CPhill

The volume of the prism  is   2.5 * 2.5 * 12  = 75 ft^3

The density  =     mass / volume    =    12000 lb / 75 ft^3    =  160 lb / ft^3

Note  that  the sum of the first  four odd positive integers  =   1  +  3  +  5  + 7    =  16 =     4^2  

So....the sum of the first N positive  odd integers  =  N^2

So

N^2  =121     take the square root

N  = 11

Let the sides of the equilateral triangle be some convenient integer, say, 6

Draw altitude  AF ......this makes DF  = 1

Using the Law of Sines we have that

AF/ sin 60  =  BA/ sin 90

AF  / sin 60  = 6

AF   =  6√3 / 2   =   3√3  =  √27  =  

And using the Pythagorean Theorem

DA  =  √[ AF^2  + DF^2]    =  √[ 27 + 1]    = √28  =  EA

And using the Law of Cosines

DE^2  =  DA^2  + EA^2   -  2(DA)(EA) cos DAE

2^2 = 28 + 28  -  (2*28) cos DAE       simplify

4  - 56 =  -56 cosDAE

-52  = -56 cosDAE

-52/-56  = cos DAE

13/14  = cosDAE

So....sin DAE  =  +√[ 1 - cos^DAE ]  = + √[ 1 -  (13/14)^2 ]   =  +√ [196 - 169] / 14  =  √27 / 14 =  (3/14)√3 

the diagram shows a sector OPQ of a circle with centre O. the radius of the circle is 18m and the angle POQ is 2π / 3 radians. 

a) find the length of the arc PQ, giving the answer as a multiple of π

b) the tangents to the cirlce at the point P and Q meet at the point T. and the angles TPO AND TQO are both right angles

i) Angle PTQ = ∝ radians. find ∝ in terms of pi 

ii)Find the area of the shaded region bounded by the arc PQ and the tangents TP and TQ

a)   arc length PQ  =  18 *  (2 pi)/3 =     36 pi / 3   =   12 pi

b) i)   Angle  PTQ   will be  supplemental to angle POQ   =   pi  - (2pi)/3   =  pi/3   = 60°

ii )    The area of the circular sector  formed by  arc PQ  =  (1/2)r^2 (2pi/3)  =  (1/2)(18)^2 (2pi)/3 = 108 pi  m^2     (1)

[PTQO]   will from a  kite   ...OQ  = 18     and  triangle TPO will form a 30-60-90 right  triangle with QO = 18 opposite the 30° angle  (angle QTO)  and QT  opposite the 60° angle  (angle QOT)....so  QT  = 18√3

So  the area of this kite =  product of the legs  of triangle OQT =  QO * QT  = 18 * 18√3  = 324√3    (2)

So...the area bounded by arc PQ  and the  tangents TP and TQ   = (2)  - (1)  =  324√3 - 108 pi ≈  221.9 m^2

11 1/12  -  2 2/3   =

133              8

___    -       ___

 12               3

133             8 * 4

___     -      _____

 12               3 * 4

133             32

___  -        ___

 12              12

101

___

 12

8   5/12

3)  x&  = 7 - x       and       &x  =  x - 7       What is   & (12&)

Note that   12&   corresponds to x& = 7 - x....so here, x  must = 12   and we have   7 - 12  =  -5

So

&(12&)  =  & (-5)

And  & (-5)  must correspond to  &x= x - 7  ...so here, x  must =  -5    and we  have  -5 - 7   =  -12

So   

& (12&) =  - 12

A toddler is stacking wooden blocks. He has 4 blocks that are all different colors (red, blue, green, and yellow). How many different towers of blocks can he make?

4 ways to choose the 1st block  * 3 ways to choose the 2nd * 2 ways to choose the  3rd * 1 way to choose the last  =  4 * 3 * 2 * 1   = 24

A committee of 3 people is to be selected from a group of 10 people. The order in which people are selected doesn’t matter. How many different committees can be formed?

We have 10 people and we  wish to choose any  3 of them....so...10C3  = 120  different  committees

An ice cream shop has 20 flavors. If Louis wants a double scoop cone, how many different flavor combinations can he get?

We  can choose   2 flavors  from 20  and order these in 2 ways...so

20C2  * 2   =  190 * 2   = 380 combos

E. 3,333.849

28.53 * 1.20   =  $34.24

  10 * 0.33  = $3.30

First one.....just add  3 + 2.58 + 1.2  + 1.3 + 0.68 + 0.5 + 2.15.....the total is your answer

Second one

9.68   = 5%  of 9.68  =

9.68  + 9.68 * .05  ≈  10.16     >  10

Not enough   !!!