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Look here.

The probability is 1/4*1/4*1/13 = 1/208.

What's an elliipse?

See the answer here:  https://web2.0calc.com/questions/help_73560#r2

1. Connect \(RG\)

2. Since \(FR=FG\), triangle \(FRG\) is isosceles with \(F\) as the vertex.

3. Angle \(GRF\) subtends arc \(GF\), which is 130, therefore angle \(GRF\) = 130/2 = \(65\)

4. Since \(FRG\) is isosceles, that means \(FRG=RGF\).

5. Angle \(GFR\) = \(50\) because all triangles have an angle sum of 180, and we know that two angles are 65 degrees. (180 - 65 * 2)

6. Angle \(PGF\) subtends arc \(FQ\), so angle \(PGF\) = 28/2 = \(14\)

7. Conside triangle \(PFG\), we know the measure of two angles \(GFR\) and \(PGF\) is  \(50\) and \(14\), respectively. That means the third angle \(GPF\) is 180 - 50 - 14 = \(116\)

8. By supplements angle \(RPG=180-116 = 64\)

Done

Just remember \(f^{-1}\) is just swapping \(x\) and \(y\). 

2)    x² + y²  =  36         and        y  =  (3x + 15)/4

       x² + [ (3x + 15) / 4 ]²  =  36

       x² + (9x² + 90x + 225) / 16  =  36

   16x² + (9x² + 90x + 225)  =  576

   25x² + 90x - 351  =  0

Quadratic formula:  x₁  =  [ -90 + sqrt[( 43 200 ) ] / 50              or         x₂  =   [ -90 - sqrt[( 43 200 ) ] / 50   

                                y₁  =  [ 480 + 3·sqrt( 43 200 ) ] / 200                     y₂  =   [ 480 - 3·sqrt( 43 200 ) ] / 200 

Distance formula:  sqrt[ ( x₂ - x₁ )² + ( y₂ - y₁ )² ]

                             =  10.392                      (calculator time)

\(x^4+\frac{1}{x^4}=23\)  (1)

\(x^2+\frac{1}{x^2}\) Square this 

\((x^2+\frac{1}{x^2})^2=x^4+\frac{1}{x^4}+2\) From (1) we know that \(x^4+\frac{1}{x^4}=23\) , substitute 

\((x^2+\frac{1}{x^2})^2=23+2=25\)

\((x^2+\frac{1}{x^2})^2=25\) Square root both sides 

\(x^2+\frac{1}{x^2}=+5 \) or \(-5\)

oh haha i didn't notice how easy the answer was 😅

thanks Melody!

angle PQR is 38^o so angle PQT is half of that, or 19^o  

since angle PQT is 19^o that means angle QPT is 71^o 

PT is half of 1.4 m, or 0.7 m 

 0.7 m

———  =  cosine 71^o  

  PQ 

             0.7 m            0.7 m

PQ  =  ——–—  =  ——–——  =  2.1500 m rounded to two decimal places is 2.15 m 

           cos 71^o        0.32557 

Since triangle(PQR) is an isosceles triangle, T will be the midpoint of PR; thereore, PT = 0.7 m

Also, QT bisects angle(PQR) so angle(PQT) = 19^o.

sin(19^o) = 0.7 / PQ   --->   PQ  =  0.7 / sin(19^o)   --->   PQ  =  .....

Does anyone have anything on number 2?

Thanks, EP!

Really brutha week 19 problem 1

22/3

OB=7OA

and

OA+AB=7OA

so

6OA=AB

so

OA=AB/6

=44/6

=22/3

problem 6

since B is reflection of A, XA=XB and P is midpoint of AB so APX=90 degrees

We can do this with the other angle, and it sort of forms a quadrilateral, where we have the two 90 degree angles plus the 34 degrees. therefore BAC=360-90-90-34=146

I'm pretty sure it is just swapping the x for the y

so  f^-1(1) = value of x when y is 1  answer  2

etc.  So I end up with 3

Hi qwertyzz,

I'm really bad with these problems so this solution is probably wrong, but I'll give it a shot anyway!

If \(23=x^4+\frac{1}{x^4}\) then what is the value of \(x^2+\frac{1}{x^2}\)?

So how I did this is: 

First, I just set \(x^2+\frac{1}{x^2}=y\)

If you square both sides of \(x^2+\frac{1}{x^2}=y\), you get \((a^2)^2 + 2a^2\cdot\frac{1}{a^2} + \left(\frac{1}{a^2}\right)^{\!2} = y^2\). 

So, this means that \(a^4 + \frac{1}{a^4} = y^2-2\).

We know that \(a^4 + \frac{1}{a^4} =23\), so \(y^2-2=23\)

So, we know \(\boxed{y= \pm5}\)

I hope this was right?

Please tell me if this is right!

:)

Slightly different method      looking for  y = mx + b

find m slope =   (9-3)/(9-6) = 2

y = 2 x  + b    sub in either point to calculate 'b'

9 = 2(9) + b     b = -9

y = 2x-9

Hi guest!

Since we know the 2 points, we can first use point-slope form and then convert it to slope-intercept form. 

Point slope form: \(y-y_1=m(x-x_1)\)

First let's find the slope. The slope is \(m=\frac{3-9}{6-9}=\frac{-6}{-3}=2\)

So, in point slope form we get \(y-9=2(x-9)\)

This is equal to \(y-9=2x-18\)

\(\boxed{y=2x-9}\)

I hope this helped you, guest!

:)

Hi guest!

So first of all, we know that the blank angle of the triangle is \(180-(w+47)^{\circ}\)since, it's a straight line.

We also know that 3 angles of a triangle sum to \(180^{\circ}\)(triangle theorem).

So, we know \(w+w+13+180-w-47=180\)

If we combine like terms and simplify, we get:

 \(w+146=180\)

\(\boxed{w=34^\circ}\)

We can check this too: 34+(34+13)+180-34-47=180

I hope this helped you, guest!

:)