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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???