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1. What is the twenty-fifth term of the arithmetic sequence with a1 = –1 and d = –10? 

Substitute a1=-1, d=-10, and n=25 in

an = a1 + (n-1)d

Answer: -241 


2.What is the twenty-ninth term of arithmetic sequence with a1 = 13 and d = –5/2? 

Substitute a1=13, d=-5/2, and n=29 in

an = a1 + (n-1)d

Answer: -57 

3. What are the two arithmetic means between –13 and 8? 

a1 - -13, a2 = ?, a3 = ?, a4 = 8

Each term differs from the preceding term by d.

a2 is a1 + d, and a3 is a4 - d, so

a1 = -13, a2 = -13+d, a3 = 8-d, a4 = 8
 
 a2 + d = a3
-13+d+d = 8-d
 -13+2d = 8-d
     3d = 21
      d = 7


 a2 = -13+7 = -6
 a3 = 8-7 = 1

-13, -6, 1, 8

They are -6 and 1



4. What is Sn for the arithmetic series with d = –4, an = 27, and n = 9? 

First substitute an = 27, d=-4, and n=9 in

an = a1 + (n-1)d

and solve for a1.

Get a1 = 59

Substitute a1 = 59, d = -4, leave n as just n in

Sn = (n/2)[a1 + (n-1)d]

Simplify

Get

Sn = n(61-2n)

Edwin

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                          x² -  3x +  2
x² + 3x - 18)x⁴+ 0x³ - 25x² + 62x - 36
             x⁴+ 3x³ - 18x²
                 -3x³ -  7x² + 62x
                 -3x³ -  9x² + 54x
                         2x² +  8x - 36
                         2x² +  6x - 36
                                2x +  0

Answer:  x²-3x+2+


Edwin

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, a divergent series

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We have to get it either in the form:

          +  = 1 

in which the ellipse will look like this ""

or this form:

          +  = 1

in which the ellipse will look like this ""

9x² + 4y² - 54x + 16y + 61 = 0

Get the x terms together, and the y terms together.

9x² - 54x + 4y² + 16y + 61 = 0

Get the 61 off the left side by adding -61 to both sides

9x² - 54x + 4y² + 16y = -61

Factor the 9 out of the first two terms on the left
(factor out just the 9, not the x)

9(x² - 6x) + 4y² + 16y = -61

Factor the 4 out of the last two terms on the left
(factor out just the 4, not the y)

9(x² - 6x) + 4(y² + 4y) = -61

Complete the square inside the first parentheses:
    Multiply the coefficient of x, which is -6 by , getting -3
    Square -3 getting (-3)² = 9
    Add 9 inside at the end of the first parentheses:

9(x² - 6x + 9) + 4(y² + 4y) = -61 + 81

Notice that to offset adding the 9 inside the parentheses,
we had to add 81 to the right side.  That's because when
we added 9 inside the first parentheses on the left, that 
actually amounted to adding 9·9 or 81 to the left side 
because of the 9 coefficient in front of the first parentheses 
on the left.
   
Complete the square inside the second parentheses:
    Multiply the coefficient of y, which is 4 by , getting 2
    Square 2 getting (2)² = 4
    Add 4 inside at the end of the second parentheses:

9(x² - 6x + 9) + 4(y² + 4y + 4) = -61 + 81 + 16

Notice that to offset adding the 4 inside the parentheses,
we had to add 16 to the right side.  That's because when
we added 4 inside the second parentheses on the left, that 
actually amounted to adding 4·4 or 16 to the left side 
because of the 4 coefficient in front of the second parentheses 
on the left.

Next we factor the two parentheses and combine the terms on the
right:

9(x - 3)(x - 3) + 4(y + 2)(y + 2) = 36

Those factorizations can be written shorter as the squares of
binomials:

         9(x - 3)² + 4(y + 2)² = 36

In fact most people skip the step before the last one.

Next we must get a 1 on the right side.  So we divide each
term by 36:

          +  = 

We simplify:

          +  = 1     

Now we compare that to

         +  = 1

and the ellipse will look like this ""

We can tell that because a² is larger than b². 

