The degree three polynomial f(x) with real coefficients and
leading coefficient 1, has 4 and 3+i among its roots. 
Express f(x) as a product of linear and quadratic polynomials
polynomials with real coefficients. This question I'm having 
trouble with but I came up witha an answer of:
f(x)=(x-4)(x^2-6x-9) is this right. thanks for looking at 
this question.

No that's not correct.  You must know the following facts about
polynomials:

1. A degree n polynomial has n roots, counting multiplicities.
2. If r is a root of a polynomial then (x-r) is a factor of the
   polynomial.
3. If a polynomial with real coefficients has p+qi as one root,
   it also has its conjugate p-qi as another root.

A, By 2 above, since 4 is a root then (x-4) is a factor.
         (You did this)
B. By 2 above, since 3+i is a root then [x-(3+i)] is a factor
   of the polynomial. 
C. By 3 above, since 3+i is a root then 3-1 is a factor of the
   polynomial.
D. By C and 2, since 3-i is a root then [x-(3-i)] is a factor 
   of the polynomial.
E. By 1, there are no more roots besides 4, 3+i, and 3-1

Therefore f(x) is the product of the three factors. It will 
have leading coefficient 1 because the c oefficients of x in 
all three factors is 1.

[Note: If it had had a leading coefficient other than 1, we 
 would have to multiply the polynomial by it, too, but thet 
 is unnecessary here.]

So we have:

f(x) = (x - 4)[x - (3+i)][x - (3-i)]

But the two factors containing imaginary numbers have to be
multiplied together, since all coefficients must be real.
We can use FOIL to multiply the last two factors together:

f(x) = (x - 4)[x - (3+i)][x - (3-i)]
f(x) = (x - 4)[x² - (3-i)x - (3+i)x + (3+i)(3-i)]
f(x) = (x - 4)[x² - (3x-ix) - (3x+ix) + (9-3i+3i-i²)]
f(x) = (x - 4)[x² - 3x + ix - 3x - ix + (9-i²)]
f(x) = (x - 4)[x² - 6x + 9 - i²]

Now since i² = -1 we substitute (-1) for i².

f(x) = (x - 4)[x² - 6x + 9 - (-1)]
f(x) = (x - 4)[x² - 6x + 9 + 1]
f(x) = (x - 4)[x² - 6x + 10]
f(x) = (x - 4)(x² - 6x + 10)

Edwin
AnlytcPhil@aol.com

 
    7
---------
 √3 – √2


To rationalize a denominator with has two terms, you must

1. Form the conjugate surd
   A.  Write the denominator √3 - √2
   B. Change the sign of the second term only √3 + √2
   

2. Form the fraction by putting the conjugate surd over
   itself. 

    √3 + √2
   ---------   (Notice that this equals the number 1]
    √3 + √2

3. Multiply the original fraction by this fraction.  The
   value won't change because this amounts to multiplying
   by 1.

      7        √3 + √2
  --------- × ---------
   √3 – √2     √3 + √2

4. Indicate the multiplication of numerators ans denominators

       7(√3 + √2)
  -------------------
   (√3 – √2)(√3 + √2)    
  
5.  Multiply the top and the bottom  

           7√3 + 7√2
   ------------------------
      3 + √3√2 - √2√3 - 2

6. The middle two terms in the bottom cancel out

    7√3 + 7√2
   -----------
      3 - 2

    7√3 + 7√2
   -----------
        1

    7√3 + 7√2
  
Edwin
AnlytcPhil@aol.com

Give the equation for the circle with center
 C(3,-2) and radius 4. Thanks!

You just have to learn the equation

(x-h)² + (y-k)² = r²

Where the center = (h,k) = (3,-2)

Substitute 3 for h and -2 for k and 4 for the radius r.

Then you have

(x-3)² + (y-(-2))² = 4²

You can simplify

(x-3)² + (y+2)² = 16

And you can multiply it out if you like

(x-3)(x-3) + (y+2)(y+2) = 16

x² - 3x + 9 + y² + 4y + 4 = 16

x² + y² - 3x + 4y + 13 = 16

x² + y² - 3x + 4y - 3 = 0

Edwin

If P(4,-5) is a point on the graph of the 
function y=f(x), find the corresponding point 
on the graph of y = 2f(x-6). Not sure but I 
think it is (6,8).  Thanks for looking it over. 

