tan3α·cotα;

It's easy if you're taking calculus, and can use
L'Hopital's rule, but not if you aren't.  Tell me 
in the thank-you note form if you are taking 
calculus or just taking trigonometry.  Then I'll
help you with it.  Is this familiar?



Numerator and denominator both approach 0, so we
can use L'hopital's rule:



Using reciplocal identity

 = 3

That's the start of the way to do it by calculus.
I'll finish it if you tell me whether you're 
taking calculus or just trig.

Edwin

You can do this either by using the general form or the standard form.
I'll first do it with the general form:

x² + y² + Dx + Ey + F = 0

Substitute (x,y) = (0,0)

0² + 0² + D(0) + E(0) + F = 0
                        F = 0
So we have

x² + y² + Dx + Ey + 0 = 0
    x² + y² + Dx + Ey = 0

Substitute (x,y) = (5,0)

5² + 0² + D(5) + E(0) = 0
              25 + 5D = 0
              25 + 5D = 0
                   5D = -25
                    D = -5

So we have:

    x² + y² + Dx + Ey = 0
    x² + y² - 5x + Ey = 0

Substitute (x,y) = (3,3)

        x² + y² - 5x + Ey = 0
(3)² + (3)² - 5(3) + E(3) = 0
         9 + 9 - 15 + 3E = 0
                  3 + 3E = 0
                      3E = -3
                       E = -1  

        x² + y² - 5x + Ey = 0
         x² + y² - 5x - y = 0

------------------------------------------
You can also use the standard form equation for a circle:

      (x - h)² + (y - k)² = r²

Substitute (x,y) = (0,0)

      (0 - h)² + (0 - k)² = r²
                  h² + k² = r²

Substitute (x,y) = (5,0)

      (5 - h)² + (0 - k)² = r²
       25 - 10h + h² + k² = r²

Substitute (x,y) = (3,3)

      (3 - h)² + (3 - k)² = r²
9 - 6h + h² + 9 - 6k + k² = r²
   18 - 6h - 6k + h² + k² = r²

So we have this system of equations:

                  h² + k² = r²
       25 - 10h + h² + k² = r²
   18 - 6h - 6k + h² + k² = r² 

If we subtract the first equation term by term from each of the
other two equations we have:

       25 - 10h = 0
   18 - 6h - 6k = 0

Solving the first for h

       25 - 10h = 0
           -10h = -25 
              h = 
              h = 

Substituting in

   18 - 6h - 6k = 0
   18 - 6 - 6k = 0
   18 - 15 - 6k = 0
         3 - 6k = 0
            -6k = -3
              k = 
              k = 

Substituting in

        h² + k² = r²
        
        
                             
                             

So the equation

      (x - h)² + (y - k)² = r² 

becomes

      (x - )² + (y - )² = 

It was harder but mainly because the center and radius were fractions.

Here is the graph.  Notice that the circle passes through the
given points and has center (,) or (2.5,.5)

The radius is  ≈ 2.55 



Edwin

eq. 1   2x+ y    =  -4
eq. 2     -2y+4z =   0
eq. 3   3x   -2z = -11

The first equation has no z-term.
So eliminate z from the other two terms.
Multiply eq. 3 by 2 and add term by term:

          -2y+4z =   0
        6x   -4z = -22
        --------------
        6x-2y    = -22

Divide that through by 2

eq. 4   3x- y    = -11

Now we have a system of two equations and two
unknowns:

eq. 1   2x+y =  -4
eq. 4   3x-y = -11

We add them term by term and y will be eliminated:

eq. 1   2x+y =  -4
eq. 4   3x-y = -11
------------------
        5x   = -15
           x = -3

Substitute x = -3 in eq. 4,

eq. 4   3x-y = -11

     3(-3)-y = -11
        -9-y = -11
          -y = -2
           y = 2

Substitute x = -3 in eq. 3 

eq. 3   3x-2z = -11
     3(-3)-2z = -11
        -9-2z = -11
          -2z = -2
            z = 1

Solution (x,y,z) = (-3,2,1)

