All Answers 358115 Answers

$$(x+4)(x+4)\\
=x(x+4)+4(x+4)\\
=x*x+x*4 \quad +4*x+4*4\\
=x^2+4x+4x+16\\
=x^2+8x+16$$

Toadstool is the princess in mario games kkkk

Who is princess toadstool?

This question does not make sense.

Well, actually, the "formula" is:

A = (b*h) / 2      .....so we have.....

72 = (18 * h) / 2      multiply both sides by 2

144 = 18 * h            divide both sides by 18

144/18 = h = 72/9 = 8 inches

But we're not taking (1122) * (1122). We're taking 1122 and raising to the power of 1122 factorial....!!! (If I read the question correctly...)

I don't know how to determine this !!!  Whatever this number is, it is HUGE!!!

The three lines include the altitude. From this, we know that ACE and EFC are right triangles. Since the next line is the median, and that the four parts are equal, we know that AE and EF are equal. EC is a shared segment. With SAS, these triangles are congruent, providing us with an isosoles triangle. The CD segment is the median, for my own reference while writing this answer. Anyway, we now know that segment a = CF. CF is the bisector.

Can you go on this?

Here's another proof:

Let's suppose that there are two positive, relatively-prime integers, call them a and b, such that a/b  = √5 with a > b.

Multiply both sides by b, so we have

a = b√5

Now square both sides

a² = 5b²

And note that we can write 5b² as 4b² + b²    Now, subtract b² from both sides....so we have...

a² - b² = 4b²

Now, note that the right side is even, and suppose that a and b are both even. But, this is impossible since they are relatively prime and thus, have no factors in common.

And neither can one be odd an the other even, because the left side would be odd, but the right side is even.

The only possibility left is that both are odd. But this can't be so for the following reason: Factor the left side as (a + b) (a - b). And if a and b are odd, then (a + b) is even and so is (a - b) - (remember that a > b, so a - b > 0).

So we have

(a + b) ( a - b) = 4b²    Now divide both sides by 4 ....... and we can write

[(a + b)/2] [(a - b)/2]  = b²

Now, if (a-b)/2 is odd, then (a+b)/2 is even ..... and vice-versa. To see this,  let  (a - b)/2 = 2k + 1.     Then a - b  = 2(2k + 1)  ....  Add b to both sides..... a = 2(2k + 1) + b....Now add b again to both sides and we have  a + b =  2(2k +1) + 2b.......Now, divide by 2 on both sides, and we have (a+b)/2 = (2k + 1) + b, and both sides are even. Likewise, it can be proved that if (a +b)/ 2 is odd, then (a-b)/2 is even.

But, an even times an odd is always even...... but b² is odd.

Thus......a and b aren't both even or both odd, and neither are they of different parity. and nothing else can be possible. Thus, our assumption that there exists two positive, relatively-prime integers such that a/b = √5 must be false. And thus, √5 must be irrational.

Ah, I was unaware of that, thank you for your correction.

There are a few general rules:

$${{\mathtt{X}}}^{{\mathtt{a}}}{\mathtt{\,\times\,}}{{\mathtt{X}}}^{{\mathtt{b}}} = {{\mathtt{X}}}^{\left({\mathtt{a}}{\mathtt{\,\small\textbf+\,}}{\mathtt{b}}\right)}$$

$${\frac{{{\mathtt{X}}}^{{\mathtt{a}}}}{{{\mathtt{x}}}^{{\mathtt{b}}}}} = {{\mathtt{X}}}^{\left({\mathtt{a}}{\mathtt{\,-\,}}{\mathtt{b}}\right)}$$

$${{\mathtt{X}}}^{{\mathtt{a}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{X}}}^{{\mathtt{a}}} = {\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{X}}}^{{\mathtt{a}}}$$

$${\left({{\mathtt{X}}}^{{\mathtt{a}}}\right)}^{{\mathtt{b}}} = {{\mathtt{X}}}^{{\mathtt{ab}}}$$

Example:

$${f}{\left({\mathtt{x}}\right)} = {{\mathtt{3}}}^{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}\right)}{\mathtt{\,\small\textbf+\,}}{{\mathtt{9}}}^{\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}}\right)}$$

$${f}{\left({\mathtt{x}}\right)} = {{\mathtt{3}}}^{\left({\mathtt{2}}{\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)\right)}{\mathtt{\,\small\textbf+\,}}{{\mathtt{9}}}^{\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}}\right)}$$

$${f}{\left({\mathtt{x}}\right)} = {\left({{\mathtt{3}}}^{{\mathtt{2}}}\right)}^{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)}{\mathtt{\,\small\textbf+\,}}{{\mathtt{9}}}^{\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}}\right)}$$

$${f}{\left({\mathtt{x}}\right)} = {{\mathtt{9}}}^{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)}{\mathtt{\,\small\textbf+\,}}{{\mathtt{9}}}^{\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}}\right)}$$

$${f}{\left({\mathtt{x}}\right)} = {{\mathtt{9}}}^{{\mathtt{x}}}{\mathtt{\,\times\,}}{{\mathtt{9}}}^{{\mathtt{3}}}{\mathtt{\,\small\textbf+\,}}{\frac{{{\mathtt{9}}}^{{\mathtt{x}}}}{{{\mathtt{9}}}^{{\mathtt{3}}}}}$$

$${f}{\left({\mathtt{x}}\right)} = {\mathtt{729}}{\mathtt{\,\times\,}}{{\mathtt{9}}}^{{\mathtt{x}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{729}}}}{\mathtt{\,\times\,}}{{\mathtt{9}}}^{{\mathtt{x}}}$$

$${f}{\left({\mathtt{x}}\right)} = \left({\mathtt{729}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{729}}}}\right){\mathtt{\,\times\,}}{{\mathtt{9}}}^{{\mathtt{x}}}$$

Something like that?