I'm afraid that's wrong.

y = 2f(x-6)

This requires two operations of y = f(x)

1. First we go from

y = f(x)

to

y = f(x-6)

That shifts the graph 6 units right.
So the point (4,-5)
 


moves right to the point (10, -5) on the graph of

y = f(x-6)

 

2. Then we go from

y = f(x-6)

to

y = 2f(x-6) 

That stretches the graph double vertically.
Imagine the graph drawn on a rubber sheet
and stretched double vertically.  The points
above the x-axis will stretch twice as high
and the points below the x-axis will stretch
twice as low.  Therefore the point (10. -5), 
being below the x-axis, will stretch twice
as low or down to (10, -10)



Edwin
AnlytcPhil@aol.com

An object is projected upward from the top of a water 
tower. Its distance in feet above the ground after t 
seconds is given by s(t) = -16t^2 + 64t + 80. How 
many seconds will it take to reach ground level? I 
came out with 4 seconds after working it out. thanks 
for checking.

Let's check your answer and see if it is correct:
We substitute 4 seconds into the equation:

                     s(t) = -16t² + 64t + 80              
                     s(4) = -16(4)² + 64(4) + 80                      
                     s(4) = -16(16) + 256 + 80
                     s(4) = -256 + 256 + 80
                     s(4) = 80 feet above ground

After 4 seconds the object is still 80 feet above 
the ground.  So 4 seconds can't be right. Let's
do it right:

Ground level is ground zero, that is, when the
height above the ground s(t) is zero,
So replace s(t) by 0

                     s(t) = -16t² + 64t + 80

                        0 = -16t² + 64t + 80

          16t² - 64t - 80 = 0

Divide every term through by 16

              t² - 4t - 5 = 0

           (t - 5)(t + 1) = 0 

          t - 5 = 0;  t + 1 = 0
              t = 5       t = -1

Answer: 5 seconds.

Discard the negative answer. It takes 5 seconds
to reach the ground.  The water tank is 80 feet high.
At the instant when 2 seconds have passed it reaches
its maximum height of 144 feet (64 feet above the 
water tower). Then it starts falling back down.  At
the instant when 4 seconds have passed, it is back
even with the tower.  Then after 5 seconds have
passed the object reaches the ground.

Edwin
AnlytcPhil@aol.com

For the set of data below, find the line of best fit.
x -2 -1  0  1  2 
y  7  6  3  2  0 



  a. y = -1.8x + 3.6    
  b. y = -0.54x + 1.95    
  c. y = 3.6x - 1.8    
  d. y = 1.95x - 0.54    
  e. None of the others    

Do you have a TI-82 or higher?

Press STAT ENTER
Put the 5 x-values under L1
Put the 5 y-values under L2 

Press STAT then RIGHT ARROW to highlight CALC
Scroll down to LinReg(ax+b)
Press ENTER

Read

y=ax+b
a=-1.8
b=3.6
···

So substitute -1.8 for a and 3.6 for b in y = ax+b

y = -1.8x + 3.6

So it's choice 1.

Were you supposed to use a formula?

Edwin

If f(x) = 5x - 1, find the following f(a-2)


f(a number or expression) = right hand side

just means to substitute that number or 
expression for x in the right hand side
and then simplify it:

f(x) = 5x - 1

Erase all the x's and replace them with blanks

f___ = 5_ - 1

Make the blanks long enough to hold (a-2)

f_____ = 5_____ - 1


Put (a-2)'s in the blanks.

f(a-2) = 5(a-2) - 1

Erase the blanks:

f(a-2) = 5(a-2) - 1

Simplify the right side:

f(a-2) = 5a - 10 - 1

f(a-2) = 5a - 11

Edwin

solve for x 
logx 25 = -2 

Use the rule that says: 

A logarithmic equation of the form

  logBA = C

is equivalent to and can be rewritten
as the exponential equation

     BC = A

So we rewrite

   logx25 = -2

as

    x-2 = 25

Then we write the x-2 as 1/x2

   1/x2 = 25

Multiply both sides by x2:

      1 = 25x2

Divide both sides by 25 

   1/25 = x2 

Take positive square roots of both sides
(logarithm bases must be positive and ¹ 1,
by definition)
    
    1/5 = x

Edwin

Give the equation for the right half of the circle. C(2,3).
I plotted it out to x=2+squrt9-(y-3)^2.  Not sure about 
the signs if I have them right or not.  thanks for checking.

