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Find the sum of the terms from the (n + 1) th to the m th term inclusive of an arithmetical progression whose

first term is a and whose second term is b.

If m = 13, n =   3 and the sum is 12a, find the ratio b: a.

Answer (m-n){a+1/2(m+n-1)(b-a)} ; 77/75

$\text{Formula arithmetical progression: }\\ \begin{array}{|rcll|} \hline a_1 &=& a \\ a_2 &=& a+d \\ \hline \end{array}$

$\text{second term is b: }\\ \begin{array}{|rcll|} \hline a_2 &=& a+d \\ a_2 &=& b \\ \hline a+d &=& b \\ \mathbf{d} & \mathbf{=} & \mathbf{b-a} \\ \hline \end{array}$

The common difference is $\mathbf{b-a}$

$\text{The sum of the member is: }\\ \begin{array}{|rcll|} \hline s_n = n\cdot a + \dfrac{n(n-1)}{2}\cdot d \\ \hline \end{array}$

$\text{The sum of the terms from the (n + 1) th to the m th term is: }\\ \begin{array}{|rcll|} \hline s_m-s_n &=& \underbrace{m\cdot a + \dfrac{m(m-1)}{2}\cdot (b-a)}_{=s_m} - \left[ \underbrace{n\cdot a + \dfrac{n(n-1)}{2}\cdot (b-a)}_{=s_n} \right] \\\\ &=& a(m-n) + \left(\dfrac{b-a}{2}\right)[ m(m-1)-n(n-1) ] \\\\ &=& a(m-n) + \left(\dfrac{b-a}{2}\right)( m^2-m-n^2+n ) \\\\ &=& a(m-n) + \left(\dfrac{b-a}{2}\right)( m^2-n^2-m+n ) \\\\ &=& a(m-n) + \left(\dfrac{b-a}{2}\right)[ m^2-n^2-(m-n) ] \\\\ &=& a(m-n) + \left(\dfrac{b-a}{2}\right)[ (m-n)(m+n)-(m-n) ] \\\\ &=& a(m-n) + \left(\dfrac{b-a}{2}\right)(m-n)( m+n-1 ) \\\\ \mathbf{s_m-s_n} &\mathbf{=}& \mathbf{ (m-n)\left[ a + \left(\dfrac{b-a}{2}\right)( m+n-1 ) \right] } \\ \hline \end{array}$

$\small{ \text{If $m = 13, n = 3$ and the sum is $12a$, find the ratio $b: a$ }\\ \begin{array}{|rcll|} \hline \mathbf{s_m-s_n} &\mathbf{=}& \mathbf{ (m-n)\left[ a + \left(\dfrac{b-a}{2}\right)( m+n-1 ) \right] } \\\\ 12a & = & (m-n)\left[ a + \left(\dfrac{b-a}{2}\right)( m+n-1 ) \right] \\\\ \dfrac{12a}{m-n} & = & a + \left(\dfrac{b-a}{2}\right)( m+n-1 ) \\\\ \dfrac{12a}{m-n} & = & a + \left(\dfrac{b}{2}\right)( m+n-1 )- \left(\dfrac{a}{2}\right)( m+n-1 ) \\\\ a\left( \dfrac{12}{m-n} - 1+ \dfrac{m+n-1}{2} \right) & = & b\left( \dfrac{m+n-1}{2} \right) \qquad m = 13 \qquad n = 3 \\\\ a\left( \dfrac{12}{10} - 1+ \dfrac{15}{2} \right) & = & b\left( \dfrac{15}{2} \right) \\\\ \dfrac{b}{a} &=& \left( \dfrac{2}{15} \right)\left( \dfrac{12}{10} - 1+ \dfrac{15}{2} \right) \\\\ &=& \left( \dfrac{2}{15} \right)\left( \dfrac{12-10+75}{10} \right) \\\\ &=& \dfrac{2\cdot 77}{150} \\\\ &=& \dfrac{2\cdot 77}{2\cdot 75} \\\\ \mathbf{\dfrac{b}{a}} &\mathbf{ =}& \mathbf{ \dfrac{77}{75}} \\ \hline \end{array} }$

In triangle
$ABC, \overline{AB}=5, \overline{AC}=6,$
and
$\overline{BC}=7.$
Circles are drawn with centers $A, B,$ and $C,$
so that any two circles are externally tangent.
Find the sum of the areas of the circles.

