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If you differentiate g(x)  with respect to x, what do you get?

Please answer, it is a simple question.

AB = x        CD = x + 9

{(2x + 9) /2}*8 = 60           x = 3

Median  M = 1/2(AB + CD)           M = 7.5

Thank you ! I got it :)

How did you get that Jugoslav?

Yes, I am aware that in algebraic notation, 2A reprents A multiplied by 2, but I really do not understand what the question is asking currently. I must be missing something because you all are answering as if there is no comprehension gap. I must be missing something obvious. I am trying to be helpful, but I cannot until I understand the problem clearly.

What is A in this particular question? Is it a length of one of the legs of the triangle? Is it the measure of an angle? Is there a hidden diagram I am missing right now? I see a guest wrote 2A = 13 without any explanation whatsoever, so I know I am missing something. I am sorry for being an inconvenience.

7*6*5*4*3*2/4/3/2/3/2 = 35 ways. 

I hope this helped. :)))

=^._.^=

That is really neat Alan   

∑ 3^(-n) n = 1/4 * 3^(-k) (-2 k + 3^(k + 1) - 3)

This question is extremely tough, and I had to think about this one for a while before I was able to come up with any answer at all.

First, we need to simplify this equation because there is a lot of simplification that is possible. I will do this by converting as many bases as possible to common bases so that the equation is easier to digest.

\(12^a*18^b=6^{a+2}*27^b\\ (6*2)^a*(9*2)^b=6^a*6^2*(9*3)^b\\ 6^a*2^a*9^b*2^b=6^a*36*9^b*3^b\\ 2^a*2^b={\color{red}36}*3^b\\ 2^a*2^b={\color{red}2^2*3^2}*3^b\\ \frac{2^a*2^b}{2^2}=3^b*3^2\\ 2^{a+b-2}=3^{b+2} \)

I simplified this equation as much as possible to see if I could make a breakthrough. With nasty equations like this one, it often best to make a key observation. This key observation will lead you to find the answer quite straightforwardly. In order to make this key observation. To make this observation, I simplified this equation a little bit more. Let's consider \(2^x=3^y\). Is there a value of x and y that will make this equation hold true? If you think about it, then the answer might jump right out at you: x = 0 and y = 0. There are no other options because the bases are different. 

If \(2^{\color{red}0}=3^{\color{blue}0}\) and \(2^{\color{red}{a+b-2}}=3^{\color{blue}{b+2}}\), then by parallel structure of these equations, \({\color{red}a+b-2=0}\text{ and }{\color{blue}b+2=0}\). This is a simple system of equations, so solving this will not be too difficult.

\(b+2=0\\ b=-2\\ a+b-2=0\\ a-2-2=0\\ a-4=0\\ a=4\)

Therefore, the only unique pair of integers is \(a=4\text{ and }b=-2\)

∑[  3/((3 n - 2) (3 n + 1)), n, 1,  33]=It converges to 99 / 100

cos fxn is shifted UP to midline 8       so   r = 8

Amplitude is multiplied by 4                so m = 4

period is changed from 2pi to 8     so n :     8n=2pi      n= 2pi/8 = pi/4

 phase shifted by one to the RIGHT      so p = 1

f(x) = 4 cos(pi/4(x-1)) +8        you should finish from here !

This is an excellent question, and I had a similar question when I was first learning. Let's investigate together! Curiosity leads to some interesting results sometimes.

\(x^2=y^2\\ x^2-y^2=0\\ (x-y)(x+y)=0\\ x-y=0\text{ or }x+y=0\\ x=y\text{ or }x=-y\)

With some algebra, we have just proven that x may equal y OR x may equal -y. This should make sense if you think about it.

I will pick an arbitrary value for y, 5.

\(x^2=5^2\\ x^2=25\)

There are two answers to this particular equation, x=5 or x=-5, even though y=5 in both cases. This is exactly what the previous algebra told us, as well.

We can take this question another step further, too! Let's think about \(x^3=y^3\). Does this mean that \(x=y?\). Let's employ the same factoring strategy to see if this is the case.

\(x^3=y^3\\ x^3-y^3=0\\ (x-y)(x^2+xy+y^2)=0\)

Once again, there are two factors. One is an irreducible quadratic, and the other is the familiar (x-y). Let's solve this equation.

\(x-y=0\\ x=y\)

This is a result that we expected. We already know that if you start with the same number that their cube will be equivalent. What is the other mysterious result, you may ask? Well, let's see.