I will assume the radius is 3.  You forgot to give that.

The equation of the circle is

(x - 2)² + (y - 3)² = 3²

(x - 2)² + (y - 3)² = 9

Since you want the right half, 
solve for x.

(x - 2)² = 9 - (y - 3)²

Since you want the right half,
take positive square roots of
both sides:
            ____________
   x - 2 = Ö9 - (y - 3)² 
                ____________
       x = 2 + Ö9 - (y - 3)² 

Yes it looks like you're right.



Edwin

A bus leaves a station at 1 p.m., traveling west at an average 
rate of 44 mi/h.  One hour later a second bus leaves along the 
same route, traveling at 54 mi/h.  At what time will the two 
buses be 274 mi apart?

Wow!!!!!!  That'll sure take a long time with the second bus 
traveling only 10 mi/hr faster than the first bus!!!!!!! At
only 10 mi/hr faster, the second bus won't even catch up to
the first to pass it until 4.4 hours later at 6:24 PM.  Then
to get 274 miles ahead will take more than a whole day.  Are
you sure you have this one copied right?  It's a ridiculous
problem, but it can be imagined. So I'll do it anyway.  Let's
assume they started on Monday.  It will be at least Tuesday 
evening before the 2nd bus is 274 miles ahead of the first bus,
going only 10 mph faster.  

Make this chart

             D      R    T   
1st bus|    
2nd bus| 

Let t be the time the first bus traveled.
Then since the 2nd bus started an hour later
its traveling time is t-1.  So fill in the 
times

             D      R    T   
1st bus|                 t
2nd bus|                t-1

Fill in the given rates (speeds):

             D      R    T   
1st bus|           44    t
2nd bus|           54   t-1

Use D = RT to fill in the distances:

             D      R    T   
1st bus|    44t    44    t
2nd bus|  54(t-1)  54   t-1

Now the distance the 2nd bus traveled
is 274 miles more than the 1st bus

So we form the equation by

2nd bus's distance = 1st bus's distance + 274

   54(t-1) = 44t + 274 

Answer = 32.8 hours

So if the first bus started at 1 p.m on Monday
the second bus will be 274 miles ahead of the
1st bus at 9:48 p.m Tuesday night.

I think you copied the problem wrong.

Edwin

Give the equation for the circle with center
C(3,-2) and radius 4. I came out with
(x-3)^2+(y-2)^2=16. But I'm not sure if I
worked it our right. thanks for looking it 
over.

You missed one sign

(x - h)² + (y - k)² = r²

h = 3, k = -2

[x - (3)]² + [y - (-2)]² = (4)²

[x - 3]² + [y + 2]² = 16

(x - 3)² + (y + 2)² = 16

To do it quickly just remember
to change both signs of the
coordinates of the center.



Edwin

A recipe calls for 1 1/3 tablespoons
of butter and serves 6 people.If you
wish to cook this recipe for yourself,
how much butter should you use?

You would divide 1 1/3 by 6.

(1 1/3) ÷ 6 =

(4/3) ÷ (6/1)

(4/3) × (1/6)

   4/18

   2/9 tablespoons

Since there are no measuring 
spoons for ninths, we use the
fact that 1 tablespoon = 
3 teaspoons

   2/9 tablespoon =

   (2/9) × (3 teaspoons) =

   (2/9)×(3/1) teaspoon =

    6/9 teaspoon =

    2/3 teaspoon  

There are no third teaspoon
measuring spoons either so you
have to measure a teaspoon and
take a litle bit out.