Edwin

If you have a TI-83 or 84,

Press ON
Press CLEAR
Press Y=

With cursor just after \Y1= 

Press the key that has X,T,θ,n printed on it
Press 2ND
Press LN
Press 3
Press X,T,θ,n
Press )
Press -
Press 1
Press ENTER

You should see  \Y1=Xe3X-1 if you have a newer calculator
                and you'll need to press the right arrow after the 
                exponent

            or  \Y1=Xe^(3X)-1 if you have the older model   

With cursor just after \Y2=

Press 3

You should see  \Y1=Xe3X-1
                \Y2=3

Press ZOOM
Press 6  

You should see a graph like this: 

Press 2ND
Press TRACE     See CALCULATE screen
Press 5

Press ENTER
Press ENTER
Press ENTER

Tou should read at the bottom of the screen
under the graph

Intersection
X=.62093895  Y=3

Solution:  x=.62093895 

Edwin

That's the probability of losing.

So the probability of winning is 1 - 0.526 = 0.474

 0.474 : 0.526

Multiply both parts of the ratio by 1000

 474 : 526

Divide both parts of the ratio by 2

 237 : 263

Edwin

You will need to know 

This quotient identity:

#1. 

and these reciprocal identities 

#2. 

#3. 

We will work ONLY with the right side:

sec(x) = tan(x)csc(x)

Use identity #1 to substitute  for tan(x).
Use identity #2 to substitute  for csc(x).

sec(x) = 

Cancel the sin(x)'s

sec(x) = 

That is equivalent to:

sec(x) = 

Now use identity #3 to substitute sec(x) for 

sec(x) = sec(x)

Edwin

If you will, in the thank-you note space, state the 
problem and instructions exactly word-for-word as it 
is stated in your book, I will answer it.

Phil

The set of marbles consists of:

{1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5} 



Edwin

So Kristin's havesting rate is 1 field per 11 hours or  or  

So Kayla's havesting rate is 1 field per 16 hours or  or  

Let the answer be x hours

So their combined havesting rate is 1 field per x hours or  or 

The equation comes from  +  = 

                        

Multiply through by LCD 176x

                    16x + 11x = 176
                          27x = 176
                            x = 
                            x =  hours

Edwin

The trick is to start with sin(3x) and simplify
and it will come out with an equation that you
can solve for  sin³(x).

sin(3x) = sin(2x+x) = sin(2x)cos(x)+cos(2x)sin(x) = 

2sin(x)cos(x)cos(x)+cos(2x)sin(x) =

2sin(x)cos²(x)+[1-2sin²(x)]sin(x) =

2sin(x)[1-sin²(x)]+[1-2sin²(x)]sin(x) =
  
2sin(x) - 2sin³(x) + sin(x) - 2sin³(x) =

3sin(x) - 4sin³(x)

So

sin(3x) = 3sin(x) - 4sin³(x)

4sin³(x) = 3sin(x) - sin(3x)

sin³(x) = sin(x) - sin(3x)

Edwin

This will likely have infinitely many solutions,
since there are more letters, variables, than equations.

2w - 3x + 4y +  z =  7
 w -  x + 3y - 5z = 10
3w +  x - 2y - 2z =  6



Swap rows 1 and 2, to get a 1 in the
upper left corner:

abbreviated R1<->R2



To get a 0 where the 2 is, multiply row 1 
by -2 and add it to row 2:

abbreviated -2R1+R2->R2



To get a 0 where the bottom left 3 is, 
multiply row 1 by -3 and add it to row 3:

abbreviated -2R1+R2->R2



To get a 1 where the -1 is on Row 2, 
multiply row 2 by -1

abbreviated -R2->R2


 
To get a 0 where the 4 is, 
multiply row 2 by -4 and add it to row 3:

abbreviated -4R2+R3->R3



To get a 1 where the -19 is on Row 3, 
multiply row 3 by -1/19



Then we change the matrix back to equations:

 w -  x + 3y -  5z = 10
      x + 2y - 11z = 13
           y -  3z =  4

Solve each equation for the first letter

                 w = 10 +  x -  3y +  5z
                 x = 13 - 2y + 11z  
                 y =  4 + 3z

Now we do back substitution:

Sunstitute the expression for y in the middle equation:

                 x = 13 - 2(4 + 3z) + 11z
                 x = 13 - 8 - 6z + 11z
                 x =  5 + 5z

Substitute the expressions for x and y in the top equation:

                 w = 10 +  x -  3y +  5z
                 w = 10 + (5 + 5z) - 3(4 + 3z) + 5z
                 w = 10 + 5 + 5z - 12 - 9z + 5z
                 w = 3 + z               

So the solution is

  (w,x,y,z) = (3+z, 5+5z, 4+3z, z)

Some teachers will tell you to use a different letter than z for z, 
such as "a", or "k".  If they use "a" the solution would be:

  (w,x,y,z) = (3+a, 5+5a, 4+3a, a)

Edwin

t = the tens digit
u = the units digit
10t+u = the number
t+u = sum of digits

So   u = t+2

We divide as below.  Multiply the partial quotient 4
by the divisor, t+u, getting 4t+4u.  Then subtract that
from the dividend, 10t+u, and get remainder 6t-3u:

            4
   t+u)10t+ u
        4t+4u
        6t-3u = Remainder

And we are given that the remainder is 3. So

6t-3u = 3

So we have to solve this system of equations:

    u = t+2
6t-3u = 3

Solve that by substitution and get t = 3 and u = 5.

So the number is 35.

Edwin

From that we have:

 1 ∈ A,   1 ∉ B
 3 ∈ A,   3 ∉ B
 7 ∈ A,   7 ∉ B
11 ∈ A,  11 ∉ B 

B - A= {2, 6, 8}, 

From that we have:

 2 ∈ B,   2 ∉ A
 6 ∈ B,   6 ∉ A
 8 ∈ B,   8 ∉ A

A ∩ B = {4, 9}

From that we have:

 4 ∈ A,   4 ∈ B,   9 ∈ A,   9 ∈ B 

So we end up with

A = {1,3,7,11,4,9}, B = {2,6,8,4,9}

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

From that we have:

 1 ∈ C,   1 ∉ D
 2 ∈ C,   2 ∉ D
 4 ∈ C,   4 ∉ D

D − C = {7, 8}, 

From that we have:

 7 ∈ D,   7 ∉ C
 8 ∈ D,   8 ∉ C

C ∪ D = {1, 2, 4, 5, 7, 8, 9}

Since 5 and 9 are not in either C - D or in D - C,
they must be elements of both sets, since they were
removed in both cases of set subtraction.  Therefore

C = {1,2,4,5,9},  D = {7,8,5,9}

Edwin

All points on the x-axis are like (2,0), (-4,0), (7,0), etc.
They all have their y-coordinates as 0.  x-intercepts are all
on the x-axis, so to find them, we let y=0.

All points on the y-axis are like (0,3), (0,-5), (0,6), etc.
They all have their x-coordinates as 0.  y-intercepts are all
on the y-axis, so to find them, we let x=0.

Edwin

There are three cases for the law of cosines and 2 cases for
the law of sines:

Case 1:
a² = b²+c²-2·b·c·cos(A)  when ∠A and ∠B are both acute angles.


Draw altitude CD ⊥ AB.  Label CD h for the height of ߡABC.
Label the left part of c, AD as x and the right part of c DB as c-x.


 
ߡADC and ߡBDC are right triangles, so by the Pythagorean
theorem,

h² = b²-x² and also h² = a²-(c-x)²

Therefore equate their right sides:

b²-x² = a²-(c-x)²
b²-x² = a²-(c²-2cx+x²)
b²-x² = a²-c²+2cx-x²
Add x² to both sides:
b² = a²-c²+2cx

From ߡADC,  = cos(A) or x = b·cos(A)

Substituting that for x:

b² = a²-c²+2c·b·cos(A)

Isolate a² on the right side:

b² + c² - 2c·b·cos(A) = a²

That is equivalent to

a² = b²+c²-2·c·b·cos(A)

or since c·b = b·c,

a² = b²+c²-2·b·c·cos(A) 

-----------------------
The law of sines for this case

 = sin(A) and  = sin(B)

h = b·sin(A) and h =a·sin(B)

So  b·sin(A) = a·sin(B)

Divide both sides by sin(A)sin(B)

     = 

That's not complete yet, for we haven't shown that those
equal to  but it will be when we finish 
the next case. 