On comparing the two we see that h=3, k=-2, b²=4 or b=2, a²=9, or a=3

The center is (h,k) = (3,-2). 

We plot the center (3,-2)



We draw the major axis vertically, which is bisected at the center.
We count a=3 units up from the center and a=3 units down from the
center to the vertices.  The vertex which is 3 units above the center
(3,-2) is the point (3,1).  The vertex which is 3 units down from
the center is the point (3,-5):


  
We draw the minor axis vertically, which is also bisected at the center.
We count b=2 units left from the center and b=2 units right from the
center to the covertices.  The covertex which is 2 units left of the center
(3,-2) is the point (1,-2).  The covertex which is 2 units right of
the center is the point (5,-2):



Now we can sketch in the ellipse:


 
All we need to do now is to find the foci. They are two points inside
the ellipse on the major axis each of which is "c" units from the center,
where c is calculated from this Pythagorean relation:

c² = a² - b²
c² = 9 - 4
c² = 5
 c = .

To find the coordinates of the upper focus, we add  to the
y coordinate of the center and get the point (3,-2+). To
get the lower focus, we subtract  from the y coordinate of the
center and get the point (3,-2-). 

We plot them:



Center: (3,-2), Vertices: (3,1) and (3,-5), Covertices: (1,-2) and (5,-2)
Foci: (3,-2-) and (3,-2+), Major axis: 2a = 6,
Minor axis: 2b = 4.  Eccentricity:  = 

Edwin

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x·x·x·x·y·y·y·z = x⁴y³z

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We can't subtract because they don't have the same 
denominator.  The LCD of 9 and 6 is 18, because 
both will divide evenly into 18. So we change 
to  and  to 



We still can't subtract because the upper fraction  
has a smaller numerator than the lower fraction .
So we borrow 1 from the 9, leaving it with 8, and that gives 
us  to add to the fraction  which just amounts 
to adding the denominator 18 to the numerator 10, getting 28, so 
we have this:

 

And now we can subtract because the numerator above is bigger
than the numerator below. So  8 minus 6 gives 2 for the whole part
of the answer, and 28-15 gives 13 for the numerator of the answer
so we subtract and get:

 

Edwin

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y = 3x + 5 

The slope is the coefficient of x, which is 3 when the equation of the line is
solved for y.  It is positive so I colored it pink

Here is the line:




y = 0.5x + 3 

That one has slope 0.5, which is also positive.  

Here is its graph:



Lines that go uphill to the right like this / have positive slopes.  
Lines that go downhill to the right like this \ have negative slopes.

Edwin

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logB(A) asks the question:

To what exponent must the base B be raised to give A?  
That exponent is the answer to what logB(A) equals. 

Because logarithms are exponents, and because we can 
add exponents of a base in order to multiply, subtract them in 
order to divide, and multiply them to raise a power to a power, 
we have these three corresponding rules for logarithms:

1.  logB(A·C) = logB(A) + logB(C)
2.  logB = logB(A) - logB(C)
3.  logB(AC) = C·logB(A)

[If the base B isn't written, it's understood to be 10.]
 
--------------------------------------

1) log(7) = 0.8
log(12) = 1.1
log(8)= 0.9
Find log

Use rule 2:

2.  logB = logB(A) - logB(C)

with A=1, B=64, and the base B understood as 10

    log = log(1) - log(64)

Now we use the definition of logarithms to find log(1).
We ask the question "To what power must the base 10 be raised to get 1?"
If you remember that 100 = 1, then you know that log(1)
is 0.  In fact the logarithm of 1 is always 0, regardless of the base.  
So now we have

    log = 0 - log(64)
or
     -log(64)

We are given the values of these three logs with understood base 10: 
 log(7) = 0.8, log(12) = 1.1, log(8)= 0.9

Now we ask:  Which of these numbers 7, 12, or 8, can we multiply
together, divide, or raise to a power, to get 64?
The answer: we can get 64 by squaring 8. That is, 8² = 64.

