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OOps....was out on a run with the doggies and realized I made a mistake here...it is now corrected...

We have 9 possibilities for the first digit

The Third digit can be 0-9 (ten possibilities)

Three possible for the second digit   0 2 4

with 0    9 of the digits (1-9) can be fourth digit

 with 2   the fourth digit can be 4 5 6 7 8 9 (six possibilities)

  and with 4   only  8 9 can be fourth digit       ( 6 or 8 cannot be the second digit because fourth needs to be 2x)

       0                           2                         4

9  * 1 * 10 * 9     +  9 * 1 * 10 * 6   +  9 * 1 *10 * 2  =   Well, I think you can do this math !

I'll provide a solution later:

 1. \(y = -\frac{1}{8}x^2+2\)

I'm thinking that the volume is supposed to be 3w^3 - 4w^2 - 15

Let the width = w

Then the height is 3w + 5

And the volume  is   Width * Heiht * Length....so we have

3w^3 - 4w^2 - 15w  = w * ( 3w + 5) * L          

w (3w^2 - 4w - 15) = w (3w + 5) * L     divide both sides by w

3w^2 - 4w - 15  = (3w + 5) * L          factor the left side

(3w+ 5) (w - 3) = (3w + 5) * L      divide both sides by (3w + 5)

(w - 3)  =  L  =  Length

Thank you!

It's 3360 I checked it.

We have ways to award gold....15 ways to award silver and 14 ways to award bronze

So

16 * 15 * 14   =

3360  ways

The arc length of QR is 120/360 of 16pi, which is the circumference. That is 16pi/3

The central angle in radians is 2/3pi. 180 degrees in radiansis pi and 360 is 2pi.

The area of the sector is 64pi/3. Like 71, we just set up a proportion again. it is just a third of the total area

The equation is (x-3)^2+(y+4)^2=4

Because the basic form is (x-h)^2+(y-k)^2=r^2 and (h,k) is the center and r is the radius.

We can do the last one by completing the square.

We take the square of half of the coefficients.

(x-3)^2+(y+2)^2=-10

Sorry if these were not detailed, I didn't have a calculator and was in a hurry.

71.  Arc length  =  2 * pi (8) (120/360)  = (16/3)pi mm ≈ 16.76 mm

72. 120 * pi  / 180  =   (120/ 180) *pi  =  (2/3) pi  rads

73.  Sector area  = (1/2) *r^2* (theta in rads)  =  (1/2) * (8)^2 * (2/3 pi)  =

(64/ 3) pi  mm^2  ≈  67.02 mm^2 

74.   (x - 3)^2  + (y + 4)^2  = 4 

75. x^2 - 6x + y^2 + 4y - 3  = 0

x^2 - 6x + y^2 + 4y  =  3            complete the square  on x and y

x^2 - 6x + 9  + y^2 + 4y + 4  =  3 + 9 + 4

(x - 3)^2 + (y + 2)^2  = 16

center ( 3,  -2)    radius  = 4 

We have 5 digits which are odd...

So....the number of "odd looking" 4 digit numbers is

(5)(5)(5)(5)  =  5^4   =  625

tan  =  sin  / cos

tan  =  ( 7/25) / (24/25)

tan  =  (7/25) * (25/24)  =  ( 7/24) * (25/25)   =  7/24 * 1   =  7/24

First one 

The measure  of   angle AED  =  (1/2)(arc AD + arc BC)

angle AED  =  (1/2)(38 + 162)

angle AED  (1/2) (200)  =   100°

Second one ....this is the intersecting chord theorem

It says  that  

DE * EB  =  AE  * EC

 3 * 16   =   4 * EC

48  = 4 * EC       divide both sides by 4

12  = EC

sin x  =  9/10  = y/r

So....cos x / r

To find x

  x  = ±√[r^2- y^2 ]  =  ±√[ 10^2  - 9^2]  =  ±√[100 - 81 ]  = ±√19

So....depending upon where x  lies

cos x  = √19 /100  ≈ .04    or    cos x  ≈  -.04  

Converting rads to degrees, we have

tan ( 60° -  A)  =  cot (60°)

And since they are co-functions....cot (60°)  =  tan(90° - 60°)  =  tan (30°)

So

tan (60 - A)  = cot(60)  = tan (30)

So

tan (60 - A)  - tan(30)

So   A  =  30°  =  pi / 6

Here's the first one

Let    a > c > b

And  since  a > b, then  l a - b l  =  a - b

And since  c > b, the l b - c l  =  c - b

And since a > c, then l c - a l  = a - c

Then  we have that

(a - b) + ( c - b) + ( a - c)  = 20  simplify

2a  - 2b  = 20       divide through by 2

a - b  =  10

And this is the max value for  l a - b l

 B.It shifts the graph down 5 units.

We have 5 photos and we want to choose only 3

So  C(5,3)  =  10

Call the pics  1, 2, 3, 4 and 5

Here are the possible sets :

(1,2,3)    (1,4,5)

(1,2,4)    (2,3,4)

(1,2,5)    (2,3,5)

(1,3,4)    (2,4,5)

(1,3,5)    (3,4,5)

For the first one....KJL  aand KML  are inscribed angles that intercept the same arc....inscribed angles that intercept the same arc are equal

So  Rarinstraw's  working is correct   !!!!

Second one...in this situation....V  will equal (1/2)  the difference  of arcs ZY and WX....so we have

V  = (1/2) (ZY - WX)

V  = (1/2) ( 130 - 50 )

V  = (1/2) (80)

V  = 40°

Hey dude! Do you mind answering a math question right now? Im having a tough time on it and almost all my class is to. Heres the question.

You have a frame that holds three pictures. You pulled out your favorite five photos. How many sets of three are there?

Make a list of all the possible combinations using the numbers 1 - 5 to represent the photos.

i think there is 10 combinations but that doesnt seemright.

                    O        A       B     AB     Total

Male (M)       210    174   74     42     500

Female (F)   315     261   111   63     750     

Total             525     435   185  105   1250

 P(F l A )  =  261/435  =  3/5

If independent  P(F and A)  = P(F) * PA) 

P(F and A)  = 261 / 1250

P(F) = 750/1250      P(A)  = 435/ 1250

So....  P(F) * P( A)  =   (750/1250) * (435/1250)  = 261/1250

So  independent

P( F and A)  = P(F) * P(A)

So...  "C"  is correct 

ok for problem #68, do 130-50 and you should get the angle for V

for 67 do the same thing but put the numbers in place of the letters

k=12 and L=8 subtract these and it should give you the value of j

k=10 and l=16 subtract these and it should give you the value of m

math is'nt my strongest subject but i beleive this is how you do it or you solve for the variables on both sides

67. 

This problem seems immensely tough, but it isn't. 

The measure of angle KJL is equal to the measure of angle KML.

12x+8=10x+16

First, get the x on one side.

12-10=2

2x+8=16

Now, subtract 8

2x=8

Now, divide both sides by 2.

x=4

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I have no idea that's why I need help lol