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

a² = b²+c²-2·b·c·cos(A)  when ∠A is acute and ∠B is obtuse.



Extend AB and draw CD ⊥ AB.  Label CD h for the height of ߡABD.
Label the left part of c, AD as x and the right part of c DB as c-x.


 
ߡADC and ߡBDC are right triangles, so by the Pythagorean
theorem,

h² = b²-(c+x)² and also h² = a²-x²

Therefore equate their right sides:

b²-(c+x)² = a²-x²
b²-(c²+2cx+x²) = a²-x²
b²-c²-2cx-x² = a²-x²
Add x² to both sides:
b²-c²-2cx = a²

From ߡADC,  = cos(A)
           c+x = b·cos(A)
             x = b·cos(A)-c

Substituting in
b²-c²-2cx = a²

  b²-c²-2c(b·cos(A)-c) = a²
b²-c²-2·c·b·cos(A)+2c² = a²
    b²+c²-2·c·b·cos(A) = a²

That is equivalent to

a² = b²+c²-2·c·b·cos(A)

or since c·b = b·c,

a² = b²+c²-2·b·c·cos(A)
 
---------------------------
The law of sines for this case

 = sin(A) and  = sin(∠CBD)

h = b·sin(A) and h =a·sin(∠CBD)

So  b·sin(A) = a·sin(∠CBD)

Divide both sides by sin(A)sin(∠CBD)

     = 

Now we use the fact that the sine of an angle 
is equal to the sine of its supplement.  Therefore
if we erase the extended part we can label ∠CBD as ∠B.

     = 

The law of sines is now complete for the first case 
was for two angles being acute, and this case is for 
when one angle is obtuse.
Since the other two angles are necessarily acute, 
the first case takes care of them and we have the
complete law of sines:

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

To prove the law of cosines when A is an obtuse angle,
we have to accept the definition from xy-plane 
trigonometry that cos(180°-A) = -cos(A).

We'll just take the above triangle and swap angles 
A and B and sides a and b:




Extend BA and draw CD ⊥ BD.  Label CD h for the height of ߡABC.
Label the extended segment, AD, as x.


 
ߡADC and ߡBDC are right triangles, so by the Pythagorean
theorem,

h² = b²-x² and h² = a²-(c+x)²

Therefore equate their right sides:

b²-x² = a²-(c+x)²
b²-x² = a²-(c²+2cx+x²)
b²-x² = a²-c²-2cx-x²
Add x² to both sides:
   b² = a²-c²-2cx

From ߡADC,  = cos(CAD),
            x = b·cos(CAD)

and since ∠CAD and ∠BAC are supplementary,

cos(∠CAD) = -cos(∠BAC)
and
           x = -b·cos(∠BAC)

   b² = a²-c²-2cx

becomes:

   b² = a²-c²-2c[-b·cos(∠BAC)]

   b² = a²-c²+2c·b·cos(∠BAC)

Isolate a² on the right

   b²+c²-2c·b·cos(∠BAC) = a²

which is equivalent to
  
a² = b²+c²-2c·b·cos(∠BAC)

or since c·b = b·c,

a² = b²+c²-2·b·c·cos(∠BAC)

and if we erase the extended 
segment x, we can write ∠BAC as
∠A and have

a² = b²+c²-2·b·c·cos(A)

---------------------------
Edwin

Central angle AOH is made up of the 3 red central angles.
Each of those three central angles measures 360°÷10 or 36°.

So the 3 of them have measure 3×36° or 108°, so angle AOH
has measure 108°.  Therefore minor arc AH which angle ADH 
subtends has measure 108°.  

Angle ADH is an inscribed angle and an inscribed angle gas 
the measure of one-half of the measure of its subtended arc.
Angle ADH subtends the same arc which central angle AOH 
subtends. Thus the measure of angle ADH is one-half of 108° 
and is therefore 54°. 

Edwin

Let x = the number of 25 paise coins, so also
x = the number of 50 paise coins.

25x + 50x = 15000
      75x = 15000
        x = 200

200 25paise coins and 200 50paise coins.