So we replace 64 by 8² and now we have:

     -log(8²)

Now we use rule 3:

3.  logB(AC) = C·logBA
with A=8 and C=2

and we have:

   -log(82) = -2·log(8)

And since we are given log(8) = 0.9

   -2·log(8)
   -2·(0.9)
     -1.8 

That's the answer, -1.8.  Try hard to follow the above.

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


2)log5(11) = 1.5
log5(6) = 1.1
log5(4) = 0.9
Find log5(264)

We are given the values of these three logs with base 5: 
log5(11) = 1.5, log5(6) = 1.1, log5(4) = 0.9

Now we ask:  Which of these numbers 11, 6, and 4, can we multiply
together, divide, or raise to a power, to get 264?

To answer that we must see if 264 can be divided evenly by one of those.
We find the 264χ11 = 24. So we write 
264 = = 11·24 and we write

   log5(264) = 
   log5(11·24)

Now we use rule 1:

1.  logB(A·C) = logB(A) + logB(C)
with A=11, B=5, C=24

    log5(11·24) = log5(11) + log5(24)

We substitute the given log5(11) = 1.5, and we have:

    1.5 + log5(24)

Now we ask:  Which of these numbers 11, 6, and 4, can we multiply
together, divide, or raise to a power, to get 24?  That answer is easy.
24 = 6·4.  So we replace 24 by 6·4, and we have:

    1.5 + log5(6·4)

Now we use rule 1 again:

1.  logB(A·C) = logB(A) + logB(C)
this time with with A=6, B=5, C=4


    1.5 + log5(6) + log5(4)
   
and we are give those logs, so we substitute:

    1.5 + 1.1 + 0.9

Answer: 3.5

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

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Since you're given all corresponding parts to be congruent, you can take your

pick of any one of these four theorems:



Side-angle-side, Angle-side-angle, Side-angle-side or Side-side-side.

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The digits are 1,2,3,4,5,6,7,8,9,and 0

The even ones are 0,2,4,6,8.  The odd ones are 1,3,5,7,9

Let's try the even digit 0.

0Χ0 = 0.  There is only one digit, so that can't be it.

Let's try the even digit 2.

2Χ2 = 4.  There is only one digit, so that can't be it either.

Let's try the even digit 8.

8Χ8 = 64.  Let's find the sum of the digits 6+4=10.  No, 10 is not 9,
so it's not 8

I skipped the even digit 6, didn't I?  

Let's try the even digit 6

6Χ6 = 36 and the sum of its digits is 3+6 or 9.

What do you know?  That must be the answer, 6.

Now was that really too difficult for you to answer all by yourself?

Edwin

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 +  = e

 +  = 

To clear of fractions, multiply through by LCD of 

 +  = 

 +  = 

da + bxc = bdxe

Get all terms in x on the right by subtracting bxc from both sides:

      da = bdxe - bxc

Factor bx out of the right side:

      da = bx(de - c)

Divide both sides by b(de - c)

 = 

 = 

 = x

x =  

Edwin

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1 + tan(A)·tan = tan(A)cot - 1 = sec(A)

We will make use of the half angle identities:

tan = , and its reciprocal cot = , also tan(A) =  and sec(A) = 

First we will prove:

1 + tan(A)·tan = sec(A)

1 +  

1 + 

Get an LCD

 + 

Combine fractions over the LCD



Cancel first and third terms on top:





Cancel the sines:





sec(A)

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

Next we will prove:

tan(A)·cot - 1 = sec(A)

 - 1 

 - 1

Get an LCD

 - 

Combine fractions over the LCD



Distribute:



Rearrange:



Use identity sin²(A)+cos²(A)=1



Cancel 1-cos(A)'s





sec(A)

Edwin

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How many pounds of gourmet candy selling for $2.80 per pound should be mixed with 7 pounds of gourmet candy selling for $1.60 per pound to obtain a mixture selling for $1.96 per pound?

