All Answers 356785 Answers

The domain of a function is the complete set of possible values of the independent variable. In plain English, this definition means: Thedomain is the set of all possible x-values which will make the function"work", and will output real y-values.

 0.7671092952

As a percent........move the decimal point two places to the right, add a "%" sign  =  76.71092952 %

As a fraction.......ignore the decimal and write all the digits after that....this will be the numerator

The denominator  will be "1"  followed by the same number of 0's as digits in the numerator, i.e., ten 0's

So we have

 7671092952                 7671092952 

___________      =       ___________  

10000000000               10000000000

This isn't as difficult as it seems

"D"   can be figured as one of three possible points

First possibility......  add A and B  and subtract C....so we have  

(-2 + 4 - 2, 3 + 2 - -1)  =  ( 0, 6)  = (x,y) =  D1

Second possibility....add B and C  and subtract A.....so we have

(4 + 2 - -2 , 2 - 1 - 3) =  ( 8, -2)  = (x,y) =  D2

Third possibility.....[ you might have already guessed it !! ]....add A and C  and subtract B

(-2 + 2 - 4, 3 - 1 - 2)  = ( -4, 0)  = (x,y) =  D3

If the coordinates have to follow a clock-wise order, then D3  will be the "correct" point  for  "D"

Here's a graph showing the possible parallelograms :

I'm doing pretty good. How about you?

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The test must contain at least 5 questions

To see why....4 questions can only be arranged  in 4!  =  4 * 3  * 2 * 1  = 24 different ways.......and this will not be enough.....no matter how she arranges the questions, some students will share the same arrangement

5 questions can be arranged  in 5!  = 5 * 4 * 3 * 2 * 1    = 120 different ways......and this will be more than enough

|x| = 9   →  l 9 l  =  9     OK

–|x| = 9 →   - l 9 l  =  -9     NO

–|–x| = 9  → - l -9 l  =  - 9   NO

–|–x| = –9   →  - l - 9 l  =  -9   OK

|x| = –9  →  l 9 l  =  9       NO

|–x| = 9  →    l -9 l   =  9   OK

|–x| = –9   →  l - 9 l  =  9   NO

Thanks, hectictar.......here's another possible approach done by completing the square

Let's write "y"  for  p(r).......we'll come back to this later

y  = 2r^2 + 2r -1        add 1 to both sides

y + 1   =  2r^2 + 2r

y + 1  =  2 [ r^2 + r]     divide both sides by 2

[ y + 1] / 2    =  r^2 + r        completet the square on r

[ y + 1 ] / 2  + 1/4  +     =    r^2 +  r  + 1/4     simplify

[ 2y + 2 + 1 ] / 4   =   ( r + 1/2)^2

[ 2y + 3 ]  / 4   =   ( r + 1/2)^2       take both roots

±  √ [2y + 3]  / 2  =   r  + 1/2

±  √ [2y + 3]  / 2   - 1/2   =  r

± ( √ [2y + 3]    - 1  )  / 2   = r          "exchange"  r and y

± ( √ [2r + 3]    - 1  )  / 2   =  y

For "y"  write  p^-1(r)

p^-1 (r)  =   ± ( √ [2r + 3]    - 1  )  / 2 

z = 6 π x y      for  x

divide both sides by  6 π y

z / [ 6 π  y ]    = x

a(b − c)= d     for c  

divide both sides by a

b - c   = d / a    

Multiply both sides by  - 1

c - b  =  - d / a   

add b to both sides

c  = b -  [d / a]

d = 5x + 5y     for y

subtract 5x from both sides

d - 5x  = 5y

dibvide both sides by 5

[ d - 5x ] / 5    = y

A=h(a+b)/2    for b

multiply both sides by 2

2a  = h (a + b)

divide both sides by h

[2a ] / h   =  a  + b 

subtract a from both sides

[ 2a ] / h   - a   =  b

P=2(l+w)    for  l

divide both sides by 2

P / 2   =  l + w

subtract w from both sides

[P / 2]  - w    =    l

∛ [ -64x^6y^9  ]  =    -4x^2y^3     [assuming x, y are positive ]

f(x)  =  [ x + 1 ] / [ x - 1]

Look at the graph :

Domain .....all values except any x that makes the denominator  = 0....thus, x = 1 isn't in the domain

Range  .....we  have the same power polynomials in the numerator as in the denominator.....thus, we will have a horizontal asymptote  given by the ratio of  the coefficient on the highest power of x's in both numerator and denominator.....the highest powers  of x have the ratio of coefficients =   1 / 1   = 1  ......thus....we will have a horizontal asymptote at  y  = 1

So   the range   is  ( -infinity , 1 ) U ( 1 , infinity)

Restrictions  - you are correct

Interval of increase   -you are correct

Intervals of decrease......  the function decreases from (-infinity, 1)  and from (1, infinity)

Roots..... - you are correct

Y intercept - you are correct

Vertices  - since we have no minimums or maximums on this graph......there are no vertexes

Thank you so much, that makes a lot of sense. For the final answer, I was wondering what exactly it would be myself. Since the original question is p(r), it only makes sense that I write it as p^-1 (r).  

Ya... What are you wanting to be solved??

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Good desc on how to make eggs.... But the main question is WHY ARE YOU ASKING HOW TO MAKE EGGS ON A MATH SITE!!!!!

