equate the two functions:

\(x^2+2x+7=6x+b\\x^2-4x+7-b=0\\\)

The discriminant must be zero if there is to be a single solution, so

\((-4)^2-4(7-b)=0\\-12+4b=0\\b=3\)

I would assume that only a straight verticle line can intercept with a parabola.

Look at a parabola. It's grows expotentially on two ends. If there is a maximum, then the growth is "downwards" (negative in the y axis), never ending meaning there is no minimum value.. If there is a minimum, then the growth is "upwards"(positive in the x axis), never ending meaning there is no maximum value.

Imagine the maximum and minimum values as the start of the road. If the road keep on growing without stopping, will there be an end?

31.52 what? We need the unit. Is it 31.52 hour? Second? Minutes? Also, what is time? We need the unit from and to.

If it's 31.52 hours to minute. Then you 31.52 * 60 = 1891.2 Minutes

If it's 31.52 minutes to hour. Then you 31.52 / 60 = 0.5253333.... hour

If your time is 00:00 and you just did 31.52 hours of something. Then the time would be:

(We want a number ranging between 0-24) so we do 31.52 - 24 = 7.52 This tells us that 24 hours has passed (ignore cause it brings us back to 00:00) and another 7 hours has passed and 0.52 hour. Then we find the minute: 0.52 * 60 = 31.2 minutes. So the final answer, the time, is 7:31 AM.

I don't know what you want.