 +  = 
     ↓      ↓       ↓      ↓       ↓ 
     ↓      ↓       ↓      ↓       ↓
  $2.80x    +    $11.20    =   $1.96(x+7)

Drop the dollar marks and move the decimals two places right

               280x + 1120 = 196(x+7)

               280x + 1120 = 196x + 1372

Subtract 1120 from both sides:

                      280x = 196x + 252

Subtract 196x from both sides:

                       84x = 252

Divide both sides by 84

                       = 
                          
                        x = 3

Answer 3 pounds.

Edwin

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A familiar identity is  tan = , and
since the cotangent is the reciprocal of the tangent, we have
cot = ,  so we will try to make
the left side into that expression. So we will multiply the left side
by , so hopefully it will have the desired denominator in the end. 

 
1 + sin(x) + cos(x)
——————————————————— = cot()
1 + sin(x) - cos(x)  


[1 + sin(x) + cos(x)]   [1 - cos(x)] 
————————————————————— · ————————————
[1 + sin(x) - cos(x)]   [1 - cos(x)]

1 - cos(x) + sin(x) - sin(x)cos(x) + cos(x) - cos²(x) 
—————————————————————————————————————————————————————
       [1 + sin(x) - cos(x)][1 - cos(x)]

1 - cos(x) + sin(x) - sin(x)cos(x) + cos(x) - cos²(x) 
—————————————————————————————————————————————————————
       [1 + sin(x) - cos(x)][1 - cos(x)]

      1 + sin(x) - sin(x)cos(x) - cos²(x) 
      ———————————————————————————————————
       [1 + sin(x) - cos(x)][1 - cos(x)]

Rearrange the terms

      1 - cos²(x) + sin(x) - sin(x)cos(x)  
     —————————————————————————————————————
       [1 + sin(x) - cos(x)][1 - cos(x)]

     [1 - cos²(x)] + sin(x) - sin(x)cos(x)  
     —————————————————————————————————————
       [1 + sin(x) - cos(x)][1 - cos(x)]
  
Use the identity sin²(x) = 1 - cos²(x)

        sin²(x) + sin(x) - sin(x)cos(x)  
       —————————————————————————————————
       [1 + sin(x) - cos(x)][1 - cos(x)]

Factor out sin(x) on the top:

         sin(x)[sin(x) + 1 - cos(x)]  
       —————————————————————————————————
       [1 + sin(x) - cos(x)][1 - cos(x)]

Then we can cancel:

         sin(x)[sin(x) + 1 - cos(x)]  
       —————————————————————————————————
       [1 + sin(x) - cos(x)][1 - cos(x)]
 
                  sin(x)  
                ——————————
                1 - cos(x)

which is the identity we showed above for cot()

                cot

Edwin

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This is one tough identity! It requires very unusual substitutions,
additions and subtractions of the same quantities.

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

                  = cos(y+y) - cos(x+x)
             
                  = cos(y+y+x-x) - cos(x+x+y-y)

                  = cos[(x+y)-(x-y)] - cos[(x+y)+(x-y)]

     Use the identities cos(A∓B)=cos(A)cos(B)±sin(A)sin(B)
     with A=(x+y), B=(x-y)

= [cos(x+y)cos(x-y)+sin(x+y)sin(x-y)] - [cos(x+y)cos(x-y)-sin(x+y)sin(x-y)]    

= cos(x+y)cos(x-y) + sin(x+y)sin(x-y) - cos(x+y)cos(x-y) + sin(x+y)sin(x-y)

= cos(x+y)cos(x-y) + sin(x+y)sin(x-y) - cos(x+y)cos(x-y) + sin(x+y)sin(x-y)

                 = 2sin(x+y)(sin(x-y)

Edwin

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I can't tell which of these you meant:

4log(4x) - 3log(5y) = 1     or   4log4(x) - 3log5(y) = 1
 log(2x) -  log(5y) = 2           log2(x) -  log5(y) = 2

I tried doing it the first way and there was no solution, 
so I think you must have meant the second way.  