$\text{Let $r_a =$ radius of circle $A$ } \\ \text{Let $r_b =$ radius of circle $B$ } \\ \text{Let $r_c =$ radius of circle $C$ }$

$\begin{array}{|lrcll|} \hline (1) & r_a + r_b &=& 5 \\ (2) & r_b + r_c &=& 7 \\ (3) & r_c + r_a &=& 6 \\ \hline (1)+(2)+(3): \\ & 2(r_a+r_b+r_c) &=& 5+6+7 \\ & 2(r_a+r_b+r_c) &=& 18 \quad & | \quad : 2 \\ & r_a+r_b+r_c &=& 9 \\ \hline \end{array}$

$\mathbf{r_c = \ ?}$

$\begin{array}{|rcll|} \hline \underbrace{r_a+r_b}_{=5}+r_c &=& 9 \\ 5+r_c &=& 9 \\ r_c &=& 9-5 \\ \mathbf{r_c} &\mathbf{=}& \mathbf{4} \\ \hline \end{array}$

$\mathbf{r_a = \ ?}$

$\begin{array}{|rcll|} \hline r_a+\underbrace{r_b+r_c}_{=7} &=& 9 \\ r_a+7 &=& 9 \\ r_a &=& 9-7 \\ \mathbf{r_a} &\mathbf{=}& \mathbf{2} \\ \hline \end{array}$

$\mathbf{r_b = \ ?}$

$\begin{array}{|rcll|} \hline r_a+r_b+r_c &=& 9 \quad & | \quad r_a+r_c = 6\\ 6 + r_b &=& 9 \\ r_b &=& 9-6 \\ \mathbf{r_b} &\mathbf{=}& \mathbf{3} \\ \hline \end{array}$

The sum of the areas of the circles:

$\begin{array}{|rcll|} \hline && \pi r_a^2 + \pi r_b^2 + \pi r_c^2 \\ &=& \pi \cdot( r_a^2 + r_b^2 + r_c^2 ) \\ &=& \pi \cdot( 2^2 + 3^2 + 4^2 ) \\ &=& \pi \cdot( 4+9+16 ) \\ &\mathbf{=}& \mathbf{29\pi} \\ \hline \end{array}$

f(x)   =   2x² + 6

f(a)   =   2a² + 6

f(a + h)   =   2(a + h)² + 6

f(a + h)   =   2(a + h)(a + h) + 6

f(a + h)   =   2(a² + 2ah + h²) + 6

f(a + h)   =   2a² + 4ah + 2h² + 6

$\quad\,\frac{{\color{blue}f(a+h)}-{\color{purple}f(a)}}{h} \\~\\ =\,\frac{({\color{blue}2a^2+4ah+2h^2+6})-({\color{purple}2a^2+6})}{h} \\~\\ =\,\frac{2a^2+4ah+2h^2+6-2a^2-6}{h} \\~\\ =\,\frac{4ah+2h^2}{h} \\~\\ =\,\frac{h(4a+2h)}{h} \\~\\ =\,4a+2h$

There are a lot of ways to think about this, but here's one way.

Also I'm assuming that   f(x)  =  1 / (ax + b) .

f^-1( x )  =  0         We want to solve for  x . Apply the function  f  to both sides of the equation.

f( f^-1( x ) )   =   f( 0 )

x   =   f(0)

x   =   1 / (a(0) + b)

x  =   1 / b          For  f  to be a function,  1/b  must be the only value of  f(0) .

So...

f^-1( 1/b )   =   0

The only solution to  f^-1(x)  =  0    is   x  =  1/b .

There might be a better way, but here is one way.

Let the radius of circle A be  a , the radius of circle B be  b , and the radius of circle C be  c .

From the given information, we can make these three equations:

a + b  =  5        solve for  a  to get        a  =  5 - b

a + c  =  6        solve for  a  to get        a  =  6 - c

b + c  =  7        solve for  b  to get        b  =  7 - c

5 - b   =   6 - c

                                 Substitute  7 - c  in for  b .

5 - (7 - c)   =   6 - c

                                 Distribute the  -1  to the terms in parenthesees.

5 - 7 + c   =   6 - c

                                 Rearrange

c + c   =   6 - 5 + 7

2c   =   8

c   =   4       Use this value of  c  to find  a  and  b .

a   =   6 - c   =   6 - 4   =   2

b   =   7 - c   =   7 - 4   =   3

area of circle A  +  area of circle B  +  area of circle C   =

2²π   +   3²π   +   4²π   =

4π   +   9π   +   16π   =

29π    sq. units

f(x)  =  3x - 8

Let's find what value of  x  makes  f(x)  be  -5 .

