All Answers 358089 Answers

First one

Let the side of the larger hexagon = S   and  the  side of the  smaller = s

Using the Law of  Cosines, we can find the relationship s / S

S^2 = s^2 + s^2 - 2(s^2)cos (120°)

S^2  = 2s^2 - 2(s^2)(-1/2)

S^2 = 2s^2 + s^2

S^2  = 3s^2

S = sqrt (3)s

s / S  =  1/sqrt (3)

These hexagons are similar figures, so the area of the smaller hexagon =   area of larger hexagon *( s/S)^2  =

(1) ( 1/sqrt 3)^2    =    

1/ 3

We have

a*0^2 + b*0 + c = 4       so  c   =  4

and

a*1^2  + b*1  + 4 =  8

a + b =  4

Since it is  monic  a = 1  and  b =3

So

f(x)  =  x^2  + 3x  + 4

If the same system = two  different  values, then we  have  no  solutions

So   K =  6

Think  about it

If  6x + 4y =  7   for some x, y

Then 6x + 4y CANNOT also =  4     for the  same  x, y   !!!!

Simplify this as

5x - 1  <  x^2 + 2x + 1 + 7x - 3

5x - 1  < x^2 + 9x - 2

0 <  x^2 - 4x - 1

So

x^2 - 4x - 1  >   0     

This is a  parabola that turns  upward......using the  Quadratic Formula, we can find  where the parabola intersects  the  x axis......the solution will lie on  (-inf , root 1) U (root 2, inf)

So

Root 1 =    [ 4 - sqrt ( (-4)^2 - 4 (1) (-1) } / 2  =  [  4 - sqrt (20) ] / 2 =   [4 - sqrt (4 * 5) ] / 2  =

[4 - 2sqrt (5) ] / 2 =  2 - sqrt 5

And by the conjugate property, Root 2 =  2 + sqrt 5

So.....the solution  is (  inf, 2 -sqrt 5)  U  ( 2 + sqrt 5, inf) 

x^2 + ?x + 10

Find all the  negative  pairs of  integers whose product = -10

(-10) (-1)  .....? =  -11

(-5)(-2)     .....?  =   -7

                sum =   -18

There are 60 pencils in 4 pencil boxes. How many pencils are in 7 boxes?     

There are 60 pencils in 4 pencil boxes.  That's 60 / 4 = 15 pencils in each box.    

How many pencils are in 7 boxes?          Then 7 x 15 = 105 pencils total.    

_.

Two circles of radius $1$ are centered at $(4,0)$ and $(-4,0).$ How many circles are tangent to both of the given circles and have radius $3$?    

Picture the circles on the x-axis.  Better yet, get some paper and draw them.      

Centers on –4 and 4 and radius 1, their edges reach to –3 and 3.  

Note that –3 and 3 are 6 units apart.    

Note that radius 3 equals diameter 6.   

So there's only 1 circle of radius 3 that will fit between them and be tangent.    

_.

Correct!

For what value of $h$ does the quadratic $7x^2 + 5x = h + 6x^2 - 3x$ have exactly one real solution in $x$?    

                                                      7x^2 + 5x = h + 6x^2 - 3x

                                                      7x² – 6x² + 5x + 3x – h  =  0

Combine units.                                  x² + 8x – h  =  0     

This is in the form                             ax² + bx + c

To have only one solution means  

that the expression is a square.  

For it to be a square, then                b² – 4ac has to equal zero.       

                                                          8² – (4)(1)(–h)  =  0    

                                                           64 + 4h  =  0  

                                                                   4h  =  –64    

                                                                     h  =  –16   

Check Answer   

Substitute –16 into the   

quadratic equation.                  x² + 8x –(–16)   

                                                x² + 8x + 16   

Will this factor?  Yes.               (x + 4)(x +  4)   

_.

Correct!

Correct!

8+8=16

Note that x^2 - x - 2  can  be  factored as   (x - 2) ( x + 1)

Multiply the equation  through by this factorization and we  get

2x + 11  - x - 13 =  A(x + 1) + B ( x -2)

x - 2  = A(x + 1) +  B ( x -2)

1x - 2  = (A + B)x  + A - 2B             equting  coefficients  on  both sides  of the eqution  we have

1 = A + B

-2 =A - 2B

Multiply the second equation  through by -1

1 = A + B

2 = -A + 2B          add these

3 = 3B

So B = 1

And

A + B = 1

So A  =   0

For the record, Post #1, signed by GA, is not by the real GA.

GA

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t=-3.5,-3.