CPhill

1) x^2 + kx - 9  

The integers we are looking for here  multiply to -9  

These  are  -9 and 1  ,  -1 and 9  ,  -3 and 3

And k will sum to these, i,e. , k = -8, k = 8 and k =0

however, if k = 0, we won't have a trinomial..!!

2) 3x^2 + kx + 7

The integers that we are looking for here can be found thusly

(3x + 1) ( x  + 7) →  3x^2 + 22x + 7    → k  = 22

(3x + 7) ( x + 1)  →  3x^2 + 10x + 7 →    k = 10

1) x^2 - 3x + k

Consider that we have

(x + m) (x + n)  =  x^2 + (m + n)x + mn

So (m + n) = -3 and k = mn

Two integers that work are -7and 4...and their product will be k = -28

And another possibility is -9 and 6.......and their product is  k = - 54

2) 3x^2 + 4x + k

To find a suitable pair, consider

(3x + m) ( x + n)  =   3x^2 + (m + 3n) x + mn

So....m + 3n will add to 4  and k will be their product

This works if m = n = 1   and their product is k = 1

Another possibility is if  m = `10 and  = -2   and their product = k = -20 

The volume of the solid cube is given by :

(4x)^3  =  3072 mm^3   simplify

64x^3  =  3072  divide both sides by 64

x^3  = 48   take the cube root of both sides

x = ∛48 mm =  2∛6 mm

So each side of the cube is   just   4x  =   4 ( 2∛6 mm )  =  8 ∛6 mm 

And the volume of the square hole is just the dimensions 2x * 2x * 4x  

(2x)^2 * 4x  =  16 x^3 =  16 ( ∛48)^3  =  16 * 48  = 768 mm^3 

222^2 - 298  =  48,986

  6q^2 - 7qr - 24r^2   

To factor this kind of problem,  first multiply the first coefficient abd last coefficient =

6 * -24   =  -144

Now....we're looking for 2 factors  the multiply to 144 and sum to the midle coefficient, -7

A little trial and error produces  -16  and 9

Now, we can write the middle term as a combo of these 

So we have

6q^2 - 16qr + 9qr - 24r^2    factoring each pair of terms, we have

2q (3q - 8r)  + 3r (3q -8r )    and (3q -8r ) is common ro both terms......so write this and then the other  two terms in a separate set of parentheses........so we have

(3q -8r ) ( 2q + 3r )

Here's how Alan may have found his answers....

[And then again, maybe not !!! ]

Let the first number be  10 a +  b

And let the other number be  10 b + a

And we know that

[10 a +  b] + [10 b + a]  = N^2

10a + a + 10b + b  = N^2

11a + 11b = N^2

11 ( a + b)  =  N^2

But.....  if the right side is a perfect square then ( a + b)  must equal 11

So.....the possibilities are

9,2

8,3

7,4

6,5

And subbing these into 10a + b  and 10b + a  will generate Alan's two digit numbers

Orient the triangle according to the following pic :

ABC is isosceles....altitude AN forms leg 12 of a 5-12-13  right triangle

The midpoint of  AB = M =  (2.5, 6)

From M, draw a perpendicular to BC.....call this  DM  = 6 

And because DM is parallel to the altitude...triangles CDM and CNX will be similar

So

 DM  / CD = NX / CN

And CD =  BC - BD  =  10 - 2.5  = 7.5

And CN = 1/2 of BC  = 5

So

6 / 7.5  =  NX / 5

(5 *  6 ) /  7.5 =  NX

30 / 7.5  = NX

4 = NX

But  AN - NX  =  AX

So 12 - 4  = AX  =  8

 (1 + x)^ -1/2  at x = 0

Using the Maclaurin Series, we have

f(0) + f ' (0)x  + f ''(0) x^2 / 2!  +   f  '''(0) x^3 / 3! + ......  + f⁽ⁿ⁾(0) x^n / n!

f(0)  =  1

f ' (x) =  ( -1/2) (1 + x)^( -3/2)   and   f ' (0)  =  (-1.2)  ( 1 + 0)^-(3/2)=   (-1/2)

f '' (x)  =  (3/4) (1 + x) ^(-5/2)  and   f '' (0)  = (3/4)  (1 + 0)^(-5/2)  =   (3/4)

f ''' (x)  =   (-15 / 8 ) ( 1 + x)^(-7/2)   and  f''' (0)  = (-15/8) (1 + 0)^(-7/2)  = ( - 15 / 8)

So we have the first four terms

1 - (1/2)x + (3/4)x^2/2   - (15/8)x^3 / 6 + ..... +  f⁽ⁿ⁾(0) x^n / n!  =

1 - x/2  + 3x^2/8  - 15x^3/48 + .....+  f⁽ⁿ⁾(0) x^n / n!  =

1 - x/2  + 3x^2/8  - 5x^3/16 + ...... +  f⁽ⁿ⁾(0) x^n / n!

Thanks, heureka......your answer is more elegant.....but hectictar and I had more fun folding paper........!!!!!!

Guest...if you're still on....check my answer again...I made a slight mistake the first time....!!!

Let's first write this as

y = 16 / [ 5 + x ]

We need to find the inverse ....the idea is to isolate x and then "exchange" x and y

Multiply both sides by [ 5 + x ]

[ 5 + x ] y =  16     now....divide both sides by y

5 + x  = 16/y        subtract  5 from both sides

x =  16/y  - 5    "  exchange " x and y

y = 16/x - 5       get a common denominator on the right

y =  16/x  -  5x / x

y =  [ 16 - 5x ] / x      and for y, write f^-1(x)

f^-1(x) =  [ 16 - 5x ] / x

Now....we want to first find  [ f^-1(2 ) ]

This means that we want to put 2 into the inverse and evaluate it.....so we have

[ 16 -  5(2) ] / 2  =   6/2  = 3

So..... [ f^-1(2 ) ] =  3

Now we want to find   [ f^-1(2 ) ] ^-2    which is just [ 3 ]^-2

Remember that a^-m  =  1 / a^m

Therefore..... 3^-2  =  1 / 3²   =   1 / 9

So......[ f^-1(2 ) ] ^-2 =   1 / 9

And that's it..!!!!