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x⁶ + 2x³ + 1

Since this is a perfect square, it can be factored as:

(x³ + 1)(x³ + 1)

Each of these factors is a perfect cube, so it can be factored as:

(x + 1)(x² - x + 1)(x + 1)(x² - x + 1)

This can be rewritten as:  (x + 1)²((x² - x + 1)²

$\sqrt{25}=5$

Since we only need to calculate the integers, the approximation for $\sqrt{140}$ does not need to be too accurate. I know that $12^2=144$ and $11^2=121$, so $\sqrt{140}$ can be rounded down to 11.

There is a formula that allows one to calculate the running total of consecutive integers from one to the last integer. It is $\frac{n(n+1)}{2}$, where n is the largest number in the set, in this case 11. However, we have a bit of an issue. 

This formula only works when the consecutive integers start at 1. In this case, the integers start at 5. Therefore, we must subtract the sum of the integers from 1 to 4. We can use the same formula of $\frac{n(n+1)}{2}$, where n=4.

$\frac{11(11+1)}{2}-\frac{4(4+1)}{2}$

Find the result of this.

2 1/4 + 3 1/2 x 1/3

Note  2 + 1/4  =  9/4

And  3 + 1/2 =  7/2

So we have

(9/4)  (7/2) (1/3)  =

(9/3) (7/2) (1/4)  =

(3) (7/8)  =

21/8    or   2 + 5/8

$\[f(x) = \left\{ \begin{array}{cl} 3-x & \text{ if } x \leq 3, \\ -x^3+2x^2+3x & \text{ if } x>3. \end{array} \right.\]Calculate $f^{-1}(0)+f^{-1}(6)$.$

f^-1 (0)    must mean   that either   3 - x  = 0       or   -x^3 + 2x^2  + 3x  = 0

However......in the second function x = 0    is the only thing that makes this true, but that function isn't defined for x  = 0

So.....it must mean that  3 - x  = 0....and x  = 3......so   f^-1 (0)   = 3

And

f^-1 (6)  must mean that either   3 - x  = 6      or that    -x^3 + 2x^2  + 3x  = 6

Looking at the second function  we have that

-x^3 +2x^2  + 3x  = 6  →    x^3 -2x^2 - 3x + 6  =  0  →   x^2( x - 2) - 3(x - 2)  = 0  →

(x^2 - 3) (x -2)  = 0  →    x  = ±√3   or x  = 2   are the x values that make this true....but....these values are all < 3  and this function isn't defined for these values

But    3 - x   = 6     when  x  = -3     which is valid

So

f^-1 (0)  =  3     and   f^-1 (6)  =  -3      

So  ....  their sum   = 0 

Thanks, guest.....that's exactly the solution that Lloyd proposed.....I just couldn't remember how he did it.........it's much more elegant than an algebraic method, but....as they say "any port in a storm".....!!!!!

 I think that you meant to write " the smallest integer value that is greater than √201 "

 P =  ceiling (√201)  ≈ ceiling (14.177)  = 15

 Q  = floor ( - √20 )  =   floor ( -4.472 )  = -5

Distance between P and Q  =  P - Q  =   15 -  (-5)   =  20 units

We don't have an "inequality" here....just a system of equations

Let P be the price of a paperback and H be the price of a hardcover book

And  we have that

10P  + 4H  =  62       ⇒  this represents Adam's situation   (1)

 7P   + 5H  =  61      ⇒   this reperesents Barb's siutation   (2)

Multiply  (1) through by  5  and (2) through by -4   and add them together

So we have

50P  + 20H  =  310

-28P - 20h  =   -244

________________

22P      =            66                     divide by  22 on both sides

P  = 3     this is the price of a paperback  [in dollars]

And using (1) to find H, we have

10 (3)  + 4H   = 62

30  + 4H  = 62         subtract 30 from both sides

4H  = 32                   divide by 4 on both sides

H  = 8    this is the price of a hradcover book [ in dollars]

I just did this in great detail and then lost it all :(

Guest's answer is correct.  :)

In how many ways can 19 students be divided into five groups, two groups of 5 and three groups of 3, so that each group studies a different topic?

