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\(2x^2=4x+9\)

\(2x^2-9=4x\)

\(2x^2-4x-9=0\)

\(x_{1,\:2}=\frac{-\left(-4\right)\pm \sqrt{\left(-4\right)^2-4\cdot \:2\left(-9\right)}}{2\cdot \:2}\)

\(x_{1,\:2}=\frac{-\left(-4\right)\pm \:2\sqrt{22}}{2\cdot \:2}\)

\(x_1=\frac{-\left(-4\right)+2\sqrt{22}}{2\cdot \:2},\:x_2=\frac{-\left(-4\right)-2\sqrt{22}}{2\cdot \:2}\)

\(x=\frac{2+\sqrt{22}}{2},\:x=\frac{2-\sqrt{22}}{2}\)

In this case, it is

\(x=\frac{2+\sqrt{22}}{2}\)

2+22+2=26

so 26 is your answer!

Draw the picture.

Let XYR be the external tangent with X on circle(P) and Y on circle(Q).

Extend PQ to point R, the point of intersection of this line with the external tangent.

Triangle(XPR) is similar to triangle(YQR).   [Why?]

This makes:  RQ / RP  =  QY / PX  

Let x represent the length of RQ; then x + 18 represents the length of RP.

Therefore:  x / (x + 18)  =  8 / 10.

You will need to solve this equation.

36792 / 72  =  511

Within the domain [-8,8], the smallest value of g(x) is 0 which occurs when x = 0.

The largest value of g(x) is 128, which occurs both when x = -8 and x = 8.

The interval has a width of 128.

yes, but just because 2xy is an integer, it doesn't say anything about x, y, or xy.

thats where im stumped

K=2/5*E

2/5*E*1.2-600=E*.45

.48*E-600=.45*E
600=.03*E

E=20000

You can work out Ken's savings in the end by yourself :)

\(x^2-5x+2 = 0\\ x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x = {5 \pm \sqrt{25-8} \over 2}\\ \)

The sum of these is 5

Just checking that those answsers are valid.  what can x not be

\(x^2-4x+2\ne0\\ x \ne {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x \ne {4 \pm \sqrt{16-8} \over 2}\\\)

Yes the answer is fine.

Hi  MathyGoo13,

I am glad that you are being persistent with asking for help, and saying you don't understand.  :)

It is what I encourage people to do.   I will take a look for you.

Now that i have done it, I don't think that this is a brain dead question at all.

I think questions like this can be really confusing. 

The temperature changed throughout a certain day at a steady rate of
- 3 ° C every hour. If it was -8° C sometime between 2 pm and 2:20
pm, then the temperature was 9° C in the time interval

2pm to 2:20pm        -8 degrees    (at some point)

1pm to 1:20pm        -8+3 = -5

12pm to 12:20pm        -5+3 = -2

11am to 11:20am        -2+3 = 1

10am to 10:20am        1+3 = 4

* 9am to 9:20am        4+3 = 7

* 8am to 8:20am        7+3 = 10

Ok, now what?

In an hour it decreases 3 degrees so in 20 minutes it decreases 1 degree.   (I am going forward in time now, its less confussing, I hope.)

so now I have

* 8am to 8:20am        7+3 = 10                  turns 10 degrees

8:20am to 8:40am        turns 9 degrees.

So it turns 9 degrees somewhere between 8:20 and 8:40 am   

-------------------

How could I have done this more quickly?

1 degree in 20 minutes

-8+ 17  = 9

17 lots of 20 minutes = 17/3hours  = 5 hours and 40 minutes

The start time was  2pm to 2:20pm

5 hours ago it was  9am to 9:20am

40 minutes before that was   8:20am to 8:40am

---------

I got the same answer 2 different ways.  That is always a good sign 

Nevermind, I understand it fully after my teacher taught me.

Nice job on your 200th answer. :))

=^._.^=

Paul = 2/5 Peter

Paul/Peter = 2/5

2x/(5x - 210) = 11/10

20x = 55x - 2310

  35x = 2310

     x = 66

(a) Peter = 5x = 330

(b) Peter throw away = 210/330

                                  = 7/11

Total number of stickers

          = 2,950

  Number of stickers stuck

    = 1/5 of 2,950 = 590

  Number of stickers with no holes

     = 33

  Number of stickers with no glue

      = 1/10 of 2,950 = 295

Number of stickers that could be used

  = 2,950 - (590 + 33 + 295)
  = 2,950 - 918 = 2,032

Darn - my only confusion is, sir - that 9 - (-8) = 17

But 17 is not a multiple of 3. 

17/3 = 5 and 2/3. 

What does that translate to?

What is '2/3'

That is what I'm looking for. 

Thanks!

Also, this is my 200th answer!!!

Draw the rhombus, its diagonals and label the points. 

You know the hypotenuse of one of the smaller triangles formed by the diagonals 15. 

The diagonals split the triangle into right triangles with one angle of 64.5 degrees, and a hypotenuse of 15. 

To find the longer side, we do \(15\times \sin(64.5) \approx13.53877927\)

Multiply this by 2, and one diagonal is: \(30 \times \sin(64.5) \approx \color{brown}\boxed{27.07755853}\)

The other length of the original right triangle is: \(15 \times \cos(64.5) \approx {6.457666452}\)

Multiply this by 2, and the other diagonal is: \({30 \times \cos(64.5)} \approx \color{brown}\boxed{12.9153329}\)

Lol laugh out loud

I want to know if the squiggly line is a function, since I already know the second and third  

Use desmos.com to draw the graph for y = sin(x) and see if that looks like your squiggly line.  

_.

Alright, I'll take a shot. 

It helps to inscribe the trapezoid within a rectangle. Using the boundaries: 

\(x = 0\)

\(y = 2\)

\(y = 34/7\)

\(x = 24/7\)

The area of this rectangle is \({20\over7} \times {24\over7} = {480\over49}\)

The area of the trapezoid is the area of the rectangle minus both triangles on the sides of the trapezoid.

The first triangle has a height of \(6 \over7\) and a width of \(24 \over7\), making the area \(72 \over49\).

The second triangle has a height of \(20 \over 7\) and a width of \(10 \over 7\), making the area \(100 \over 49\). 

This means that the area of the triangle is: \({480\over49}-{72\over49}-{100\over49}=\color{brown}\boxed{308\over49}\)

Here image: 

Ok, the sum will equal 9 in that case. But I think there are other cases too, for example when x^2-4x+2 = 1. Can you help me think of other ones?