4log4(x) - 3log5(y) = 1
 log2(x) -  log5(y) = 2

I will change the first term in the first equation so 
it will be a log to the base 2, like the first term in
the second equation.  Using the change of base formula 
on that term:

4log4(x) =  =  =  =  = 2log2(x) 

The system of equations is now:

2log2(x) - 3log5(y) = 1
 log2(x) -  log5(y) = 2

let u = log2(x) 
let v = log5(y)

The system of equations is now:

2u - 3v = 1
 u -  v = 2

Solve that system of equations by substitution or elimination
and get (u,v) = (5,3).  I'm sure you can do that.

But we don't want u and v, we want x and y.
So we substitute back:

u = log2(x), which is equivalent to the exponential equation:

x = 2u = 25 = 32

v = log5(y)

y = 5v = 53 = 125

So the solution is

(x,y) = (32,125)

Edwin

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10x²y³z³ – 6x²y²z – 4xy³z²

10, 6, and 4 can all be divided by 2, so we can take out a 2 factor

x², x², and x can all be divided evenly by x, so we can take out an x factor

y³, y², and y³ can all be divided evenly by y², so we can take out a y² factor 

z³, z, and z² can all be divided evenly by z, so we can take out a z factor

So we can take out 2xy²z. So we write this:

2xy²z(

To get the first term we divide 10x²y³z³ by 2xy²z and get 5xyz², so we
write that first in the parentheses:

2xy²z(5xyz²

To get the second term we divide –6x²y²z by 2xy²z and get -3x, so we
write that second in the parentheses:

2xy²z(5xyz² - 3x 

To get the second term we divide –4xy³z² by 2xy²z and get -2yz, so we
write that third in the parentheses and close the parentheses:

2xy²z(5xyz² - 3x - 2yz)

Edwin

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Since AB is congruent to AC, we know that ᐃABC is
isosceles.  Therefore we know that its base angles are
congruent and thus have the same measure.  Therefore we
will label the measure of the other base angle C, m∠A = 40°
as well:
 

Since we know that the sum of the measures of the three angles of any 
triangle must always equal to 180°, we can write

m∠A + m∠B + m∠C = 180°

m∠A + 40° + 40° = 180°

      m∠A + 80° = 180°

            m∠A = 180° - 80°

            m∠A = 100°



Edwin

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450 divided by 270

Here are two ways:

First way:

      1 
270)450
    270
    180

 =  = 

Second way:



Since both end with a 0, they can both be divided by 10, so:



To divide by 10, just drop off the 0's on the end and you get



Now if we remember our 9's multiflication facts, we
remember that 5Χ9 = 45 and 3Χ9 = 27, so



and you get



Then you change that to a mixed number and get:



Edwin

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Substitute the value of the radius for r in

x² + y² = r²

For instance, to get the answer to 

find the equation of a circle centered at the origin and having radius 5

just substitute 5 for r and get

x² + y² = 5²

and then do one more step, change 5² to 25

x² + y² = 25

That's the equation.

That's all there is to it!

Edwin

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Three friends a,b,c can do a piece of work in T hours working together. working alone, a can do the work in 6 hours more, b in 1 hour more and c in twice the time if all of them were working together. How long would it take to finish the work if all of them were working together (find the value of T)?

Make this chart.

                        number of      number of    time in            
                        pieces of        hours      pieces of
                          work         required     work/hour
a working alone

b working alone

c working alone

all 3 working together 

In each case exactly 1 piece of work is completed, so we put 
1 for the number of pieces of work in all 4 cases.  We also put
in T for the time for all three working together, T+6 hours for 
a's time, T+1 hours for b's time, and 2T for c's time.

                        number of      number of     rate in            
                        pieces of        hours      pieces of
                          work         required     work/hour
a working alone             1             T+6

b working alone             1             T+1

c working alone             1              2T

all 3 working together      1              T

We fill in the rates in pieces of work/hour by dividing the
number of pieces of work by the hours required:

                        number of      number of     rate in            
                        pieces of        hours      pieces of
                          work         required     work/hour
a working alone             1             T+6        
b working alone             1             T+1        
c working alone             1              2T         
all 3 working together      1              T          

The equation comes from:

        +  +  =  

              +  +  = 


Solve that equation and get T =  of an hour or 40 minutes.