Edwin

Evaluate h(-2/3) if h(x) =[[x-2]] 

Double brackets indicates the "floor function".
It means the largest integer that does 
not exceed the number between the double 
brackets:

h(x) =[[x-2]]

h(-2/3) = [[-2/3-2]]

h(-2/3) = [[-2/3 - 6/3]]

h(-2/3) = [[-8/3]]

h(-2/3) = [[-2.888888888888···]]

The largest integer that does not exceed
-2.888888888888···  is -3.

So h(-2/3) = -3

Edwin
AnlytcPhil@aol.com

One number is 5 times another...the sum of their
reciprocals is 6/35...what are the 2 numbers?

Smaller number = x

>>...one number is 5 times another...<<

Larger number = 5x

Reciprocal of smaller number = 1/x

Reciprocal of larger number = 1/(5x)

>>...the sum of their reciprocals is 6/35...<<

  1/x + 1/(5x) = 6/35

Can you solve that? If not post again.
Hint: multiply thru by LCD = 35x

Answer: x = 7, So smaller number = x = 7
Larger number = 5x = 5(7) = 35

Edwin

The approach pattern to an airport requires pilots
to set an 11o angle of descent toward a runway. If
a plane is flying at an altitude of 9,500 m, at what
distance (measured along the ground) from the airport
must the pilot decend 

                                 P
                                  ·
                      ·           | 9500 m 
A        ·     11°                |
·---------------------------------
              x                   B

The plane is at P, the Airport is at A,
B is a point on the ground directly under the plane.
We want to find AB, the length of which we call x

9500 is the opposite side to 11°
x is the adjacent side to 11°

  tan(11°) = opposite/adjacent

  tan(11°) = 9500/x

x·tan(11°) = 9500 

         x = 9500/tan(11°)

         x = 48873.26315 m or about 48900 meters.

Edwin
AnlytcPhil@aol.com

Can anyone solve the following?

                27 < 4|y + 9| - 9

You must learn two principles for 
inequalities to remove the absolute
value bars:
-----------------------------------
1. If C > 0
              |Ax + B| < C

   can be rewritten as

         " -C < Ax + B < C " 
-----------------------------------
2. If C > 0

              |Ax + B| > C

   can be rewritten as

   "  Ax + B < -C OR  Ax + B > C  "

------------------------------------  
Note the above two principles also 
hold for < and >          

    -4|y + 9| + 27 < -9

         -4|y + 9| < -9 - 27

         -4|y + 9| < -36

Divide both sides by -4, which reverses
the inequality:

           |y + 9| > 9 

This is principle 2 above

    y + 9 < -9  OR y + 9 > 9

        y < -18 OR y > 0

Plot on a numberline:

 <======o-----------o======>   
      -18           0

Interval notation:

      (-¥, -18) È (0, ¥)  

Edwin
AnlytcPhil@aol.com

Choose the correct function value for: csc 23°20'

csc(23°20') = 1/sin(23°20') 

If you have a TI-82 or better calculator

Press MODE to make sure "Degree" mode is higlighted,
not "Radian"

To get ° press 2nd ANGLE 1
To get ' press 2nd ANGLE 2

Answer 2.524743968

Edwin
AnlytcPhil@aol.com

I know I make common mistakes when solving 
equations can you help with the following: 
The equation: 2(x -2) =x+3 
First step in solving: 2x-2 =x+3 
Can you tell me a clear explanation of what 
error has been made. What could be done to 
avoid this error?

Your error is in the distributive law or 
principle. You have

2(x - 2) = x + 3

Then you have

2x - 2 = x + 3

That is wrong. It should be

2x - 4 = x + 3

That's because when you have 2(x - 2) you multiply 
the red 2 first times the x, which you did,  but ALSO
you must multiply the red 2 times the blue - 2 as well.
What you did was to multiply the red 2 by the x and
then just copy down the blue - 2 without multiplying 
the red 2 by it to get - 4.

Edwin
AnlytcPhil@aol.com

I am working on factoring trinomials. The thing is 
I am so confused with some problems. 

                   2x² + x - 6

can you help me????

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

To factor Ax² ± Bx ± C when there is no common factor 
of |A|, |B|, and |C| and A is positive.

