All Answers 356032 Answers

3/12=5/x

Multiply each denominator by the the numerator of the other fraction.

12*5=60

3*x=3x

Set them equal to each other.

60=3x

Divide by 3 to leave only x.

x=20

Find the common denominator: 4

Set both denominators to that: 2/4 +1/4

Now, like regular addition, 2+1=3

3/4

Gena wants to estimate the quotient of –21.87 divided by 4.79. Which expression shows the best expression to estimate the quotient using front-end estimation?

-21.87 is around -22

4.79 is around 5

-22/5=-4.4

m_f(x)=1/m_g(x)

g(0)=0

f(0)=3

203.14

A(x₁, y₁) B(x₂, y₂)

M_AB=([x₁+x₂]/2, [y₁+y₂]/2)

=([-3+2]/2, [5+4]/2)=D(-1/2, 9/2)

d_za=sqrt([x₂-x₁]²+[y₂-y₁]²)

d_CD=sqrt([3/2]²+[17/2]²)=sqrt(9/4+289/2)=sqrt(298/4)=sqrt(298)/2

M_BC=E(3/2,0)

d_AE=sqrt([-3/2]²+5²)=sqrt(109/4)=sqrt(109)/2

M_AC=F(-1, 1/2)

d_BF=sqrt(3²+[7/2]²)=sqrt([36+49]/4)=sqrt(85)/2

I think it almost always pays to plot a graph to get a handle on these sorts of questions.

See my latest comment under your original post.

This is why I said the trick is often in choosing the right terms for u and v.  If you choose the wrong ones you can easily make things worse, as you have demonstrated.  When this happens it's generally a good idea to go back and choose differently!

That said, there are occasions when the process needs to be repeated.  However, my advice is that, if you get something that is even more complicated than you started with, go back and make a different choice for u and v.

2^27 + 3^27   =  7625731702715  

This number is not divisible by 2

It is also not divisible by 3 because the sum of its digits is not divisible by 3

It is divisible by 5     because   it ends in  a "5" 

To test whether it is divisible by 7, double the last interger  and subtract the remaining part.....if the result is divisible by 7, then the number itself is divisible by 7....so we have.....

[2* 5 -  762573170271] / 7  = -108939024323

So.......5   and 7 are the smallest distinct prime divisors  of  7625731702715  

And their sum is 12

1 day x 24 hours x 60 minutes x 60 seconds =86,400 seconds in 1 day.

It means that you have to take that operation inside the parenthesis FIRST.

There are thousands of formulas to calculate Pi. Here is one that has been used for about 300 years:

Pi/4 =4 arctan(1/5) - arctan(1/239). Here is another pretty one found by Euler:

(Pi^2)/6 = 1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 + 1/5^2...........to infinity.

(a) Factor 3xy-4x^2-36y+48x.

(x  - 12)    (3y - 4x )

HERE IS A VERY LONG AND DETAILED SOLUTION. GO THROUGH IT LINE BY LINE SLOWLY AND CAREFULLY. GOOD LUCK:

Solve for x over the real numbers:
abs(x-9) = abs(x-5)+abs(x-1)

Split the equation into two possible cases:
x-9 = abs(x-5)+abs(x-1) or x-9 = -abs(x-5)-abs(x-1)

Subtract abs(x-5)+abs(x-1) from both sides:
-9+x-abs(x-5)-abs(x-1) = 0 or x-9 = -abs(x-5)-abs(x-1)

Subtract -9+x-abs(x-1) from both sides:
-abs(x-5) = 9-x+abs(x-1) or x-9 = -abs(x-5)-abs(x-1)

Multiply both sides by -1:
abs(x-5) = -9+x-abs(x-1) or x-9 = -abs(x-5)-abs(x-1)

Split the equation into two possible cases:
x-5 = -9+x-abs(x-1) or x-5 = 9-x+abs(x-1) or x-9 = -abs(x-5)-abs(x-1)

Subtract -9+x-abs(x-1) from both sides:
4+abs(x-1) = 0 or x-5 = 9-x+abs(x-1) or x-9 = -abs(x-5)-abs(x-1)

Subtract 4 from both sides:
abs(x-1) = -4 or x-5 = 9-x+abs(x-1) or x-9 = -abs(x-5)-abs(x-1)