Edwin

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Let the coordinates of the circumcenter (or "circumcentre", 
as you spell it in the UK) be O(h,k).  Then the circumradius
r = AO = BO = CO.

We use the distance formula d = 

AO = 
BO =  = 
CO =  = 

 =  = 

Squaring all three expressions:

(h-4)² + (k-3)² = (h+3)² + (k-2)² = (h-1)² + (k+6)² 

Equating the first two:

(h-4)² + (k-3)² = (h+3)² + (k-2)²

Rearrange terms to create the difference of squares:

(h-4)² - (h+3)² = (k-2)² - (k-3)² 

Factor (or as they say in UK, "factorise")

[(h-4) - (h+3)][(h-4) + (h+3)] = [(k-2) - (k-3)][(k-2) + (k-3)]

[h - 4 - h - 3][h - 4 + h + 3] = [k - 2 - k + 3][k - 2 + k - 3]

[-7][2h-1] = [1][2k-5]
-14h + 7 = 2k - 5
-14h = 2k - 12
 -7h = k - 6

Equating the lst and 3rd

(h-4)² + (k-3)² = (h-1)² + (k+6)²

Rearrange terms to create the difference of squares:

(h-4)² - (h-1)² = (k+6)² - (k-3)² 

Factor ("factorise")

[(h-4) - (h-1)][(h-4) + (h-1)] = [(k+6) - (k-3)][(k+6) + (k-3)]

[h - 4 - h + 1][h - 4 + h - 1] = [k + 6 - k + 3][k + 6 + k - 3]

[-3][2h-5] = [9][2k+3]
-6h + 15 = 18k + 27
-6h = 18k + 12
  h = -3k - 2
  
So we solve this system of two equations:

 -7h = k - 6
   h = -3k - 2

by substitution and get 

-7(-3k - 2) = k - 6
   21k + 14 = k - 6
        20k = -20
          k = -1

   h = -3(-1) - 2
   h = 3 - 2
   h = 1

So the circumcenter (or "circumcentre") is O(h,k) = O(1,-1)

The circumradius r is

r = AO =  =  =  =  =  = 5

Edwin

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A goat is tied to one of the corners of a rectangular barn on a rope that is 50
feet long.  The dimensions of the barn are 40 feet by 30 feet.  Assuming that
the goat can graze wherever its rope allows it to reach, what is the square
footage of the grazing area for the goat?



The area of a circle is pr²

The grazing area consists of 

1. three quarters of a big 50ft-radius circle, which has area

    ·p(50)² = ·2500p square feet = 1875p square feet. 

2. one quarter of a 20ft-radius circle on the left, which has area

    ·p(20)² = ·400p square feet = 100p square feet 

3. one quarter of a small 10ft-radius circle on the top, which has area

    ·p(10)² = ·100p square feet = 25p square feet.  

Total = 1875p + 100p + 25p = 2000p square feet of grazing area.

Edwin McCravy aka AnlytcPhil

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The triangle below is isosceles. Look how short the base is and look how long the legs are!

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There are many answers.

You could have 3 nickels like this:

17 quarters = $4.25
 4 dimes        .40
 3 nickels  =   .15
-------------------
24  coins   = $4.80 
           
Or you could have 6 nickels like this:

18 quarters = $4.50
 6 nickels  =   .30
-------------------
24  coins   = $4.80

Or you could have 0 nickels at all

19 quarters = $4.75
 5 pennies  =   .05
-------------------
24  coins   = $4.80

If half dollars or dollar coins are allowed there are many many more ways.

Here is a way with 4 nickels if you use 8 half dollars.

 8 half dollars = $4.00
 2 quarters     =   .50
 4 nickels      =   .20
10 pennies      =   .10
-----------------------
24 coins        = $4.80

If you can use dollar coins, here's a way to have 17 nickels.

