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Draw radii from O and A to the point of tangency with the common external tangent, as shown. Let this point be T. Since OT and AT are tangent to the same circle, they are congruent.

Since PQ​ is a common external tangent, PT=QT. Therefore, right triangles OPT and AQT are congruent. In particular, ∠TPO=∠TQA.

Let x=QR. Since ∠ROQ is a straight angle,

[\angle TPO + \angle TQA = 180^\circ,]

so ∠TPO=∠TQA=90∘.

Then triangles ORP and OAQ are similar, so [\frac{4}{x + 4} = \frac{16}{4 + 10}.]

Then 4⋅14=x⋅16+64, so x=2​.

Draw radii from O and A to the point of tangency with the common external tangent, as shown. Let this point be T. Since OT and AT are tangent to the same circle, they are congruent.

Since PQ​ is a common external tangent, PT=QT. Therefore, right triangles OPT and AQT are congruent. In particular, ∠TPO=∠TQA.

Let x=QR. Since ∠ROQ is a straight angle,

[\angle TPO + \angle TQA = 180^\circ,]

so ∠TPO=∠TQA=90∘.

Then triangles ORP and OAQ are similar, so [\frac{4}{x + 4} = \frac{16}{4 + 10}.]

Then 4⋅14=x⋅16+64, so x=2​.

Draw radii from O and A to the point of tangency with the common external tangent, as shown. Let this point be T. Since OT and AT are tangent to the same circle, they are congruent.

Since PQ​ is a common external tangent, PT=QT. Therefore, right triangles OPT and AQT are congruent. In particular, ∠TPO=∠TQA.

Let x=QR. Since ∠ROQ is a straight angle,

[\angle TPO + \angle TQA = 180^\circ,]

so ∠TPO=∠TQA=90∘.

Then triangles ORP and OAQ are similar, so [\frac{4}{x + 4} = \frac{16}{4 + 10}.]

Then 4⋅14=x⋅16+64, so x=2​.

Draw radii from O and A to the point of tangency with the common external tangent, as shown. Let this point be T. Since OT and AT are tangent to the same circle, they are congruent.

Since PQ​ is a common external tangent, PT=QT. Therefore, right triangles OPT and AQT are congruent. In particular, ∠TPO=∠TQA.

Let x=QR. Since ∠ROQ is a straight angle,

[\angle TPO + \angle TQA = 180^\circ,]

so ∠TPO=∠TQA=90∘.

Then triangles ORP and OAQ are similar, so [\frac{4}{x + 4} = \frac{16}{4 + 10}.]

Then 4⋅14=x⋅16+64, so x=2​.

Draw radii from O and A to the point of tangency with the common external tangent, as shown. Let this point be T. Since OT and AT are tangent to the same circle, they are congruent.

Since PQ​ is a common external tangent, PT=QT. Therefore, right triangles OPT and AQT are congruent. In particular, ∠TPO=∠TQA.

Let x=QR. Since ∠ROQ is a straight angle,

[\angle TPO + \angle TQA = 180^\circ,]

so ∠TPO=∠TQA=90∘.

Then triangles ORP and OAQ are similar, so [\frac{4}{x + 4} = \frac{16}{4 + 10}.]

Then 4⋅14=x⋅16+64, so x=2​.

Use the same method as was given here:  https://web2.0calc.com/questions/graphing_51126#r2

f(c)  = -1

This  means that some  value of c  = x   produces  -1

This  value is  x =  c =   -1

What we want  is the   intersection of the  circle with the  equation

(x -2)^2 + (y - 7)^2 =  25      (1)

with the line  y = 13 - x     (2)

Sub  (2)  into (1)

(x - 2)^2  +  (13 - x - 7)^2  =  25

(x - 2)^2  + (6 -x)^2   = 25

x^2 - 4x + 4 + x^2 -12x + 26 = 25

2x^2 -8x + 30  = 25

2x^2 -8x + 5  = 0

The  x values are

[ 8 + sqrt ( 8^2 - 4*2 *5) ] / 4   =   [ 8 + sqrt (24) ] / 4  =  2  +  sqrt (6) / 2 

And 2 - sqrt (6) / 2

The y values are   13 - [ 2 + sqrt (6)/2]  =   11 - sqrt (6)/2

And   13 - [2 - sqrt (6)/2] =    11 + sqrt (6) / 2

So   the points are

( 2 + sqrt (6)/2 , 11 - sqrt (6) /2)     and ( 2 -sqrt (6)/2 , 11 + sqrt (6) /2)