By definition, absolute value is positive. Therefore abs(x-1) = -4 has no solution:
x-5 = 9-x+abs(x-1) or x-9 = -abs(x-5)-abs(x-1)

Subtract 9-x+abs(x-1) from both sides:
-14+2 x-abs(x-1) = 0 or x-9 = -abs(x-5)-abs(x-1)

Subtract 2 x-14 from both sides:
-abs(x-1) = 14-2 x or x-9 = -abs(x-5)-abs(x-1)

Multiply both sides by -1:
abs(x-1) = 2 x-14 or x-9 = -abs(x-5)-abs(x-1)

Split the equation into two possible cases:
x-1 = 2 x-14 or x-1 = 14-2 x or x-9 = -abs(x-5)-abs(x-1)

Subtract 2 x-1 from both sides:
-x = -13 or x-1 = 14-2 x or x-9 = -abs(x-5)-abs(x-1)

Multiply both sides by -1:
x = 13 or x-1 = 14-2 x or x-9 = -abs(x-5)-abs(x-1)

Add 2 x+1 to both sides:
x = 13 or 3 x = 15 or x-9 = -abs(x-5)-abs(x-1)

Divide both sides by 3:
x = 13 or x = 5 or x-9 = -abs(x-5)-abs(x-1)

Add abs(x-5)+abs(x-1) to both sides:
x = 13 or x = 5 or -9+x+abs(x-5)+abs(x-1) = 0

Subtract -9+x+abs(x-1) from both sides:
x = 13 or x = 5 or abs(x-5) = 9-x-abs(x-1)

Split the equation into two possible cases:
x = 13 or x = 5 or x-5 = 9-x-abs(x-1) or x-5 = -9+x+abs(x-1)

Subtract 9-x-abs(x-1) from both sides:
x = 13 or x = 5 or -14+2 x+abs(x-1) = 0 or x-5 = -9+x+abs(x-1)

Subtract 2 x-14 from both sides:
x = 13 or x = 5 or abs(x-1) = 14-2 x or x-5 = -9+x+abs(x-1)

Split the equation into two possible cases:
x = 13 or x = 5 or x-1 = 14-2 x or x-1 = 2 x-14 or x-5 = -9+x+abs(x-1)

Add 2 x+1 to both sides:
x = 13 or x = 5 or 3 x = 15 or x-1 = 2 x-14 or x-5 = -9+x+abs(x-1)

Divide both sides by 3:
x = 13 or x = 5 or x = 5 or x-1 = 2 x-14 or x-5 = -9+x+abs(x-1)

Subtract 2 x-1 from both sides:
x = 13 or x = 5 or x = 5 or -x = -13 or x-5 = -9+x+abs(x-1)

Multiply both sides by -1:
x = 13 or x = 5 or x = 5 or x = 13 or x-5 = -9+x+abs(x-1)

Subtract -9+x+abs(x-1) from both sides:
x = 13 or x = 5 or x = 5 or x = 13 or 4-abs(x-1) = 0

Subtract 4 from both sides:
x = 13 or x = 5 or x = 5 or x = 13 or -abs(x-1) = -4

Multiply both sides by -1:
x = 13 or x = 5 or x = 5 or x = 13 or abs(x-1) = 4

Split the equation into two possible cases:
x = 13 or x = 5 or x = 5 or x = 13 or x-1 = 4 or x-1 = -4

Add 1 to both sides:
x = 13 or x = 5 or x = 5 or x = 13 or x = 5 or x-1 = -4

Add 1 to both sides:
x = 13 or x = 5 or x = 5 or x = 13 or x = 5 or x = -3

abs(x-9) ⇒ abs(-9-3) = 12
abs(x-5)+abs(x-1) ⇒ abs(-5-3)+abs(-1-3) = 12:
So this solution is correct

abs(x-9) ⇒ abs(5-9) = 4
abs(x-5)+abs(x-1) ⇒ abs(5-5)+abs(5-1) = 4:
So this solution is correct

abs(x-9) ⇒ abs(13-9) = 4
abs(x-5)+abs(x-1) ⇒ abs(13-5)+abs(13-1) = 20:
So this solution is incorrect

The solutions are:
Answer: | x = 5  or   x = -3

Find the maximum p such that 2x⁴y² + 9y⁴z² + 12z⁴x² - px²y²z²>0 for all real x,y,z .