Edwin

1 + 2 – 3 + 4 – 5 + 6 – 7 + … + 2000

That has 2000 terms.  Let's write in the next to
the last three terms:

1 + 2 – 3 + 4 – 5 + 6 – 7 + … + 1998 - 1999 + 2000

We put parentheses in like this:

1 + (2 – 3) + (4 – 5) + (6 – 7) + … + (1998 - 1999) + 2000 

Only the 1 on the left and the 2000 on the right are not in
parentheses.  Those are the only two numbers that are not
in parentheses so the other 1998 of the 2000 numbers are in
parentheses.  Since there are two numbers in each set of 
parentheses there are one-half of 1998 or 999 sets of 
parentheses.

Every one of those 999 sets of parentheses has -1 in it.

So we have 999 negative one's which makes -999 plus the 1 on the left 
and the 2000 on the right.  That makes the sum -999+1+2000 = 1002

Edwin

Since they are mutually exclusive, they cannot both happen, so P(A and B) = 0

P(A or B) = P(A) + P(B) - P(A and B)
P(A or B) = .27  + .31  -    0
P(A or B) = .58

Edwin

We could also do it using algebra:

Let x = the number of ordinary bicycles, 
Let y = the number of tandem bikes, and 
Let z = the number of tricycles


there are 135 seats,

There are x seats on the ordinary bicycles, 2y seats on the tandem bikes,
and z seats on the tricycles.  So that gives us the equation:

x + 2y + z = 135

118 front handlebars, 

There are x front handlebars on the ordinary bicycles, y front handlebars on the tandem bikes, and z front handlebars on the tricycles.  So that gives us the equation:

x + y + z = 118

and 269 wheels. 

There are 2x wheels on the ordinary bicycles, 2y wheels on the tandem bikes,
and 3z seats on the tricycles.  So that gives us the equation:

2x + 2y + 3z = 269

So the system of equations is

 x + 2y +  z = 135
 x +  y +  z = 118
2x + 2y + 3z = 269

Solve that and get x=68 regular bikes, y=17 tandem bikes, and z=33 tricycles.

Edwin

We memorize the way to factor the sum or the difference of cubes:

A³ ± B³ = (A ± B)(A² ∓ AB + B²)

That is, the sum of two cubes factors like this:

A³ + B³ = (A + B)(A² - AB + B²)

And the difference of two cubes factors like this:

A³ - B³ = (A - B)(A² + AB + B²)

Be sure to memorize these. 

Your problem is

64x³ - 27y³

since 64 = 4³ and 27 = 3³, we can write that as

(4x)³ - (3y)³

So we substitute A = 4x  and B = 3y  in

A³ - B³ = (A - B)[A² + AB + B²]

(4x)³ - (3y)³ = (4x - 3y)[(4x)² + (4x)(3y) + (3y)²]

Then we simplify the bracket expression:

              = (4x - 3y)[16x² + 12xy + 9y²]
 
              = (4x - 3y)(16x² + 12xy + 9y²)

Edwin

|x| + |y| = 2

We find the x-intercepts by setting y=0

|x| + |0| = 2
      |x| = 2
        x = ±2

So the x-intercepts are (2,0) and (-2,0)

We find the y-intercepts by setting x=0

|0| + |y| = 2
      |y| = 2
        y = ±2

So the y-intercepts are (0,2) and (0,-2) 

We plot those:



We find some more points.

Let x=1 and solve for y

|1| + |y| = 2
  1 + |y| = 2
      |y| = 1
        y = ±1  so we have the points (1,1) and (1,-1)

Let x=-1 and solve for y

|-1| + |y| = 2
   1 + |y| = 2
       |y| = 1
         y = ±1  so we have the points (-1,1) and (-1,-1)

We plot those four points:



And we sketch in the graph:



Edwin

Here's another way to do the problem:




Suppose (x,y) is the center of the circle.  Then the distance from
(x,y) to (2,1) must equal the distance from (x,y) to (3,5), so
we use the diatance formula and set them equal:



 = sqrt( (x-3)^2 + (y-5)^2)}}}

(x-2)² + (y-1)² = (x-3)² + (y-5)²

Rearrange to get difference of squares on each side:

(x-2)² - (x-3)² = (y-5)² - (y-1)²

[(x-2)-(x-3)][(x-2)+(x-3)] = [(y-5)-(y-1)][(y-5)+(y-1)]

[x-2-x+3]{x-2+x-3] = [y-5-y+1][y-5+y-1]

[1][2x-5] = [-4][2y-6]

2x-5 = -8y + 24

2x + 8y = 29

So we have the system of equations:



Solve that and get the center, and do the rest like the
other solution.

