All Answers 358335 Answers

1.  See the image

X  = (12, 5)

Length of SX  = sqrt (12^2 + 5^2 ) = sqrt (169) = 13

Length of PX  = sqrt (12^2 + (10 - 5)^2 )  = sqrt (12^2 + 5^2) = sqrt (169) = 13

Using the Law of Cosines

RS^2 = PX^2 + SX^2 - 2(PX)(SX)cos(PXS)

10^2 = 13^2 + 13^2 - 2(13)(13) cos (PXS)

100 = 2(169) - 2(169)cos(PXS)

100 - 338

_________  = cos(PXS)

     -338

-238

____  = cos (PXS)

-338 

238                            119

___  = cos(PXS)  =   ____

338                            169

A^2 + B^2  = C^2            subtract A^2

B^2 = C^2 - A^2           take the square root    

B = √ [ C^2 - A^2 ]

Feb  =  8125(1)/4 + 2000 = $4031.25

Mar = 8125(2)/4 + 2000 =  $6062.50

Apr = 8125(3)/4 + 2000 = $8093.75

[5! / 5!.0!.2^5] x 100 =1 / 32 =3.125%

Take c^2 minus a^2 and then find the sqaure root of whatever number you get

\(\text{Linear growth is a straight line with a single slope}\\ \text{It has the form }L(t) = m t + L_0 \text{ here we have}\\ L_0=2000\\ L(4)=10125 = 4m + 2000\\ m = \dfrac{8125}{4}\\ L(t) = \dfrac{8125}{4}t+2000,~t \text{ in months}\)

I'll let you fill out the table

To answer their question take a look at the difference in earnings between adjacent months.

1)  Nickolas, if you increase 6 by 2 what do you get?

2)  How did you get that?

3)  So if you increase x by 2 what do you get?   

Please answer those 3 questions for me here. on this thread.

I think this thread is a bit old so after you have done it can you please message me with the address of the question so I can come and have a look at what you have done.   (Otherwise I might not see your answer)   

Two points on the dotted line  are 1,-2    and   3,1      slope =  (-2-1)/(1-3) = 3/2

y = 3/2 x + b

sub in one of the points to find b

1 = 3/2 (3) + b

b= -7/2

so    y > 3/2 x -7/2

Area of rhombus =   d₁ d₂ / 2

    840 = 40 d₂ /2

       d₂ = 42                       1/2 of each of the diagonals form the legs of a right triangle with the side length as the hypotenuse

20^2 + 21^2 = s^2           s = 29

A = 20 000 (given)

a.)  for t = 3    P =  20 000 (1.3)^(-.234 (3))

                          =16600 (rounded)

b.) population is now 20 000   1/2 of this is 10 000    solve for  t    

                                      10 000 = 20 000 (1.3)^(-.234t)

                                       10 000 / 20 000 = 1.3^(-.234t)

                                              .5 = 1.3^(-.234t)       take log of both sides

                                             -.301 = -.234t  log (1.3)

                                                t =11.3 yrs                                           

Sorry...I don't know the answer to this one......maybe someone else does.....

What is the x-intercept of the line perpendicular to the line defined by 3x-2y=6 and whose y-intercept is 2?

The slope of the given line  =  (3/2)

The slope of the perpendicualr line  = -2/3

If the y intercept of this line is 2, we have

y = (-2/3)x + 2     the x intercept occurs when y = 0

0 = (-2/3)x + 2

-2 = (-2/3)x

x = 3  = the x intercept

Not sure about the first...but I believe that the least possible value will occur at the intersection of the two given lines...so...

2x + 5y = 10    ⇒   6x + 15y  = 30     (1)

3x + 4y = 12    ⇒   -6x - 8y = -24      (2)      add (1) and (2)  and we have that

7y = 6

y = 6/7

And we can find x as

2x + 5(6/7) = 10

2x = 10 - 30/7

x = 40/14  = 20/7

So ( x , y )  =  ( 20/7, 6/7)

So  8(20/7) + 13(6/7)  =  34 = the least possible value

See the graph here : https://www.desmos.com/calculator/orm3rbjxft

1 N/F1 = (5/1)^2

1 / (5/1)^2 = F1

1/25 N = F1

Which of the following represents P(B|A)?

A. https://i.imgur.com/OXci63e.jpg

B. https://i.imgur.com/L3gN7xr.jpg

C. https://i.imgur.com/bYknc5k.jpg

D. https://i.imgur.com/WONpmWn.jpg

I don't believe any of these are correct, MS

The sample space is a listing of all possible outcomes

This should be  2^8  =  256

a ) If we draw a perpendicular to the base from the top right vertex....we have a 30-60-90 right triangle

The distance, D, from the bottom right vertex to the inersection of the perpendicular and the base = 

cos 60” =  D /2

2(1/2)  = 1

So.....by symmetry.....the length of the bottom base = 1 + 2 + 1  = 4

And the perimeter of the trapezoid is  bottom base + top base + sides  =

4 + 2 + 2 + 2  =         10 units

b) The height of the right triangle we created  = √3

The area =  (1/2)(sum of bases) (height)  =  

(1/2)(4 + 2)√3  =

(1/2) (6)√3 =

3√3  units^2