12 min= .2 degrees

So we have

cot(85.2°) = 0.08397 = .0840 rounded

Actually, this theorem was proved by an English mathematician, Andrew Wiles, in the mid 1990's. It was one of the most famous "unsolved" problems in mathematics - remaining unproven for 358 years. Interestingly, Wiles himself almost gave up on his effort to prove it!!!

Here are some links to the theorem and to Wiles himself.

Fermat's last theorem states that for natural numbers X,Y,Z,n with n>2 there are no solutions to the equation:

$${{\mathtt{X}}}^{{\mathtt{n}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{Y}}}^{{\mathtt{n}}} = {{\mathtt{Z}}}^{{\mathtt{n}}}$$

The first difficulty which occurs is that there are an infinite amount of natural numbers which means you can't check them all to prove this theorem. To make matters worse 'n' can also reach infinity which changes the entire nature of the equation. To the power three is a whole different equation than to the power four et cetera. It's practically impossible to fit them all in a neat little box like with a fixed 'n'. Hence most mathematical proofs which relate to this theorem only prove the theorem for a certain 'n'. Still it is seemingly impossible to get an overall proof which works for every 'n'.

In short infinite possibilities no known way of having an 'overall rule'.

x*57 = 294

so 57/294 = 5.1578947368421053

x = 5.1578947368421053

**I do not know which side has the even or odd, only that both exist.**

Huh?  I am afraid you talk in riddles David. 

Any square number has to have an odd number of factors because all factors come in pairs but the squared one is doubled.  Multiply a square number by 5 and it has to have an odd number plus one more. That is it has to have an even number of factors.

Isn't what I said right?   

The first problem can be solved with trigonomy.

To sustain the weight each cord will need to deliver a power of 50N on the y-axis.

You can represent the forces in a triangle like this, in which |AC| is the power delivered by tension in the cord:

|BC|=50 N so |AC|=|BC|/cos(50)=50/cos(50)=77.79. This means each cord delivers a power of 77.79N due to tension. Which would mean that the correct answer is D.

The second problem can be solved with the formula:

$${\frac{\left({Y}{\left({\mathtt{A}}\right)}{\mathtt{\,-\,}}{Y}{\left({\mathtt{B}}\right)}\right)}{\left({X}{\left({\mathtt{A}}\right)}{\mathtt{\,-\,}}{X}{\left({\mathtt{B}}\right)}\right)}} = {\mathtt{slope}}$$

We take the points A (-4,0) and B (0,-6) and insert them in the formula:

$${\frac{\left({\mathtt{0}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}\right)}{\left({\mathtt{\,-\,}}{\mathtt{4}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0}}\right)}} = {\mathtt{\,-\,}}{\frac{{\mathtt{3}}}{{\mathtt{2}}}} = -{\mathtt{1.5}}$$

The correct answer is A.

Composite numbers cannot have both an even and odd number of prime factors because all composite numbers have a unique prime factorization.

That is true. This solution assumes that it is, and proves it is not a composite number.

This is impossible because they are the same number and must have the same number of factors.

The “x” and “Y” values have the same number of factors, the equation does not. They cannot because the total factors are odd. Only at an infinite, diminishing point where the “constant” --the additional factor, converges does the equation balance. That value, in this case, is the SQR( 5).

I do not know which side has the even or odd, only that both exist. The total factors are odd, because adding all the factors, except for the 5 is the same as multiplying by 2, making the total an even amount, then add the one additional factor (the 5), to the count, makes it odd.

Therefore there are no integers x and y that meet this criterion.

That is essentially correct.

This is conjecture, but the mathematics that communicate this may be related to the maths that relate why the set of all integers is equal in number to the set of all even (or odd) integers.

~~D~~

Monday 1/9/14

1) This one looks interesting.

@@ End of Day Wrap:    Sun 31/8/14       Sydney, Australia      Time 1:00am (Really Monday morning)     ♬

Hi all,

Our great answerers today included Honga, DavidQD, AzizHusain, Will85237 and CPhill.  Thanks all.  

I'd like to extend a very big welcome to Honga who is new to our web2.0calc forum.  We hope you really like it here.  

Now for a few interesting posts:

* 1) Proove that $${\sqrt{{\mathtt{5}}}}$$ is irrational.

I don't understand the question.  

Just use the site calc.  It is on the home page.  

I only just realised that Honga had answered this.  thanks Honga.

Alan answered a similar question only a couple of days ago.  This is the thread.