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The editing portion, to clarify.

Use the formula area = (base)(height)/2.

\(\text{Area} = \dfrac{6\text{ in.} \times 7 \text{ in.}}2 = 21 \text{ sq. in.}\)

Dang, I thought for a moment I would be featured in a sticky topic 

All jokes aside though, Im glad you were only trolling - I legit Wouldve left this site if not. Ive had some really bad experiences with being called an idiot on forums. (Im a DIY handyman for my family).

Thank you for checking Max!!!

-Vinculum

2y = 2f(x) + 2 is just y = f(x) + 1. That's the image when the graph of y = f(x) is translated 1 unit upwards.

So, translate (9, 7) one unit upwards, and you will get (9, 8). That point must be on the graph of 2y = 2f(x) + 2.

Therefore, the answer is 17.

The system is equivalent to \(\begin{cases} 2x + ky &=& 9\\ kx + 3y &=& 13 \end{cases}\).

As long as the lines represented by the equations are not parallel, the system would have a unique solution.

\(-\dfrac2k \neq -\dfrac k3\\ k^2 \neq 6\\ k \neq \pm \sqrt 6\)

For \(k \neq \sqrt 6\) and \(k \neq -\sqrt 6\), the system has a unique solution.

P.S. You need to convert it to inches per second.

Yes, your answers are correct.

The factored polynomial is of the form \((ax + b)(cx + d)\). Since 13 and 17 are primes, either a or c is 13, and either b or d is 17 or -17.

Listing all possibilities:

\(\begin{array}{rcl} (13x - 1)(x + 17) &=& 13x^2 + 220x - 17\\ (13x + 1)(x - 17) &=& 13x^2 -220x - 17\\ (13x + 17)(x - 1) &=& 13x^2 + 4x - 17\\ (13x - 17)(x + 1) &=& 13x^2 - 4x - 17 \end{array}\)

The possible values of n are -220, -4, 4, 220.

Hey Max, sorry for the confusion, I accidentally called the point (2, 1), P instead of A. I meant...

Let's say you got a vector, v = (1, 1)

We also have a point, A, at (2, 1). (This will be the tail of our vectors).

Then you got points X, Y, and Z (unknown points), which will be the head of our vectors.

Lets says that: 

Vectors AX = v, AY = -v, and AZ = 2v. What are points X, Y, and Z. 

I got: X: (3, 2) Y: (1, 0) and Z: (4, 3). Though I feel confident with my answer, I would just like to confirm with anyone else to make sure I got it right.

Let a be the number of badges Aaron has, b be the number of badges Ben has, and c be the number of badge Calvin has.

Then, \(\begin{cases} a &=& \dfrac34 b\\ b &=& \dfrac25 c\\ a &=& b - 8 \end{cases}\)

Plugging in the first equation into the third equation gives \(\dfrac34b = b - 8\). Solving gives b = 32.

Can you take it from here? You can calculate the value of c and calculate c - b to be the answer.

But are the coordinates of A given?

The maximum of f(x) + g(x) is always not greater than the maximum of f(x) + the maximum of g(x) because if the maximum point of f(x) and the maximum of g(x) point is not aligned on a vertical line, the maximum of f(x) + g(x) is going to be smaller.

That means, \(\max(f(x) + g(x)) = b \leq \max(f(x)) + \max(g(x)) = 3 + 1 = 4\).

Then \(b \leq 4\).

I did?

It would be nice if the function keeps being 3x^2 + 2 throughout.

Since the graph of y = 3x^2 + 2 is a parabola, we know that it is continuous.

Then a = 3 is the answer.

Alternative solution:

Since we know the roots are a and b, we know that \(7x^2 + x - 5 = 7(x - a)(x - b)\).

Expanding the right-hand side gives \(7x^2 + x - 5 = 7x^2 - 7(a + b)x + 7ab\).

Comparing the coefficients of x on both sides: \(1 = -7(a + b)\).

Therefore, \(a + b = -\dfrac17\).

Now, \((a - 4) + (b - 4) = (a + b) - 8\), using this gives \((a - 4) + (b - 4) = -\dfrac17 - 8 =\boxed{ -\dfrac{57}8}\).

Why did you like your comment when you got it wrong?

Ok so we have 8 students and we want to make 2 teams.

Each team is 4 students.

Let Alice and Andrew be in the same team. Then, there are two more students needed in this team.

Now, we have 6 students and we want to choose 2, so: \(6C2\)

Now, there are 4 students remaining, and the other team is to be filled.
Thus, we want to choose 4 students from 4 students, This is done in one way only: \(4C4=1\)

Therefore, for both events to occur, we multiply: \(6C2*4C4=15*1=15\)

Thus, there are 15 ways.

sumfor(n, 1, 2015, 10^n) mod 11==4- the remainder.

Ok I guess I found a "valid" solution.

I do not think this is the way the question was intended to be solved.

However, this is a correct mathematical proof that there does not exist integers that satisfy the given equation.

Consider what we found earlier:

\(x=\frac{y+\sqrt{5y^2+204}}{2}\)

The point is,  consider: \(2x-y=\sqrt{5y^2+204}\)  (*)

Let's proof this by contradiction:
Suppose x and y are positive integers.

Then the L.H.S is an integer.

Now, we will prove that the R.H.S is not an integer. It is in fact, an irrational number! For all integers y.

For the R.H.S to be an integer, \(\sqrt{5y^2+204}\) must be the square root of a "Perfect square" 
Consider:   \(y^2\) This is a perfect square for all y. As, \(\sqrt{y^2}=y\) an integer.

However, consider:  \(\sqrt{5y^2}=\sqrt{5}*\sqrt{y^2}=y\sqrt{5}\) . Clearly, for all integers ,y, the \(\sqrt{5}\) will always be there.

Hence, \(\sqrt{5y^2}\) Is not an integer, it is always an irrational number.

Now, obviously, adding +204 does not make it an integer either.

Hence, \(5y^2+204\) is not a perfect square; thus, \(\sqrt{5y^2+204}\) is not an integer.

So the R.H.S of (*) is irrational, but the L.H.S was supposed to be an integer by our assumption.

This contradicts our assumption, and therefore, there does not exist integers x and y such that (*) is satisfied.

(By the way, this proof can even be extended to integers! (Need not to be only positive!))

hope this helped!

This question has already been answered

https://web2.0calc.com/questions/help-please_52172#r3

-Vinculum

Hi Hipie,

\(\displaystyle \int_{0}^{3} x^2 dx\\~\\ =\left[ \frac{x^3}{3} \right]_0^3=\frac{3^3}{3}-0=9u^2 \)

Yes it is 9. 

I like this post you have just written Ginger.

However, I think it would be a bit hard to justify putting the whole thread into the sticky note list.

I don't think many people look at the sticky notes anyway.  

Pity, there is some good posts in there.

I at least refer to some of them.... Mostly the 'reference material'.

Very occasionally someone will reference one or more of them, that always does impress and please me.   

Just as well we have you to keep track of our posts, especially our more contraversial posts.  You can always pull them from the archives upon command.

And now my mind has taken me to a elsewhere,  :)

Enjoy.

\(\displaystyle \int_{0}^{3} x^2 dx = 3^2 - 0, \mathrm{which\space is} \ \textcolor{red}{9}\), so the area under the curve from 0 to 3 is 9.