1. Multiply |A| by |C|, getting AC

2. If the last sign is +, think of two positive integers
   which have product AC and which have SUM |B|
   If the last sign is -, think of two positive integers 
   which have product AC and which have DIFFERENCE |B|

3. Rewrite the middle term ±Bx using those two integers 
   found in the preceding step, attaching appropriate
   signs.

4. Factor by grouping.  


In your problem, 2x² + 1x - 6, I placed the 1 coefficient
beside the middle term for emphasis.

1. Multiply |A| by |C|, getting AC

   Multiply |2| by |-6| or 2×6 getting 12

2. The last sign is -, so think of two positive integers 
   which have product AC and which have DIFFERENCE |B|

   So we think of two positive integers which have product
   12 and difference of 1.  These are 4 and 3, because
   4×3=12 and 4-3 = 1 

3. Rewrite the middle term ±Bx using those two integers 
   found in the preceding step, attaching appropriate \
   signs.

   We rewrite +1x as +4x - 3x.  So now we have

   2x² + 4x - 3x - 6


4. Factor by grouping.

   2x² + 4x - 3x - 6

Factor 2x out of the first two terms

  2x(x + 2) - 3x - 6

Factor -3 out of the last two terms.  Notice
that you factor out a negative when the next
to the last term is preceded by a minus sign.
Also when factoring out a nagative, the sign
of the last term changes:

 2x(x + 2) - 3(x + 2)

Notice the common factor (x + 2) which I will
color red for emphasis:

 2x(x + 2) - 3(x + 2)
   
We factor out the common red factor and leave 
the black factors inside parentheses:

  (x + 2)(2x - 3)

  (x + 2)(2x - 3)

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

Here is another example:

Factor  6x² - 19x + 10.

1. Multiply |A| by |C|, getting AC

   Multiply |6| by |10| or 6×10 getting 60

2. The last sign is +, so think of two positive integers 
   which have product AC and which have SUM |B|

   So we think of two positive integers which have product
   60 and sum of 19.  These are 15 and 4, because
   15×4=60 and 15+4 = 19  

3. Rewrite the middle term ±Bx using those two integers 
   found in the preceding step, attaching appropriate signs.

   We rewrite -19x as -15x - 4x.  So now we have

   6x² - 15x - 4x + 10

4. Factor by grouping.

   6x² - 15x - 4x + 10

Factor 3x out of the first two terms

  3x(2x - 5) - 4x + 10

Factor -2 out of the last two terms.  Notice
that you factor out a negative when the next
to the last term is preceded by a minus sign.
Also when factoring out a nagative, the sign
of the last term changes:

 3x(2x - 5) - 2(2x - 5)

Notice the common factor (2x - 5) which I will
color red for emphasis:

 3x(2x - 5) - 2(2x - 5)
   
We factor out the common red factor and leave 
the black factors inside parentheses:

  (2x - 5)(3x - 2)

  (2x - 5)(3x - 2)

Edwin

what can be said of points with equal abscissa? 

"Abscissa" is a fancy name for the 1st coordinate,
or "x-coordinate"

Let's plot some and see.  Below I have plotted (3,7),
(3,3), (3,-4) and (3,-8), all having the same 
abscissa, 3.



Notice that they all lie on the same vertical line.



Incidentally the equation of that vertical line is 
simply this:  

            x = 3 

----------------------------------------------------
with equal ordinates?

"Ordinate" is a fancy name for the 2nd coordinate, 
or "y-coordinate"

Let's plot some and see.  Below I have plotted (-6,-4),
(3,-4), (4,-4) and (7,-4), all having the same 
ordinate, -4.



Notice that they all lie on the same horizontal 
line.



Incidentally the equation of that horizontal line is 
simply this:  

            y = -4 

Edwin

Find the exact value of sin 2θ if
cosθ = -(√5)/3 and 180° < θ < 270°. 
Thank You 

We need to use the formula

sin(2θ) = 2sinθ·cosθ

We know cosθ but not sinθ.  We have to find sinθ.