2x⁴y²+9y⁴z²+12z⁴x²-px²y²z²=0

p(x²y²z²)=x²y²(2x²)+y²z²(9y²)+x²z²(12z²)

p=(2x²/z²)+(9y²/x²)+(12z²/y²), which is the maximum, for all real x,y,z

Yep.....I goofed this one ......40 lashes with a wet spaghetti noodle for me.........

This proves that CDD sometimes triumphs over reading comprehension........

 |x-1| + |x-5| = |x-9|         we can write this as

√[(x -1)^2] + √ [(x - 5)^2]   = √ [(x - 9)^2 ]         square both sides

(x - 1)^2 + 2√[ (x -5)^2 * (x -1)^2] + (x -5)^2   =  (x - 9)^2           simplify

x^2 - 2x + 1 + 2√[ (x -5)^2 * (x -1)^2]  + x^2 - 10x + 25   =  x^2 - 18x + 81

 2√[ (x -5)^2 * (x -1)^2]  =   -x^2 - 6x + 55

 2√[ (x -5)^2 * (x -1)^2]  =   - [x^2 + 6x - 55]   factor the right side

 2√[ (x -5)^2 * (x -1)^2] =  - (x + 11)(x - 5)        square both sides

4 (x - 5)^2 * (x - 1)^2   = (x + 11)^2 (x - 5)^2

4(x - 5)^2 * (x - 1)^2  - (x - 5)^2 (x + 11)^2   = 0         factor out (x - 5)^2

(x - 5)^2   [ 4(x - 1)^2 - (x + 11)^2 ]   =  0        simplify

(x - 5)^2 [ 4x^2 - 8x + 4  - x^2 - 22x - 121]   = 0

(x - 5)^2 [ 3x^2 - 30x - 117]  = 0

(x - 5)^2 [ x^2 - 10x - 39] * 3  = 0        divide through by 3   and factor

( x - 5)^2 [ (x - 13) ( x + 3) ]  =  0

So.........setting each factor to 0 we have that    x = 5  or x = 13   or x = -3

x = 13   is an extraneous solution produced by squaring both sides........reject it

The other two solutions x = 5   or x = -3      are good

Wow, you are all wrong.
total seats = 465 + 155 = 620
465 = empty
percentage of empty = 465 / 620 = 75%

My error was that I didn't change -4x^2 sign. Sorry, that was a no brainer... meh

Huge thanks!

See if this helps, Tony

                x^4 + 2x^3 + 2x^2 + 9x  + 10

x  - 2    [   x^5 + 0x^4 - 2x^ 3 + 5x^2 - 8x - 20 ]

                x^5  - 2x^4

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

                         2x^4  - 2x^3

                         2x^4 -  4x^3

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

                                    2x^3  + 5x^2

                                    2x^3  - 4x^2

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

                                               9x^2 -  8x

                                               9x^2 -18x

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

                                                        10x  - 20

                                                        10x -  20

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

                                                                0

               x^3 + 2x  + 5

x + 2   [   x^4 + 2x^3 + 2x^2 + 9x  + 10 ]

               x^4 + 2x^3

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

                                   2x^2 + 9x

                                  2x^2  + 4x

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

                                              5x  + 10  

                                              5x +  10

                                              ________

                                                      0

The remaining polynomial, x^3 + 2x + 5      doesn't have any rational roots........just one "real" root and 2 complex roots

That's what I just realized haha!

Well, I start with: 

      x^5+0x^4-2x^3+5x^2-8x-20             / x-2

    -(x^5-2x^4)                                         x^4

=           2x^4-2x^3+5x^2-8x-20

           -(2x^4-4x^3)                                2x^3

=                    2x^3+5x^2-8x-20

                    -(2x^3-4x^2)                       2x^2

=                               x^2-8x-20 

                               -(x^2-2x)                   x

=                                     -6x-20

So from here if I keep going I'll be left with -32... Can't seem to find my error...

CPhill: Can you please explain this result using Wolfram/Alpha? Where am I going wrong?

∏ [ 2^(2n) + 1], n=0 to 1023

=product_(n=0)^1023(1+2^(2 n))≈1.016900708110399312582356305677699335401×10^315,345

If you have two rational roots, you shouldn't end up with any remainder......after dividing by (x  - 2)   and (x + 2), you should have either another polynomial [or 0]  as a remainder

Can you give us an example of what you have???