hectictar

Let's draw a height of parallelogram ABEF so that BE and AF are the bases, and call it  h .

sin( m∠ABE )  =  h / AB
sin( 15° )  =  h / x

x sin 15°  =  h

sin( m∠PBE )  =  h / BP

sin( m∠PBE )  =  x sin 15° / 10

m∠PBE  =  arcsin( x sin 15° / 10 )

And  m∠PBE  =  (m∠PBQ)/2

Thank you!!

Hmmmm....ohh...I think you are right 

This is a true statement:

For  ∠A to be the largest angle, it must be that  0 < x < 100

So there is more than one possible value of  n - m .

(  And it is also true that if  2.9999 < x < 3 ,  then ∠A  is the largest angle,

   But it is not true that it  must  be that  2.9999 < x < 3  for  ∠A  to be the largest angle.  )

(25 + ki)(3 + 2i)  =  75 + 50i + 3ki + 2ki²   =   75 + 50i + 3ki - 2k

Let's find what value of  k  makes  50i + 3ki  =  0

50i + 3ki  =  0

i(50 + 3k)  =  0

50 + 3k  =  0

3k  =  -50

k  =  -$\frac{50}{3}$

When  k  =  -$\frac{50}{3}$  ,    (25 + ki)(3 + 2i)   is a real number.

Let the radius of the smaller circle  =  s

Let the radius of the bigger circle  =  b

And let's label the point of tangency " P ".

Side  s  is drawn from the center of the circle to  P ,  so side  s  meets  AB  at a right angle.

And by the hypotenuse-leg theorem, the two triangles are congruent.  And  AB =  80  So...

AP + PB  =  80

                               And we know  AP  =  PB

PB + PB  =  80

2 * PB  =  80

PB  =  40

By the Pythagorean theorem....

40²  +  s²  =  b²

1600 + s²  =  b²

1600  =  b² - s²

area of shaded region  =  area of bigger circle - area of smaller circle

area of shaded region  =   π b²   -   π s²

area of shaded region  =  π( b² - s² )

area of shaded region  =  1600π     (sq units)  

Here are all the restrictions we can place on  x  (that I can think of!):

∠A is the largest angle, so the side opposite ∠A is the largest side, so...

x + 9  >  3x

9  >  2x

x  <  $\frac92$

x + 9  >  x + 4

9  >  4               (Any value of  x  makes this inequality true.)

The sum of any two sides of a triangle must be greater than the third side, so...

(3x) + (x + 4)  >  x + 9

4x + 4  >  x + 9

3x  >  5

x  >  $\frac53$

(3x) + (x + 9)  >  x + 4

4x + 9  >  x + 4

3x  >  -5

x  >  -$\frac53$

(x + 9) + (x + 4)  >  3x

2x + 13  >  3x

x  <  13

For A to be the largest angle, it must be that:   x < $\frac92$  and  9 > 4  and  x > $\frac53$  and  x > -$\frac53$  and  x < 13

We can simplify that to say...

For A to be the largest angle, it must be that:   x  <  $\frac92$   and   x  >  $\frac53$

In other words...

For A to be the largest angle, it must be that:   $\frac53$  <  x  <  $\frac92$

And       n - m   =   $\frac92$  -  $\frac53$   =   $\frac{27}{6}$  -  $\frac{10}{6}$   =   $\frac{17}{6}$

But it is confusing that it says, "What is the least possible value of n - m," when it seems like there is only one possible value of  n - m .

Let the number of pink marbles  =  p

Let the number of yellow marbles  =  y

The ratio of pink marbles to yellow marbles is 3:5 . This means...

the number of pink marbles / the number of yellow marbles  =  3 / 5

p / y  =  3 / 5

There are a total of 72 marbles in the bag. So...

the number of pink marbles  +  the number of yellow marbles  =  72

p + y  =  72

y  =  72 - p

Now we can use this value of  y  in the first equation and solve for  p .

p / (72 - p)  =  3 / 5

                                    Multiply both sides of the equation by  5 .

5p / (72 - p)  =  3

                                    Multiply both sides by  (72 - p)

5p  =  3(72 - p)

5p  =  216 - 3p

                                    Add  3p  to both sides.

8p  =  216

                                    Divide both sides by  8 .

p  =  27

the number of pink marbles  =  27

I can't say for sure that this hasn't been done in a comic or movie before, and I know that I am not the first person to wish for this "superpower," but sometimes I think it would be cool if I could "save" real life like a game, and then I could reload to a previous save if something goes wrong without losing knowledge of what happened.

Every edge of SPQR has length 18 or 41, but no face of SPQR is equilateral. So every face of SPQR is a triangle with sides of length   41, 41, and 18   or   18, 18, and 41 .  However, it is not possible to have a triangle with sides of length 18, 18, and 41 because  18 + 18  <  41 .  So all of the faces of SPQR must be triangles with sides of length 41, 41 and 18 .

And I'm pretty sure this is possible because you can imagine folding a piece of paper in half and cutting an isosceles triangle out of the paper so that the base is the folded edge and is 18 units long, and the two legs are 41 units long. Then if you unfold the paper, you get two congruent isosceles triangles connected at the base. Unfold the paper so that the distance between the two vertex angles is 18 units. Then if you connected those two vertex angles together with a line, you would get SPQR.

The surface area of SPQR is  4  times the area of one of these faces.

area of one face  =  (1/2)(base)(height)

Let the base be  18 .

And from the Pythagorean theorem we can say...

height  =  √[ 41² - 9² ]

height  =  √[ 1681 - 81 ]

height  =  √[ 1600 ]

height  =  40

area of one face  =  (1/2)(18)(40)

area of one face  =   360

surface area of SPQR  =  4 * 360

surface area of SPQR  =  1440     sq units

∠AEB  ≅  ∠CED   because they are vertical angles

∠EAB  ≅  ∠ECD   because they are alternate interior angles

So by the AA similarity theorem, △EAB ~ △ECD

Since △EAB ~ △ECD,

CD / AB  =  DE / BE

100 / 150  =  DE / BE

2 / 3  =  DE / BE

∠BCD  ≅  ∠BFE   because they are corresponding angles

∠BDC  ≅  ∠BEF   because they are corresponding angles

So by the AA similarity theorem, △BCD ~ △BFE

Since △BCD ~ △BFE,

DC / EF  =  BD / BE                       And    DC  = 100     and     BD  =  BE + DE

100 / EF  =  (BE + DE) / BE           Distribute  1 / BE  to the terms in parenthesees.

100 / EF  =  BE / BE + DE / BE

100 / EF  =  1 + DE / BE                And  DE / BE  =  2 / 3

100 / EF  =  1 + 2 / 3

100 / EF  =  5 / 3                Multiply both sides of the equation by  3 .

300 / EF  =  5                     Multiply both sides of the equation by  EF .

300  =  5 * EF                    Divide both sides of the equation by  5 .

60  =  EF

The number of centimeters in the length of EF is  60 .