Edwin

To have an average of 5, the sum of the three must be 15.

Here is a fact about positive numbers we can use:

If two positive numbers have a certain sum, their
product is largest when they are closest together
and smallest when they are farthest apart.

However, here we have three integers with that sum, not 
just two.  So that complicates matters
a bit.

But, we can consider only two numbers and use that
fact above. Here's how:

Suppose the three integers are a, b, and c, where 

a < b < c.  We can consider just two numbers,

(a+b) and c.

We want to choose (a+b) and c as far apart as possible.

To do that we choose (a+b) as small as possible, which is 3,
when a=1 and b=2, and that choice will make c as much larger 
than 3 as possible.  In order to have sum 15, c must be 12,
and 12 is as far away from 3 as we can get.  So the smallest 
possible product is 1*2*12 = 24.

Edwin

By dividing 161 by odd prime numbers, trying to find one that
will give a whole number, when we come to dividing 161 by 7 
we get the integer 23.  So

161 = 7×23

So the two-digit number "a" with the digits reversed was 23,


Therefore a = 32, and b = 7.

So (ab)² = (32×7)² = 224² = 50176

Edwin

The procedure is: 

1. Put parentheses around the denominator:
2. Create the conjugate of the denominator and put it in parentheses
3. Put that conjugate over itself (which has value 1)
4. Multiply by that.  (multiplying it won't affect the value since it equals 1)

1. Put denominator in parentheses: 

2. To create the conjugate of , just change the
   sign of the second term, making the conjugate .
   Put it in parentheses .
3. Put that conjugate over itself, like this 
4. Multiply by that:






FOIL out the bottom:



The two middle terms cancel out on the bottom:





On the bottom,  is just 5 and  is just 3







Divide top and bottom by 2



Edwin

             

Factor the numerator on the right side as the sum of 
two cubes, using 

                    

                   

                   

Use the identity  on the 1st and 3rd terms:

                   

Use the identity  multiplied
through by  or 

                   

                   

Edwin

First we'll develop a polynomial equation with
solutions 1,1,-5/2, and -1.

  x = 1,    x = 1,      x = -5/2,    x = -1
x-1 = 0   x-1 = 0   x+5/2 = 0      x+1 = 0

         (x-1)(x-1)(x+5/2)(x+1) = 0

To avoid fractions multiply through by 2

      (x-1)(x-1)(2)(x+5/2)(x+1) = 0
          (x-1)(x-1)(2x+5)(x+1) = 0
            (x²-2x+1)(2x²+7x+1) = 0
               2x4+3x³-7x²-3x+5 = 0

That's a quartic equation with those roots.
Therefore a quartic function with those zeros is:

   p(x) = 2x4+3x³-7x²-3x+5

To graph it

1. the coefficient of the leading term 2x4 is positive
and therefor the graph goes upward on the extreme right side.
2. the degree is 4, which is an even number, so the graph
also goes up on the extreme left. 
The first zero on the left is -5/2 or -2.5, so the curve
comes downard from the left and since -2.5 has odd multiplicity,
1, it cuts through the x-axis at -2.5 going downward to the right.
2. The next zero is -1, so the graph must turn around and go
upward.  Since -1 has odd multiplicity, 1, it cuts through the 
x-axis at -1 going upward to the right.
3. The next and final zero is 1, so the graph must turn around and go
downward.  Since 1 has even multiplicity, 2, the graph bounces off 
the x-axis at 1 going back upward to the right. 
You might plot these points:
(-2.6,4.1), (-2.5,0), (-2,-9), (-1.3,-3.8), (-1,0), (0,5), (.5,2.25),
(1,0),(1.5,5)    
   


Edwin