I will state right off that i do not know how to do this.

This would be my guess but it is probably wrong.

Choose 10 from the 19  This can be done in 19C10 ways

Then chose 5 from those 10. This can be done in 10C5 ways

But if you have one group of 5 you automatically have the other group of 5 so I halved this.

So far that is

19C10 * 10C5 / 2!

Then choose 3 from what is left   9C3 ways

Then choose 3 from what is left  6C3 ways

but it doesn't matter what order these are chosen in so that is divided by 3! ways

so that is   

9C3 * 6C3 / 3!

Put that together and I get

(19C10 * 10C5) / 2!  *  (9C3 * 6C3) / 3!

(nCr(19,10)*nCr(10,5)/2)*(nCr(9,3)*nCr(6,3)/3! = 325 9095 840 ways

It is probably wrong. 

If you find out how do do it ,for certain, could you please let me know :)

The word problem is asking you to create an inequality that accuratly represents the amount that they charge for hardcovers compared to paperbacks.

The answer to the square root of 52 is between 7 and 8 and is closer to 8. Therefore the answer will be between 7 and 8. Using a calculator we can find that the precise answer is 7.211102551

Work from the back to the front, three applications of the difference between two squares leads you to sqrt(33^2 - 8).

Actually, I figured it out,thanks!

$\sqrt{52}=\sqrt{4*13}=\sqrt{4}\sqrt{13}=2\sqrt{13}\approx7.2111$

$\sqrt{33+\sqrt{8}}*\sqrt{6+\sqrt{3+\sqrt{8}}}*\sqrt{3+\sqrt{3+\sqrt{3+\sqrt{8}}}}*\sqrt{3-\sqrt{3+\sqrt{3+\sqrt{8}}}}\approx32.87856$

YIPPEE! I have 1111 hearts now!

Cartography and astronomy rely heavily on trigonometry. 

With trigonometry one can accurately represent the locations of the landmasses of the Earth or calculate the distance in between two stars.

An irrational number is a real number that cannot be represented as a ratio of $\frac{a}{b}$such that a and b are both integers.

Another way to know if a number is irrational is to notice if the decimal expansion does not terminate or repeat indefinitely. If it is, that number is irrational.

$\pi$, for example, is irrational. $\sqrt{2}$ is also irrational. $\phi$ (known as the golden ratio) is irrational, too.

$\frac{1}{3}$ is not irrational because it is a ratio of $\frac{a}{b}$ where a=1 and b=3.

$2.7551$ is not irrational because the decimal terminates.

$1.3\overline{949494}$ is not irrational because the decimal repeats indefinitely. 

An Irrational Number is a real number that cannot be written as a simple fraction.

Eg: 1.13577894561...

       7.89463148723...

| x - 3 | < 7
How to solve the inequality?
We are looking for the solution.
Please also specify the individual steps for the solution.

DO NOT DELETE YOUR QUESTIONS GAYMER !!

It is extremely rude and Admin has been known to deregister members of this behaviour!

Trigonometry would be used in many different ways. It is used for surveying and for all types of engineering. 

For example:

If a bridge is not to fall down than all the stresses etc on it need to be calculated accurately. Trigonometry is an essential tool here. It is used in many other fields as well.  

Hi, 

Your question was not answered because it is so vague. :/

Trigonometry always has to do with angles but it is not just useful for right angled triangles.

It is used to solve many different problems. 

It is used in surveying and engineering and constrution and in so many other ways too.

Perhaps you could post a more specific question / problem?

| x - 3 | < 7

The thing to notice here is that x-3 must be less than 7 units from 0 on the number line .. so...

-7 < x-3 < 7

-4 < x < 10

I like it too :)