This  graph is  its own  inverse

So

f(x)  = f^-1 (x)

So

f^-1 (4)  =  f(4)   =   4

The slope of the line = 3    (y changes by 3 when x changes by 1)

The point (1,2) is on the line

The equation of the line  (except at x = 2) is     

y = 3 ( x -1) + 2

y = 3x - 1 =   f(x)

When x = 7      y = 20

So    f(7)  = 20

And  f (f(7)) =  f(20)

When x =20 , y =  59

So

f (f(7)) = f(20)  = 59

The sophomores  can  occupy  any of  4  positions

And the can  be arranged in 4! = 24 ways

And for each of these the sophomores can be arranged in 3!  = 6 ways

So

4 * 24 * 6  =   576  ways

1,2,3,4,5,6,6

Number of possible identifiable sequences = 

(Number of rolls) !  / (Number of any repeats ) !   =      7! / 2!  =    2520

f(-2)  means  that we are looking  for  the y value on the  graph  when  x =  -2

Can you find it  ???

Since all rotations are the  same, let A be any point on  the circle

AC = 1.5  = AB

DA = 1 = DB

Using the Law of Cosines

BA^2 = DA^2 + DB^2   - 2 (DA * DB)cos (ADB)

1.5^2 = 1^2 - 1^2 - 2 ( 1 * 1) cos (ADB)

[ 2.25 - 1 -1 ]/ [ -2 ]  = cos (ADB)

.25 / -2 = cos (ADB)

-(1/8) = cos (ADB)

arccos (-1/8) ≈ 97.18°  =  ADB = ADC

Probability ≈   [  2 measure ADB ]   / 360  =   [2 * 97.18 ] / 360   ≈  54%

x^3 + 4x > 5x^2 + 28x + x^3      simplify

5x^2 + 24x  <  0 

x ( 5x + 24)  <  0

Setting the equation equal to 0  and  solving for  x produces   x =  -24/5  and x =  0

The solution comes from either (-inf, -24.5)  or (-24.5, 0) or ( 0 ,inf)

Testing a point in each interval :

The solution comes from      ( -24.5 , 0)

45 chocolates

The 7 people had 6 chocolates reach and  3  were left over

The 5 people had  9 each with none  left over

x^2 - 2x + 1  = 0

(x - 1) (x - 1)  =  0 

This has a repeated root of 1

The sum of any  row of entries =  2^n , where n = 0,1,2,3........

So the average =   2^n / ( n + 1)     where n = 0,1,2,3.....

So

2^n / ( n + 1)  =  4/3

Note that this is true when  n =  2

x^2 + nx + 15 < -21 - 7x + x^2      simplify

(-7 -n) x >  36

x >   36 / - (7+n)       {the denominator  cannot =  0 }

This will have no  real solutions when n =  -7

A                   B

                     5

                     P

                     2

D     4   Q  3 C

Let A = (0,7)

Let P = (7,2)

Let Q = (4,0)

AP  = sqrt ( 7^2 + 5^2) =  sqrt ( 74)

AQ = sqrt ( 4^2 + 7^2)  = sqrt (65)

PQ = sqrt (3^2 + 2^2) = sqrt (13)

Law of Cosines

PQ^2  = AP^2 + AQ^2  - 2 ( sqrt [AP * AQ ]) cosPAQ

13 = 74 + 65  - 2 sqrt (74 * 65) cos PAQ

cos PAQ =  [ 13 - 74 - 65 ] / [ -2 *sqrt( 74 *65) ]  =  63 / sqrt (4810)

sin PAQ =  sqrt [4810 - 63^2 ] /  sqrt [ 4810 ] = sqrt [841 / 4810 ] 

x² + x² + 7x - 9x - 4 - 6 = 0

2x² - 2x - 10 = 0

Divide both sides by 2:

x² - x - 5 = 0

Now, find the roots using the quadratic formula:

x = 1 +/- √1 - 4(1)(-5)                   1 +/-   √21

      ------------------------         =     -----------------

                  2                                       2

So, a = (1 + √21) / 2,  b = (1 - √21) / 2

We are trying to find what is (a + 3) + (b + 3), so we just plug in the numbers:

     1 + √21                 1 - √21                    1      √21         1       √21

(   ----------- +  3) + (   ---------- + 3) = 3 --  +  -----  +  3 ---  -  -----  = 3.5 + 3.5 = 7

         2                           2                         2        2           2         2