 3 dollar coins = $3.00
 1 half dollar  =   .50
 1 quarters     =   .25
 2 dimes        =   .20
17 nickels      =   .85
-----------------------
24 coins        = $4.80 

Edwin

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The number "N" is a 5 digit natural number. The 6 digit number N1, formed by placing the digit 1 after N, is 3 times as large as the 6 digit number 1N, formed by placing the 1 in front of N. What is the original 5 digit number N? Write each as an algebraic expression: N1 = ? and 1N = ? Write the equation to solve for finding "N"=

 N = ↏↏↏↏↏
N1 = ↏↏↏↏↏1 = 10N + 1
1N = 1↏↏↏↏↏ = 100000 + N

10N + 1 = 3(100000 + N)
10N + 1 = 300000 + 3N
     7N = 299999
      N =  42857

     N1 = 428571
     1N = 142857

Edwin

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Form the conjugate of the denominator by the rule:

the conjugate of A + B is A - B.  You just change the sign
of the second term but keep the sign of the first term.

So the conjugate of the denominator  is
.

Put this conjugate over itself like this: .

Since that red fraction just equals 1 we can multiply our express
by it and not change its numerical value, but change its form. So we multiply
by it:

Χ.

We'll put parentheses around the binomials:

Χ.


And multiply numerators and denominators:
















To get the negative sign off the bottom, bring it to the top:


 
We can multiply the negative sign into the parentheses so it won't
be sticking out in front:



Finally we will reverse the terms in the parentheses:



That's the simplest form.  Or this way



Edwin

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 -6x²y + 4xy – 7y

Get rid of the exponents by writing the square x² as the product
of x and x.

-6xxy + 4xy - 7y

Now count the number of letters multiplied by the coefficient in 
each of the three terms:


-6xxy  has a row of 3 letters multiplied by the coefficient -6,
so that term has degree 3.


+4xy  has a row of 2 letters multiplied by the coefficient +4,
so that term has degree 2.


-7y has just one letter multiplied by the coefficient -7, 
so that term has degree 1.

Now whichever term has the largest degree, we take that to 
be the degree of the ENTIRE polynomial.  So since the first 
term has the largest degree, which is 3, the whole polynomial 
has that same degree, which is 3.

In a nutshell, the degree of a polynomial is the most number 
of letters multiplied together by the coefficient in any of its 
terms.  An exponent is considered to be a string of the same 
letter repeated to be multiplied the same number of times as the 
exponent.  

The longest string of letters in any of the three terms is 3,
so the polynomial, which is called a "trinomial" because it 
has three terms, has degree 3.

But don't get those mixed up just because they're both 3.  It's
a trinomial because it has three terms, but it has degree 3 
because the first term has the most number of letters 
multiplied together by the coefficient, which is 3 letters.

Edwin

You can put this solution on YOUR website!
= = = °

You can put this solution on YOUR website!

Since the first two digits are 12 in that order, there is only 1 way
to choose the first two digits.  That's 1 way to choose the first
two digits.

Since the third digit is bigger than 6, it can only be 7,8, or 9.

So that's 1Χ3 choices for the the first three digits.

Since the 4th digit is 3 times the 5th digit, the number can only end in 
31, 62, or 93

So for each of the 1Χ3 way to choose the first 3 digits, there are 3
ways to choose the last two digits.

That's a total of 1Χ3Χ3 = 9 possible numbers.  Here are all 9:

1.  12731
2.  12762
3.  12793
4.  12831
5.  12862
6.  12893
7.  12931
8.  12962
9.  12993

Edwin

You can put this solution on YOUR website!
Candy Mixtures - Someone wants to mix some candy that is worth 45 cents per pound. Some is worth 80 cents per pound to make 350 lb of a mixture worth 65 cents per pound. How much of each type of candy should be used?

Suppose we mix x pounds of the 45’ candy with y pounds of the 80’ candy.
Then we have two equations, a candy equation and a money equation:

The candy equation comes from this:

          +  =   


                          x + y = 350

The money equation comes from
    
          +  =   

                        (45’)x + (80’)y = (60’)(350)  or
                              45x + 80y = 21000

So solve this system of equations:

                         x +   y =   350
                       45x + 80y = 21000

(If you can't solve that system of equations, post again asking how)

Answer: Mix 200 pounds of the less expensive candy and 150 pounds of the cheaper candy.

Edwin