\(x^2+xy+y^2=0\)

This quadratic is not factorable unfortunately, so we will have to employ creative algebra to understand this factor. I will multiply this equation by 2 to achieve

\(2x^2+2xy+2y^2=0\\ {\color{red}{x^2+2xy+y^2}}+x^2+y^2=0\\ {\color{red}{(x+y)^2}}+x^2+y^2=0\)

Notice how the lefthand side of this equation are all quantities that are squared and added together. Squared quantities are always nonnegative, and the only way the addition of nonnegative value can yield 0 is if the quantity that was squared was 0. Therefore, the only possible solution to this equation is that x = 0 and y = 0. That's it! In this particular case x = y, so we already had this situation covered anyway.

Therefore, if \(x^3=y^3\), then \(x=y\). We should note that this property only holds for the real number set.

We can go further once more to \(x^4=y^4\), but I think you understand the idea. I hope this aroused your curiosity!

There are 16 possible rotations of the octagon.

POSSIBLY   

rember when yo take the square root of some thing you get   a plus and a minus root

  sqrt 4 = ±2

so x² = y²     sqrt both sides results in

±x  = ± y      which MIGHT be true......but not all of the time !

Center is in between the points @    2,2

   diameter is distance between @   sqrt( (5- -1)^2 + (1-3)^2) ) = sqrt40

Stadard equation of a circle becomes    (x-2)^2  + (y-2)^2 = 40

so no because if, lets say x is 2. y could be -2 and the equation would still work

oh wait nvm from the main page it sayd x2 y2

do you mean 2x = 2y? then yes, just divide both sides by 2

The equation of the line is y = -3/2*x + 17/2.

40,000

The general equation of a standard parabola is \(f(x)=ax^2+bx+c\). Given the vertex and another arbitrary point that lies on the parabola, it should be possible to generate the specific equation of this parabola.

If the problem had not given the coordinates of the vertex, then we know that the coordinates of any vertex is given by \(V=\left(-\frac{b}{2a},f\left(-\frac{b}{2a}\right)\right)\). The x-coordinate of the vertex is given as 2.

\(-\frac{b}{2a}=2\\ b=-4a\)

Let's apply this substitution for our arbitrary function.

\(f(x)=ax^2-4ax+c\)

We know two points that lie on this parabola, (2, 4) and (1, -11). Let's use both of these to solve a system of linear equations.

\(f(2)=4a-8a+c\\ \text{Eq1: }-4a+c=4\)

\(f(1)=a-4a+c\\ \text{Eq2: }-3a+c=-11\)

I will use the elimination method to solve this system.

\(\quad -4a+c= \;\,4\\ +(\;\,3a-c=11)\\ \text{______________}\\ \quad -a \quad\quad = 15\\ a=-15\\ b=60\\ c=-56\)

Equation of Parabola: \(f(x)=-15x^2+60x-56\)

bump halp pls

First, I calculated the angle ADE (see my diagram), and then I subtracted that angle plus 90º angle (∠CDE) from 180, and I got 22.5

Value of 22.5 is the sum of angles BAO and AEC.

The ratio of these 2 angles is ∠BAO : ∠AEC = 2 : 1

And the rest is easy

I think I will do one of these, and you should be able to complete the rest of these because the process does not change significantly with each problem anyway.

\(f(x)=A\left(\sin\left[B(x-C)\right]\right)+D\\\) is the general form of a sine curve where \(|A|\) is the amplitude, \(\frac{2\pi}{|B|}\) is the period (also known as frequency), C = horizontal shift, and the equation of the midline is \(y=D\).

Let's put all of this information to use to do all these problems.

The amplitude is specified to be 6, so let's solve for the amplitude.

\(|A|=6\\ A=\pm6\\ A=6\)

We only need one sine curve that satisfies the conditions, so I will just favor the positive answers, if I can help it.

The period (or frequency is \(\frac{18}{\pi}\)), so let's figure out what \(B\) should be.

\(\frac{2\pi}{|B|}=\frac{18}{\pi}\\ |B|=\frac{\pi^2}{9}\\ B=\pm\frac{\pi^2}{9}\\ B=\frac{\pi^2}{9}\)

Once again, I will just get rid of the negative answer as we do not need it.

The midline occurs at y = 3, so \(D=3\).

Just by chance, the sine function with its current parameters already intersects the y-intercept at (0, 3), so there is no need to do a horizontal shift of any kind.

Therefore, the equation of this particular sine curve is \(f(x)=6\sin(\frac{\pi^2}{9}x)+3\)

To be honest, I am a tad puzzled with "the function is not a reflection of its parent function over the x-axis" requirement. I am not sure what this means, but I am pretty sure that this curve satisfies this cryptic condition.