CPhill

Call Ian's rate  =  R

And call the time he drove in hours  ,T

So

Han drove   at  (R + 5)  for a time of (T + 1)

And

Jan drove  at (R + 10)  for a time of (T + 2)

We have  this  equation

(R + 5) (T + 1)  -  RT  =  70

RT + 5T + R  + 5  - RT  =70

5T + R  =  65

R =  65 - 5T

So  the distance that  Ian drove =  RT  =   (65 - 5T)(T)   =  65T - 5T^2      (1)

And the distance  that Jan drove  =  (R + 10) (T + 2)  =   (65 - 5T + 10) (T + 2)  =  (75 - 5T)(T + 2)  = 

75T - 5T^2 + 150 - 10T  =   65T - 5T^2  + 150    (2)

Subtract  (1)  from (2)

65T - 5T^2 + 150  = 65T - 5T^2  =   150

So  Jan drove 150 miles farther than Ian

8^(x^2)*4^x = 2

We can write

(2^3)^(x^2)*(2^2)^x  = 2^1

[2^(3x^2) ] * [ 2^(2x) ]  =  2^1

2^ [ (3x^2 + 2x) ]  = 2^1             we have the bases equal....solve for the exponents

3x^2 + 2x  =  1      complete the square

3(x^2 + (2/3)x + 1/9)^2  = 1  + 1/3

3 ( x + 1/3)^2  = 4/3                      multiply both sides by 1/3

(x + 1/3)^2  =  4/9              take both roots

x + 1/3 =   2/3                or        x + 1/3  =  -2/3

x   = 2/3 - 1/3                            x = -2/3 - 1/3

x = 1/3                                      x =  -1

csc B  =   hypotenuse /  opposite   =   65 / 33

We are in Quad 4

r =  sqrt [ 9^2 + 4^2 ]   = sqrt  97

a)   sin =   -9/sqrt 97

      cos =  4/sqrt 97

      cot = -4/9

       sec =  sqrt (97)  / 4

       csc =  -sqrt (97) /4

b)   arctan (-9/4)   =  -66.04°

So....the angle is  360  -  66.04  =  293.96°

a)    r = 10   x = -6  y  =  -8

sin A  =  -8/10  =   -4/5  

cos A =  -6/10 =  -3/5

tan A  =  -8/-6 =  4/3

cot A =  3/4

sec A  =  5/-3

csc A =  -5/4

b)   arcsin (-4/5)  = -53.1°

However, this  angle lies in Quad 3   because both x and  y are  negative

So....the angle is  180 + 53.1 =  233.1°

y^2 - 4y + 4  factors as    (y -2)^2

So

(y -2)^2 = 0

So  when y = 2, the denominator   = 0  and  the expression is undefined

You will never be able to win 90%  of your games  because  you are only winning 2/3 = 67%  of the games you play after the 5th week......you would have to win > 90% of the games you play after the 5th week to  make this possible

The slope of  the line perpendicular to the given line  is   -2

A line with this slope going through A  has the equation

y  = -2 ( x -3) + 1

y = -2x + 7

Setting the two equations   equal  to  find the x intersection of  these lines we have

(1/2)x + 2 =  -2x + 7

(5/2)x  =  5

x =  5 ( 2/5)   = 2

And the y intersection = -2(2) + 7   =   3

So the intersection pt  is (2,3)

Using the midpoint rule we can find the coordinates of B as

[ 3 + x] / 2 = 2   ⇒    3+ x = 4   ⇒    x  =1

[1 + y] / 2  =  3   ⇒ 1 + y = 6 ⇒   y =  5

So

B = ( 1 , 5)

In this problem, any x value that makes the denominator =  0   will produce an asymptote at that x value

We can write the denominator as

(x - 1) ( x - 2)   =

x^2  - 3x  +  2

a  = -3      b =  2        their sum is  -1

The diagonal length  =   20sqrt 2  cm

[The diagonal of a sqaure is  always   side length * sqrt 2 ]