So if you look at the question x=9. All of it is positive. By this information you are looking for the possible ones that are positive. Another usage we can use is take out the slashes to read them normally.

So the first one is x = 9 so that's a keep

-x = 9 has a negative x so not it

If you remember the rules of positive and negatives, 2 negatives make a positive so it will read as x = 9. That's a keep

For the fourth, it has 3 negatives. The first to make the x positive but because there's not another negative it reads as x = -9. Not a keep

The fifth isn't it because of the -9

Not the sixth because of the -x

I'm not sure of the 7th because of the negatives can cross out to make them both possitive but at the same time it can read as -x= -9 on it's own.

So for your answer it is:

|x| = 9

–|–x| = 9

And maybe |–x| = –9 but I'd make sure

I followed your work..I think you've got it right so far!

(Except  c = (-1 - y) , but you had it right in the problem.)   
 

You are very very close to the answer. Maybe this will nudge you in the right direction.

\(x=\frac{-2\,\pm\,2\sqrt{2y\,+\,3}}{4} \\~\\ x=\frac{2(-1\,\pm\,\sqrt{2y\,+\,3})}{2(2)}\)    Factor a 2 out of the numerator and factor a 2 out of the denominator.

Below is the rest of the answer if you need more help.

----------

\(x=\frac{2(-1\,\pm\,\sqrt{2y\,+\,3})}{2(2)}\)

                                                Divide the numerator and denominator by  2  .

\(x=\frac{-1\,\pm\,\sqrt{2y\,+\,3}}{2}\)

                                                Rearrange the terms in the numerator.

\(x=\frac{\pm\,\sqrt{2y\,+\,3}\,-\,1}{2} \\~\\ \begin{array}\ x=\frac{\sqrt{2y\,+\,3}\,-\,1}{2} \qquad\text{ or }\qquad &x=\frac{-\,\sqrt{2y\,+\,3}\,-\,1}{2}& \\~\\ & x=\frac{-1(\,\sqrt{2y\,+\,3}\,+\,1)}{2}&\\~\\ & x=-\,\frac{\,\sqrt{2y\,+\,3}\,+\,1}{2}& \end{array}\)

Also...for your final answer, do you need to say  p^-1(r)  for  x  and  r  for  y  ?

Google it !!!.

Simplify the following:
2 sqrt(54) - 4 sqrt(24)

Hint: | Simplify radicals.
sqrt(54) = sqrt(2×3^3) = 3 sqrt(2) sqrt(3):
2×3 sqrt(2) sqrt(3) - 4 sqrt(24)

Hint: | For a>=0, sqrt(a) sqrt(b) = sqrt(a b). Apply this to sqrt(2) sqrt(3).
sqrt(2) sqrt(3) = sqrt(2×3):
2×3 sqrt(2×3) - 4 sqrt(24)

Hint: | Multiply 2 and 3 together.
2×3 = 6:
2×3 sqrt(6 ) - 4 sqrt(24)

Hint: | Simplify radicals.
sqrt(24) = sqrt(2^3×3) = 2 sqrt(2) sqrt(3):
2×3 sqrt(6) - 42 sqrt(2) sqrt(3)

Hint: | For a>=0, sqrt(a) sqrt(b) = sqrt(a b). Apply this to sqrt(2) sqrt(3).
sqrt(2) sqrt(3) = sqrt(2×3):
2×3 sqrt(6) - 4×2 sqrt(2×3)

Hint: | Multiply 2 and 3 together.
2×3 = 6:
2×3 sqrt(6) - 4×2 sqrt(6 ) 

Hint: | Multiply 2 and 3 together.
2×3 = 6:
6 sqrt(6) - 4×2 sqrt(6)

Hint: | Multiply -4 and 2 together.
-4×2 = -8:
6 sqrt(6) + -8 sqrt(6)

Hint: | Combine like terms in 6 sqrt(6) - 8 sqrt(6).
6 sqrt(6) - 8 sqrt(6) = -2 sqrt(6): 
 = -2 sqrt(6)

I think you must have 5 questions.

1!  =  1  way to arrange 1 question

2!  =  2  ways to arrange 2 questions.

3!  =  6  ways to arrange 3 questions.

4!  =  24 ways to arrange 4 questions.

5!  =  120 ways to arrange 5 questions.

120 is more than 27. So there must be at least 5 questions on the quiz.

They could go in this order:

1, 2, 3, 4 ,5

1, 2, 3, 5, 4

1, 2, 4, 3, 5

1, 2, 4, 5, 3

1, 2, 5, 3, 4

1, 2, 5, 4, 3  ...etc.

put butter in the pan, crack the eggs in the pan, cook until no longer slimy, then add salt and pepper if you want....... never heard a dumber question, lol

5) 

(1/4) / (1/12) = x / 1 cross multiply

1/4 = 1/12 x                divide both sides by 1/12

x =1/4 / 1/12

x =1/4 x 12/1

x =12/4

x = 3 ounces of nuts.   

6 - do exactly the same as in 5 above and you should get 3 ounces of butter.

7- do exactly the same here. You should get 1 2/7 liters of water.

n****r

Suppose the questions were numbered from: 1, 2, 3, 4, 5.........etc. In order for each of the 27 students to get the different order, or beginning with a different number, you would have to have a minimum of 5 questions in the puzzle. Anything less would have to be duplicated.