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tysm 

<3

Remember, distance=rate*time

                          Distance                 Rate              Time

First part of trip        30                       9                  30/9

Last Part of trip        10                       x                  10/x

Whole trip                40                       10                  4

(Time for the first part of the trip) + (time for the second part of the trip) = (total time of trip)

\(\frac{30}{9}+\frac{10}{x}=4 \\30x+90=36x \\-6x=-90 \\x=15\)  

so the cyclist's rate must be 15 kph in the next 10 km in order to bring the overall speed up to 10 km per hour

JP

Interior  angle of a pentagon  =   108°

Ext  angle  =  180 -  108   = 72°

108  /  72    =        

3 * 36

_____  =  

2 *  36 

3 / 2

sqrt 600  =   10sqrt 6

sqrt 150  =    5sqrt 6

4 sqrt 54   =   4*3 * sqrt (6)  = 12sqrt 6

So.....the  numerator =  27sqrt 6

6sqrt 32  =    6 * 4 sqrt 2  =      24 sqrt 2

-3sqrt 50  =   - 5*3 sqrt 2 =       -  15 sqrt 2

-sqrt 72    =    - 6sqrt (2)

So.....the  denominator =  3sqrt 2

So

27sqrt 6

_______  =       9  sqrt  ( 6/2)    =      3  sqrt 3

3 sqrt 2

Let's simplify: (10√6 + 5√6 + 12√6)     /       (24√2 - 15√2 - 6√2)      

Can you take it from here? 

nvm you deleted the question

No prob   !!!!

Let  two  of  the  sides    =   x    and  the  remaining side  = y

Perimeter   = 2x + y

360  =  2x +  y

y = 360   -2x

Area  =   xy

Area =  x  ( 360 -2x)

Area =    -2x^2  + 360x

The  area will be  maximized when   x =  -360 /  (2 * -2) =   90

And the  max area  =

xy   =

90  ( 360  - 2 *90)   =

90  ( 180)  =

16200  ft^2

i can't see the question...basically the whole thing is missing

thanks chris :DDDD

We  can  write

2 - x            (4x - 28)

____   *    _____________    =

x - 7            x^2 + 5x  - 14

2 - x                4 ( x  - 7)

_____  *      ____________   =

x  -7              (x - 2) ( x + 7)

- (x - 2)                4 ( x - 7)

_______ *   _______________  =

(x - 2)             (x - 7)      (x + 7)

-1    *      4

          ____   =  

          x + 7 

   -4

______

   x + 7 

This  is  a parabola  that  turns  upward.......the  vertex  is  (-3, 5)

h  will have  a  negative  rate  of   change  on the time interval  (-inf, -3) 

See the graph here  :   https://www.desmos.com/calculator/cmdoy1jrdp

(x  = t  , and    y =  h )

there's three UNDEFINED terms we have: point, plane, and line.

point: no size, but has location

plane: infinite size, extends infinitely in two dimensions.

line: no width, extends infinitely in one dimension.

answer is plane

Triangles  ABC  and  ADE  are  similar

Which implies  that

AB  / AD   =  BC /DE

AD  =  AB  + BD  = 15

So

10   /15  =  BC  /   12

2 / 3    =  BC / 12

12 * 2/3   =   BC  =     8

The smallest area of a triangle is when k = 2      A = 3.653 u²

                                                               k = 1      A = 8.956 u² 

We  have  a 30 -60 -90  right triangle

The  side opp  the 60°  angle   =   sqrt (3) * side opp the 30°  angle

So......the perimeter is

21  +  21sqrt 3  +  42  =

63  + 21sqrt 3  =

21  ( 3  + sqrt 3   )    ≈  99.37

Thx, SWest    !!!

Here's  another way

By Vieta's theorem  :

sum of  the  solutions =    -5/3  =  u + v

Square  both sides    25/9   = u^2  + 2uv + v^2      (1)

product of the  solutions  =  7/3 =  uv

So   2uv   =   14/3   

Sub this into (1)  and we  have

25/9  = u^2  + (14/3)  +  v^2

25/9  - 14/3   = u^2  + v^2

25/9   -  42/9   =  u^2  +  v^2

-17 / 9   = u^2  + v^2

First, you need to find the solutions.

3x²+5x+7=0

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(x = {-5 \pm \sqrt{5^2-4*3*7} \over 2*3}\)

x = -(5-i√59)/6, and -(5+i√59)/6

We can rewrite as:

v = -(5+i√59)/6

 u = -(5-i√59)/6

u²+v² = (-(5-i√59)/6)²+ (-(5+i√59)/6)²

= -17/9

SWest

Call the original price  P

A markdown of  55%  means  it is  worth  45%  of  P

And  a markdown of another  60%   means it is worth 40%  of    P(.45)

So

P (.45)  ( .40)   =  .18P

You end up paying 18% of the original price

Second one....it will have  equal  roots  ( a double root)  when the discriminant   =  0 

So  all  we  need to change in the previous problem is 

k^2  -  14k   -  39  =  0

(9k + 13) ( k - 3)   =  0

9k + 13  = 0                   k  - 3  =  0

9k  =-13                          k  = 3

k = -13/9

thats wrong

x^2  + (3k - 1)x  + 10  + 2k   =  0

We will have  real roots  as long  as  the  discriminant is ≥  0

So

(3k - 1)^2    -   4 (1) ( 10 + 2k) ≥  0

9k^2  - 6k  + 1  -  40 -  8k  ≥  0

9k^2  -  14k  -39  ≥   0

(9k +13) (k -  3) ≥ 0

This will be  true for  k  on the intervals   (-inf  , -13/9 ]    and  [3 , inf )

I'm assuming you want this simplified.

So 5x^2-3x-2 + x^2+2-5

= 6x^2-3x-2-3

= 6x²-3x-5

SWest

In the diagram below, WXYZ is a trapezoid such that WX||ZY and WY ⊥ ZY. If YZ=12, tan Z=1.5, and tan X=2, then what is XY?

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

WY = tanZ * 12 = 18

WX = 18 / tanX = 9

XY = sqrt(WX² + WY²)

a = 9    and  b  = 8

.989898.......  =    98 / 99 =    x