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tan(x) = -1        $x=\frac{3}{4}\pi + k*\pi$           $k\epsilon Z$

Thank you. Seriously, you were very helpful, you didn't understand the question.

That is a good answer mathhemathh   

but QED means 'it is proven' so it is only appropriate to use it at the end of proofs 

First, note that $\tan({\pi\over4})=1$. Imagine this on the unit circle. Since we need to find $x$ when its tangent is negative 1, we have to get to the 2nd and 4th quadrant while retaining the tangent's absolute slope. To do that, we add $90^o$ or ${\pi\over2}$ and $270^o$ or ${3\pi\over2}$ to the angle ${\pi\over4}$. When we do that, we get the angles ${3\pi\over4}$ and ${7\pi\over4}$.

Q.E.D. 

Let $A$ be the first pack of 400g, and $B$ be the second pack of 400g+25%.

First, we calculate the total weight of pack $B$ with the 25%:

$400g*25\%=100g$

$400g+100g=500g$ in total

Then, we calculate the unit price (price per gram) for each pack: $unitPrice={price\over quantity}$

A). ${\$2.98\over400g}=\$0.00745$

B). ${\$3.28\over500g}=\$0.00656$

As we can clearly see, the unit price of $A$ is greater than the unit price of $B$, therefore pack $B$ is a better buy, but only by a fraction of a cent!

Nevermind again. I found the answer.

factorise 4(x+3)+x(x+3)

Think about this:

4A+XA

This is 4 lots of A plus X lots of A

so when you factored it you will have  A(4+X)     or    (4+X) *A

WELL

your question is the same. 

4(x+3)+x(x+3)

This is 4 lots of (x+3)  plus  x lots of (x+3)   

when you put it together you have  (4+x) lots of (x+3)  which is just  (4+x)(x+3)

this woul normally be reorganised as 

(4+x)(x+3)= (x+4)(x+3)

i dont understand the last two steps can u pls explain

$\begin{array}{|rcll|} \hline && \dfrac{{\color{red}(x-2)(x+2)} + 2x} {3x(x+2)-12} \quad & | \quad {\color{red}(x-2)(x+2)} = x^2 + 2x-2x- 2\cdot 2 = x^2 -4 \\\\ &=& \dfrac{x^2-4 + 2x } {3x^2+6x-12} \\\\ &=& \dfrac{x^2-4 + 2x } {{\color{red}3}x^2+2\cdot {\color{red}3}\cdot x-({\color{red}3}\cdot 4)} \\\\ &=& \dfrac{x^2-4 + 2x } { {\color{red}3}\cdot ( x^2+2\cdot x-( 4))} \\\\ &=& \dfrac{x^2-4 + 2x } { {\color{red}3}\cdot ( x^2-4+2x)} \\\\ &=& \dfrac{(x^2-4 + 2x) } { {\color{red}3}\cdot ( x^2-4+2x)} \quad &| \quad \dfrac{(x^2-4 + 2x) } { ( x^2-4+2x)} = 1 \\\\ &=& \dfrac{1} { {\color{red}3}} \cdot 1 \\\\ &=& \dfrac{1} { {\color{red}3}} \\\\ \hline \end{array}$

You are not given acceleration    m/s is a velocity NOT an acceleration.

So perhaps the acceleration is 0 in which case the net force is 0 etc etc etc.

I can only think of 1 subset that will meet this restriction. That is the the set of even numbers from 2 to 100 inclusive.

So I think the answer is 1

It divides evenly one time into 17.

SOLUTION: $1$

If I am understanding this correctly, to find y you simply need to multiply -89 by 78 and then subtract 657.

SOLUTION: $-89*78-657= -7599$

So, Y= -7599

 <---- You

Here is the formula that I learned for this type of equation: $F=M x A$ Force=Mass x Acceleration.

To solve this just multiply Mass (66kg) by Acceleration (2m/s) which equals 132 Newtons (N).

  <--- You

This is an AOPS question. This presentation is an adaptation of Lancelot Link’s solution for a closely related question.

--------

This is question has a subtle contingency and requires two probability equations to solve.

For the Giants to win the series in game six, they need to win three games in five trials. This is a binomial distribution.

$P(G\text{ win series in 6 games}) = \\ P(G \text{ wins 3 in 5})*P(G\text{ wins 6th game}) =\\ \binom{5}{3}(0.5)^3(0.5)^2 *(0 .5) = 0.15625$

Further ... For the series to end in six games, either the Giants or the Royals must win the series in six games.

$P(\text{Series ends on game 6})=P(G\text{ wins series on game 6})+P(R\text{ wins series on game 6}) \\ = 2P(G\text{ wins series on game 6}) \\ = (2)\binom{5}{3}(0.5)^3(0.5)^2 * (0.5) \\= \frac{5}{16}$

The probability the Giants win the series in game 6  is 5/16.

GA

Edit: changed ASOP (the fabulist) to AOPS, (the fabled academy of higher mathematics).

Your rearrangement is correct if  side "c"  is opposite "theta"

In my answer, I arbitrarily made  "a"  the opposite side to "theta"

I am so sorry, but I still don't understand. Would you please mind elaborating on what I did wrong?

Angles                                      45       45      90

Sides opposite the angles       √2      √2        2

Angles                                      30      60      90

Sides opposite the angles         1      √3        2

Notice from this point 

a^2 =  b^2 + c^2 - 2bc *cos (theta)  

I subtracted a^2  from both sides and added 2bc *cos (theta)   to both sides giving us

2bc * cos(theta)  =  b^2 + c^2 - a^2 

Wouldn't C be subtracted by B and A and then divided by 2AB? Or is my new formula, after some deriving and algebra I did in the time it took to answer this question (Very impressed by your quick response! Thank you!!!) wrong?

C^2 - B^2 - A^2 / 2AB = Theta

HAHAHA~~~I'm typing in the dark while watching the baseball game....!!!!

P.S. - I'll make the correction    !!!!!

Let  the sides be  a, b and c

So we have that

a^2 =  b^2 + c^2 - 2bc *cos (theta)     rearrange as

2bc * cos(theta)  =  b^2 + c^2 - a^2      divide both sides by   2bc

cos (theta)  =  [  b^2 + c^2 - a^2 ] /  [2bc ]

Take the cosine inverse

cos^-1 (  [  b^2 + c^2 - a^2 ] /  [2bc ] )  =  theta

Remember that "theta"  is the angle opposite  side "a"

I wondered the same ning, nyself.  It seems CPhill mas many halents.   

CPhill, how do you "Nultiplt" both sides by 13.4???