CPhill

1.  Sum  =   ( First term + Last term) * (number of terms) / 2  =

(44 + 26) * (7/2)  =

(70) / 2 * 7 =

35 * 7  =

245

2. Similarly

(7.6 +   -1.5) * 8 /2  =

(6.1) * 4  =

24.4

Area of trapezoid  =

height / 2  * (sum of base lengths) =

(2/10) / 2 * ( 8 + 2)  =

(2/20) * 10   =

20/20  =  

1 square unit   = Total Area

P ( x  ≤ 3)  

We have a trangle with  a height of (2/10)  and base = 3

Area  = (1/2)(3)(2/10)  = 3 / 10

So....P ( x  ≤ 3)    =   ( Area of triangle) / (Total Area )  =

(3/10) / 1  =   3/10 =   0.3

First one....we have a trapezoid.....

The area of a trapezoid  is

height/ 2 * ( sum of base lengths)

So we have......

.25 / 2  *  ( 6 + 2)  =  8/8   =    1  unit^2

And we are looking for areas =  .25 of this  =1/4 of this  =  .25   units^2

P( x ≥ 5)  =   area of a triangle with base = 1  and height  .25/2

Area  (1/2)base * height  =  (1/2) * 1 (.25 /2)  = .25/ 4  =  .0625  No

P ( 3 ≤ x ≤ 4)  = Area of rectangle = .25 * 1  = .25    Yes     (1)

P (  x ≤2 )  = Triangle with base of 2 and height of .25  =

(1/2)(2) * .25  =   .25  Yes

P ( 2 ≤ x ≤ 3)....same area as (1)   =  .25

So....only  the first answer is incorrect 

3.

Note NSS  that the distances  are  26, 44, 62, 80.....  each successive term is 18 more than the succeeding one

The 7th term  is

26 +  18 ( 7 - 1)   =   26 + 18 * 6   =  134

For an arithmetic sequence.....the sum is given by :

 ( First term + Last term ) *  number of terms / 2  =

(26 + 134) * 7 /2    =   160 * 7 / 2   =  160 / 2 * 7   =  80 * 7  =  560 ft

EP's  answer to the second one is good

And for his third answer     5 + 6 + 7 + 8  =

[ 5 + 8] * 4 / 2   =   13 * 2    = 26  

First one

P ( 2.5   ≤ x   ≤ 4 )  =

Width of the interval of interest

_________________________     =

    Total width of interval

(4 - 2.5)                   1.5

______     =        ________    =   .375

  4 - 0                       4

Second one

P( < 180)  =

180 - 177                  3

_________  =         ___    =   .60

182  - 177                 5

We have

4px  =  y^2         p  = distance between the origin and  the directrix

4(5/2)x = y^2             

10x = y^2

x  = (1/10)y^2

a₁ = 6; a_n = 1/4_an−1

The explicit rule is

aⁿ  =  6(1/4)^n-1

(2 + 2i)^8  =

r  = √ [ 2^2 + 2^2 ]  =  2√2

tan  θ  =  2/2   =  1 

 θ  = pi/4

(2 + 2i)^8  =  (2√2)^8* e^(i * 8pi /4 )  =   (2√2)^8 *e^( i * 2pi)  =

4096 ( cos 2pi + i * sin (2pi) )  =

4096 ( 1 + 0i )  =

4096

Let two vertexes of the square lie on  (a,0) and (0, b)

Where a < 8    and b < 6

Let side BC be the hypotenuse of the right triangle

And the equation of BC is:

y  =  -(6/8)(x - 8)

8y = -6x + 48

6x + 8y - 48 = 0

Now....using the formula for the distance between a point and a line we have two outcomes

Distance between (a,0)  and the line  =

abs [6(a) + 8(0) - 48 ]           abs[    6a  -  48]          abs[ 3a  - 24]

_________________      =    ____________      =   _________

     10                                        10                                    5

Distance between (0,b) and the line =

 abs[6(0) + 8(b -48) ]         abs[  8b - 48]              abs [ 4b - 24]

 ________________      =  ___________    =      ___________

     10                                       10                                5

And these distances are equal

Since a < 8  and b < 6 we can write

(24 - 3a) / 5  = (24 - 4b) / 5

24 - 3a  = 24 - 4b

-3a  = - 4b

3a = 4b  ⇒  b = (3/4) a

And the distance between the two points = the distance from one of the points to the hypotenuse 

Thus...

√[a^2 + b^2]  =  abs (3a - 24 )/ 5

Since  a < 8,  and the left side must be > 0...we can write...

√[a^2 + b^2]  =   (24 - 3a ) / 5

√ [a^2 + (9/16)a^2 ]  =  (24 - 3a) / 5

√ [ 25a^2] / 4  =  (24 - 3a)/5

5a/4  =  (24 - 3a) / 5

25a/4 = 24 - 3a

25a  = 96 - 12a

37a = 96

a = 96/37     

b =  (3/4 a)   = (3/4)(96/37) =   72/37

So.....the length of the side of the square  =

√[(96/37)^2 + (72/37)  ^2]  =

√[96^2 + 72^2] / 37   =  √14400 /  37    =   120 / 37 cm 

Here's a pic.... D  = "a"  = (96/37, 0)   and  E = "b"  = (0, 72/37)

We have 5 equal segments on EF ...and we are looking for the point that is 3/5 of the way between E and F in that directon

So we have

( -1 + (3 - (-1) )(3/5) , 6 + (9 - 6)(3/5) )  =

(-1 + (4)(3/5) , 6 + (3)(3/5) )  =

(-1 + 12/5 , 6 + 9/5 )

(7/5, 39/5)