CPhill

See the  following Venn Diagram  : 

The only difficult part is  trying  to find the  number who only take Physics   and the  number who only take  Chemistry

We know that  twice as many take Chemistry  as take Physics

When  we  add  all the  known   elements in the  diagram  we get that 90 students are unaccounted for

If we call x  the number  who only take Physics....then 90 - x  must  be the number who only take  Chemistry 

So we  can solve this equation

2 times the number taking Physics  = the number taking Chemistry

2  ( 20 + 15 + 75  + x) =  85 + 15 + 75 +  ( 90 - x )

2 ( 110 + x )  =  265 - x

220 + 2x  = 265 - x

3x  = 265 - 220

3x  = 45

x  =15 =   number who only take  Physics

90 - 15  =75   =  number who only take  Chemistry

Number taking Physics  =  75 + 15 + 15 + 20   =  125

(a) The distance  around the  cross-section  is the  circumference  of a circle and we  can  find the radius thusly:

47.1  = 2 pi * r

47.1 / (2pi)   = r   ( in cm )

(b)  Volume of a  cylinder = 

pi  * r^2  * height   =

[ Note that .7 m =   70cm]

pi  [ 47.1  / (2pi) ] ^2  * 70        ≈  12357.5  cm ^3

Let   the integers in increasing order be  a, b, c, d 

All we know  for sure is that

a + b =   10

a + c  =  18

b + d  =  21 

c + d =   29

Subtracting the    first equation  from the second  we  have that     c - b =  8    ⇒  c =  8 + b

Either    a +  d   = 19    or b +  c  = 19

Let's assume the  second  is true

Then  2b + 8   =  19          but b isn't an integer  here

So

a +d    = 19

And

b + c =  20

So

b + 8 + b =  20

2b + 8 = 20

2b =12

b = 6

a  = 4

c=  14

d = 15

So

a   b    c    d    

4   6   14  15

See the following image

The slope of the line  containing  AN  is  (1/.5)  =  2

And the equation of the  line containing this segent is   y = 2x      (1)

The slope of the  line containing DM  is   ( 1 - .5) /( 0 -1)  =  -.5 =   -1/2

And the  equation of the line containing  DM is

y = (-1/2)x + 1               (2)

Equating (1)  and (2)  we have 

2x =(-1/2)  x + 1

2.5x  = 1

x =1/2.5  = 2/5  = .4    =the xcoordinate of O

And the y coordinate of O  is   y = 2(.4)  =  .8

So triangle DCM    has a base of 1  and a height of .5 = 1/2

So its area =  (1/2)(1) (1/2)  =  1/4

And triangle  DOA  as a base of 1  and a height of  .4 = 4/10  = 2/5

So its area is (1/2) (1)(2/5)  = 1/5

So  [ABMO ] =  area of the square  - [DCM ] - [ DOA ]  = 

1 - 1/4   -1/5  =

1 - .25 - .20  =

1 - .45  =

.55  =   55/100  =  11/20

An employee at a construction company is ordering interior doors for some new houses that are being built. There are 3 one-story houses and 6 two-story houses on the west side of the street, which require a total of 93 doors. On the east side, there are 8 one-story houses and 2 two-story houses, which require a total of 94 doors. Assuming that the floor plans for the one-story houses are identical and so are the two-story houses, how many doors does each type of house have?

Call the number of doors on the  one story house =  O

Call the number of doors on the two story house  =  T

So

3O  + 6T   =  93     dividie this trough  by 3  ⇒   O  + 2T  = 31     (1)

8O  + 2T =    94     (2)

Subtract (1)  from (2)   and we get that

7O  =  63

O  = 63/7  = 9    doors on the  one story house

And using (1)

9  +  2T  = 31

2T  =31   - 9

2T  =22

T =22/2    = 11  doors on the two story house

1.  (  z - ( -7+ 2i) )   ( z  - ( -7 - 2i) )  =

z^2  -  z ( -7 + 2i)  - z(-7 -2i)  +  (-7+2i)(-7 - 2i)   =

z^2  + 7z - 2iz  + 7z + 2iz  +   49  -  4i^2   =

z^2  +  14z  +  49  +  4   =

z^2 + 14z  + 53

a = 14  b  = 53

a +  b  =   67

2.    x + y  =10   ⇒  y =  10  - x     (1)

       x^ 2 + y^2  =  56     (2)

Sub (1) into (2)

x^2  + (10 - x)^2   = 56

x^2 + x^2  - 20x + 100  = 56

2x^2  -20x + 44  = 0     divide  through  by 2

x^2 -10x + 22    =   0      complete the square on x

x^2 - 10x + 25 = -22 + 25

(x - 5)^2  =  3     take  both  roots

x  - 5  = ± √3

x = ± √3 +  5

When x  = √ 3  + 5   y =   - √3 + 5

When x    =  -  √3 +  5   y =   √3+ 5

Let the number of people that a sedan can carry   = S

Let  the number of people that a minivan can carry =  M

So

1S + 2M  = 16

3S + 2M  = 24         sbtract the first equation  from the second

2S =  8

S = 4

Using the  first equation  to find   M

1(4)  + 2M  =16

4 +  2M  =16

2M = 16 - 4

2M  =12

M  = 6

A sedan can hold 4   people ....a minivan can hold  6

1/x  + 1/y  =  3

(x + y)   / xy   =  3

x + y =   3xy

So   

 xy  + x + y =  4

xy + 3xy =  4

4xy  =  4

xy   =1

x^2y + xy^2  =

xy (y + x)  =

xy ( x + y)  

1 ( 3xy)  =

1 ( 3(1)) =

3

"a"  cannot be   -1     because   x^2 -  x + 1    = 0    has no  real roots  (the discriminant is <  0  )

Correct.....