3x - 8   =   -5

3x   =   -5 + 8

3x   =   3

  x   =   1

When   x  =  1   ,  f(x)  is  -5 . That is,  f(1)  =  -5 .

g( f(x) )  =  2x² + 5x - 3

g( f(1) )  =  2(1)² + 5(1) - 3

g(  -5  )  =  2(1)² + 5(1) - 3

g(  -5  )  =  2(1) + 5(1) - 3

g(  -5  )  =  2 + 5 - 3

g(  -5  )  =  4

x^3  + 7x^2 + 7x - 15

Possible  roots   =  ±  [ 1, 3, 5, 15 ]

1 is a root

-3 is a root

-5 is a root

f(x)   =   x⁶

Notice that a negative number raised to an even power is the same as the positive version of the number raised to the power. For example...

f(1)   =   (1)⁶   =   1⁶   =   1

f(-1)   =   (-1)⁶   =   1⁶   =   1

and

f(2)   =   (2)⁶   =   2⁶   =   64

f(-2)   =   (-2)⁶   =   2⁶   =   64

This is true for all numbers and their negatives. This makes the graph symmetrical about the y-axis.

We can rule out the second and fourth option because on those graphs, for example,  f(1)  ≠  f(-1)

We just found that  f(2)  =  64 , so the graph of  f(x)  should pass through the point  (2, 64) .

The first graph passes through the point  (2, 4) , so it can't be right.

The correct graph must be the third one.

Notice that  f(x)  factors like this...

f(x)   =   x³ + 4x² - 9x - 36

                                              Factor  x²  out of the first two terms; factor  -9  out of the last two terms.

f(x)   =   x²(x + 4) - 9(x + 4)

                                              Factor  (x + 4)  out of both remaining terms.

f(x)   =   (x + 4)(x² - 9)

                                              Factor  x² - 9  as a difference of cubes.  x² - 9   =   (x + 3)(x - 3)

f(x)   =   (x + 4)(x + 3)(x - 3)

So the factors of  f(x)  are  (x + 4) ,  (x + 3) ,  and  (x - 3) .

x^4  - 2x^3  - 103x^2   + 200x  + 300

Write  -103x^2   as  -3x^2   -  100x^2

So we have

x^4 - 2x^3  - 3x^2  -  100x^2 + 200x +  300

x^2 (x^2 - 2x - 3) - 100 (x^2 - 2x - 3)        take out the common factor

(x^2 - 2x - 3)  (x^2 - 100)

(x - 3) ( x + 1) (x + 10) (x - 10)

Note that the zeroes are  x =  3 , -1, -10, 10

$(2x^3 - 5x^2 + 7)\div(x - 2) = 2x^2 - x - 2 + \frac{3}{x-2}$

Use synthetic division to find  (x⁴ - 3x³ - 23x² + 75x - 50)  /  (x - 2) .

(x⁴ - 3x³ - 23x² + 75x - 50)  /  (x - 2)   =  x³ - x² - 25x + 25

So

x⁴ - 3x³ - 23x² + 75x - 50   =   (x - 2)(x³ - x² - 25x + 25)

Now we can factor   x³ - x² - 25x + 25  .

x³ - x² - 25x + 25   =   x²(x - 1) - 25(x - 1)   =   (x - 1)(x² - 25)   =   (x - 1)(x - 5)(x + 5)

So

x⁴ - 3x³ - 23x² + 75x - 50   =   (x - 2)(x - 1)(x - 5)(x + 5)

Ah....thanks, Melody.....it's funny how it's sometimes hard to see....and then when someone else points it out.....it seems really easy.....LOL!!!!!

I think that my drawing of this maybe wasn't quite correct.....Melody's made it easy to see.....

f(x)   =   -$\frac12$x²

f(0)   =   -$\frac12$(0)²   =   -$\frac12$(0)   =   0          So the graph of  f(x)  passes through the point  (0, 0) .

f(1)   =   -$\frac12$(1)²   =   -$\frac12$(1)   =   -$\frac12$        So the graph of  f(x)  passes through the point  (1, -$\frac12$) .

f(-1)   =   -$\frac12$(-1)²   =   -$\frac12$(1)   =   -$\frac12$     So the graph of  f(x)  passes through the point  (-1, -$\frac12$) .

The graph of  f(x)  passes through the points  (0, 0) ,  (1, -$\frac12$) ,  and  (-1, -$\frac12$) .