We are given 180° < θ < 270° so we know θ is in 
quadrant III.  So we draw the picture of θ in the
third quadrant

    _ |
  -√5 |
-------------
  |  /|
  | /3|
  |/  |

Since the cosine is x/r or adjacent/hypotenuse, we
                       _             _ 
put the numerator of -√5/3, namely -√5 the on the x-side
(or adjacent side) and 3 on the r, the radius vector (or
hypotenuse).

We use the Pythagorean theorem to find the y-side (or the
opposite side).
      _______     ____________      ___     _
y = ±√r² - x² = ±√(3)² - (-√5)² = ±√9-5 = ±√4 = ±2

We know to take the negative sign since y goes down
from the x-axis, so we have

    _ |
  -√5 |
-------------
  |  /|
-2| /3|
  |/  |
                        
Now we know sinθ = -2/3

So
sin(2θ) = 2sinθ·cosθ 
                     _ 
sin(2θ) = 2(-2/3)·(-√5/3)
                     _ 
sin(2θ) = 2(-2/3)·(-√5/3)
            _ 
sin(2θ) = 4√5/9

Edwin
AnlytcPhil@aol.com

Find the exact value of sin θ/2 if cos θ = 2/3 and 270° < θ < 360°.

Use the formula 
             __________
sin(θ/2) = ±Ö(1-cosθ)/2

Where the sign of the sine is taken according to which quadrant x
is in.

It is unfortunate that the words "sine" and "sign" are homophones,
because it sounds funny and is often confusing to speak of "the
sign of the sine".  Some teachers use "positiveness or
negativeness of the sine" instead of "sign of the sine".

Anyway, substitute 2/3 for cosθ in the formula:

             __________
sin(θ/2) = ±Ö(1-2/3)/2 

Change the last "/" to a "÷"

             _______ 
sin(θ/2) = ±Ö(1/3)÷2 

Change the 2 to 2/1
             ___________
sin(θ/2) = ±Ö(1/3)÷(2/1)

Invert and multiply to perform the divion under
the radical:
             ___________   
sin(θ/2) = ±Ö(1/3)×(1/2) 
             ___   
sin(θ/2) = ±Ö1/6

Now we have to determine whether to use the + or
the - .

We are givem:

270° < θ < 360°

so we whii divide all three sides of that by 2

(270/2)° < θ/2 < (360/2)°

 135° < θ/2 < 180°

This tells us that θ/2 is in quadrant II.
The sine is positive in the second quadrant
Therefore we know that  sin(θ/2) is positive,
So
            ___
sin(θ/2) = Ö1/6

Your teacher may want you to rationalize the
denominator and get
            _   
sin(θ/2) = Ö6/6
 
Edwin

what do points on the line parallel to the y-axis in 3 units to the left of it
have in common?with a summary.

Draw the line. Then mark some point arbitrarily and compare their coordinates
to see what they have in common.

 

I have marked some points arbitrarily on that line.  The top one is (-3,7), the
next one down is (-3,3), the bottom one is (-3,-8).  Notice that they all have
the same x-coodinate, namely -3.

That's the answer.  All the points on that line have in common the fact that
their x-coordinates are -3.  

Edwin
AnlytcPhil@aol.com

I am lost on this problem... 
radical -50 times radical -2
 ___  __
Ö-50 Ö-2

Don't fall for the trap of always multiplying under radicals first.
Don't do that!  Negative numbers under square roots must be taken
out of the radical before multiplying under them:
 ___  __      __    _        __  _        ___ 
Ö-50 Ö-2 = (iÖ50)(iÖ2) = i² Ö50 Ö2 = (-1)Ö100 = (-1)(10) = -10 


If you multiply under the radical before taking negative signs
out of square root radicals as i's on the outside, you will get
the wrong answer.

Edwin
AnlytcPhil@aol.com

In which year did the greatest percent increase in filings occur? 



Percent increase is figured by the formula

(New value minus old value)÷(old value)×100%
	
In 1997 the number of bankruptcies was 14 thousand
	
In 1998 the number of bankruptcies was 16 thousand, which was an
increase over the 14 thousand in 1997 of 16-14 or 2 thousand. The
percent increase was 2÷14×100% = 14.3%, approximately 

In 1999 the number of bankruptcies was 21 thousand, which was an
increase over the 16 thousand in 1998 of 21-16 or 5 thousand. The
percent increase was 5÷16×100% = 31.3%, approximately 

In 2000 the number of bankruptcies was 24 thousand, which was an
increase over the 21 thousand in 1990 of 24-21 or 3 thousand. The
percent increase was 3÷21×100% = 14.3%, approximately 

In 2001 the number of bankruptcies was 30 thousand, which was an
increase over the 24 thousand in 2000 of 30-24 or 6 thousand. The
percent increase was 6÷24×100% = 25%, exactly.

So the greatest percent increase in filings was in 1999 with 31.3%
approximate percentage increase over the 16 thousand filed the
year before in 1998.

Edwin

Simplify: (3 1/2) ÷ (2 3/5) × (1 1/4) ÷ (7/13)

(show your work)

(3 1/2) ÷ (2 3/5) × (1 1/4) ÷ (7/13)

Change the mixed fractions to improper fractions:

(7/2) ÷ (13/5) × (5/4) ÷ (7/13)

Invert the fractions that are divided and change the
division to a multiplication

(7/2) × (5/13) × (5/4) × (13/7)

or

 7      5     5     13
——— × ———— × ——— × ————
 2     13     4      7

Cancel the 7 numerator in the first fraction
and the 7 denominator in the last fraction:
 
 1
 7      5     5     13
——— × ———— × ——— × ————
 2     13     4      7
                     1

Cancel the 13 denominator in the second fraction
and the 13 numerator in the last fraction:

 1                  1 
 7      5     5     13
——— × ———— × ——— × ————
 2     13     4      7
       1             1

So you have

 1     5     5     1
——— × ——— × ——— × ———
 2     1     4     1

Multiply the numerators 1×5×5×1 = 25
Multiply the denominators 2×1×4×1 = 8

You get

 25
————
  8

Finally we can change that to a mixed number

3 1/8

Edwin
AnlytcPhil@aol.com

Write the equation sin(y) = x in the form of an inverse function.

Thank you

The equation in words would be

"The sine of y is x"

That sentence can be written this way:

"y is what we must take the sine of to get x"

The mathematical symbol for the words

"what we must take the sine of to get"

is 

sin-1

That is not really a "negative one exponent", it's
only written that way because a negative
one exponent is what inverts a fraction, and that 
reminds us that we are writing another kind of 
inverse somewhat analogous to that, but far from
meaning that.

So from the beginning:

sin(y) = x

translates as

"The sine of y is x"

which translates as 

"y is what we have to take the sine of to get x"

which translates as

y = sin-1(x)

Edwin
AnlytcPhil@aol.com

which number is a factor of 20? 

A) 1

B) 12

C) 40

D) 60

You must learn that the words

"IS A FACTOR OF"

are exactly equivalent to the words

"WILL DIVIDE EVENLY INTO WITHOUT
LEAVING A REMAINDER (OR A REMAINDER
OF 0)."

So pretend the problem were stated this way:

which number WILL DIVIDE EVENLY INTO 20 WITHOUT 
LEAVING A REMAINDER (OTHER THAN 0)? 

A) 1

B) 12

C) 40

D) 60

Let's see. We'll start with the last one.

Try (D).  WILL 60 DIVIDE EVENLY INTO 20 WITHOUT
LEAVING A REMAINDER OTHER THAN 0? Let's see:

    0
60)20
    0
   20

No that leaves a remaider of 20

Try (C).  WILL 40 DIVIDE EVENLY INTO 20 WITHOUT
LEAVING A REMAINDER? Let's see:

    0
40)20
    0
   20

No that also leaves a remaider of 20
   
Try (B).  WILL 12 DIVIDE EVENLY INTO 20 WITHOUT
LEAVING A REMAINDER? Let's see:

    1
12)20
   12
    8

No that leaves a remaider of 8 

Try (A).  WILL 1 DIVIDE EVENLY INTO 20 WITHOUT
LEAVING A REMAINDER OTHER THAN 0? Let's see:

  20
1)20
  2
   0
   0
   0

YUP, that only leaves a remainder of 0,
so that's the answer.

Edwin
AnlytcPhil@aol.com