Only one of the graphs in the options passes through these points.

x + y  =  -1    subtract x from both sides ⇒  y    =  - 1 - x   (1)

3x  =  4 - 3y      (2)

Note that we can put (1)   into (2)...and we have that

3x  =    4  -  3 [ 1 - x ]     simplify

3x  =  4  - 3  + 3x     if we subtract  3x from both sides we have

0  =  4 -3

0  = 1    This is never true...so.....strangely.....we have no solutions  for x and y !!!

3x  - 4y  + 2  =  0

10 - 10 y   =  10y  - 15x

Divide the second equation through by 5  and we get

2 - 2y = 2y - 3x

Add 3x to both sides....subtract  2, 2y from both sides

3x - 4y   =  - 2    (1)

Subtract 2 from both sides of the first equation  and we have

3x - 4y   =  -2     (2)

Notice that the transformed equations are exactly the same......whenever this occurs, we have  infinite solutions for x and y  !!!!

One way to express this is to let x be "fixed"  and  solve any of the equations for y

So......

3x - 4y = -2 

3x + 2  = 4y

[ 3x + 2 ] / 4  = y

So....the solution set can be expressed as  {x , y }   =  { x , [3x + 2] / 4 }

Don't appologise, your repost request was done exactly as I always asked for it to be done.

I had not looked at your question before then. :)

Thanks Chris,

The bit that follows is exactly the same as Chris said and then i have continued the proof   

I am going to call the angles in the original triangle 

All angles in the 3 other triangles are all 60 degrees because they are equilateral triangles.

Consider

  $\triangle ACX\;\; and \;\; \triangle YCB \\ AC=b, \;\;CX=a\;\; and \;\; \angle ACX=\angle C+60^\circ\\ YC=b, \;\;CB=a\;\; and \;\; \angle YCB=\angle C+60^\circ\\ \therefore \triangle ACX\cong \triangle YCB\qquad \text{(2 sides and included angle test)} \\ \therefore AX=BY \qquad \text{Corresponding sides in congruent triangles.}$

$\text{Now consider} \triangle ABY\;\; and \;\; \triangle AZC \\ AB=AZ=c\\ AY=AC=b\\ \angle YAB=\angle CAZ=\angle A+60^\circ\\ \therefore \triangle ABY\cong \triangle AZC\qquad \text{(2 sides and included angle test)} \\ \therefore BY=ZC \qquad \text{Corresponding sides in congruent triangles.}\\ $

$\therefore AX=BY=ZC\\~\\ QED$

Not as tricky as it seems....!!!

Note we can factor this as

x ( x^4   + 4x^2  - 5 )  

And the second polynomial can be factored as

x (x^2 + 5) (x^2 - 1)   =   

And we can continue

x ( x^2 + 5) (x + 1) ( x - 1)

We have thre real roots   x = 0, x = -1 and x  = 1

The  "x^2 + 5"  will produce two  complex roots

So.....2  is the number of complex roots (zeroes)

1. If the function is continuous at x = -1, f(-1)  and  lim as x → -1  from  both sides  must have the same value

f(-1)  =  -1 √ [ - 1 + 3 ]   =   - √2

And

lim              x √ [ x + 3 ]   =   (-1) √ [-1 + 3]  =   -√2

x → -1

So this is true

2. Also    f(a)   must  =  the limit as it approaches "a"  from the right and the left

These are going to look similar

lim              x √ [ x + 3 ]   =   (-1) √ [-1 + 3]  =   -√2       =  f(-1)

x → -1⁺

And

lim              x √ [ x + 3 ]   =   (-1) √ [-1 + 3]  =   -√2   =  f(-1)

x → -1^-

So...we have proved that

1. The limit - from both sides - is the same as the function value at  x  = -1

2. The limit from the right side is the same as the function value at x  = -1

3.  The limit from the left side is the same as the function value at x  = -1

I'm not sure what you're taught.....but....I see that 2 is a root....so....we can use synthetic division to find the other roots

2  [    1      -1     0     -2      -4  ]

                   2    2      4       4

     ___________________

         1      1      2      2      0

Our remaining polynomial is  

x^3  + x^2  + 2x  +  2

And -1 is a root...so...performing synthetic division  again, we have

-1  [  1     1        2      2   ]

              -1        0     -2

      _________________

        1     0        2      0

The remaining polynomial is  

x^2  +  2.....  setting this to 0 to find the other roots, we have

x^2  +  2  = 0

x^2  = -2       take both roots

x = ±  √2  i

So.....the factorization is :

f(x)   =  (x - 2) (x + 1) (x -   √2 i )  ( x +   √2 i  )