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how many matches does each team play in total?

 A.The range for f(x) is all real numbers. (true)

B.The domain for f(x) is the set {-2, -1, 0, 1, 2, 3, 4, 5, 6}. (true)

C.f(5) = -2 (false) f(5) = 12

D.f(-1) = 6 (true)

 f(x)=(3+x)

         (x-3)

f(a + 2)=(3+a+2)

              (a+2-3)

f(a + 2)=(5+a)

              (a-1)

ln(i) =1.57079633 i

i believe so, it says f(x)=(3+x) , what is f(a+2)

                                     (x-3)

do you have to expand it?

Let's find M thusy

Note that there are 5 equal sements  in PQ.....M is  2/5 of the way between these....so we have..

[  -9  + (11 - -9)(2/5)  ,  7 + (12 - 7) (2/5)  ]  =

[ -9 + (20)(2/5) , + 7 + (5)(2/5) ]=

[ -9 + 8 , 7 + 2 ]  =

(-1, 9)   =  M

The slope of PQ  is    [ 12 - 7 ] / [ 11 -  - 9 ]  =  5/20  =  1/4

So...the slope of the perpendicular line through M  =  -4/1  =  -4

So...the equation of the line L is

y = -4 (x - -1) + 9

y = -4( x + 1) + 9

y = -4x - 4 + 9

y = -4x + 5

Here's a graph :   https://www.desmos.com/calculator/axxvuzl6js

Using logs and solving for b, you get b = 7

Log[sqrt(4)(8/25)] / Log(7) =

Log[16/25] / Log(7) =-0.2293462.....

The difference between terms is 9

Consider  "C"

a₁  = 12(1)  +  9   =  21  

So.....this must be the correct choice

y  = pq^x

Note that when  x  = 0,   then y  = 5

This suggests that

5 = pq^0

5  = p (1)

5  = p

And the point  (4, 405)  is on the graph..so

405  = 5q^(4)    divide both sides by 5

81  = q^4        take the 4th root of each side and we get that

3  = q

So....to find  k   when  x  =2, we have

k  =  5*(3)^2   =  5 * 9   =  45

height of hovercraft  x  seconds after takeoff  =  h(x)  =  -(x - 11)(x + 3)

We want to know the height of the hovercraft  0  seconds after takeoff.

height of hovercraft  0  seconds after takeoff  =  h(0)  =  -(0 - 11)(0 + 3)  =  -(-11)(3)  =  33

So...

The height of the hovercraft at the time of takeoff is  33  meters.

low

By the way, it just occurred to me that:

1/4 of the area of the circle which includes the unshaded segment =4Pi

The area of the rectangle which includes the unshaded area = 4

Then the shaded area of the 1/4 circle - shaded area of the rectangle =4Pi - 4

p(adult and male)=12/45

p(adult and male)=0.27

true statements:

p(Female and Adult)=0.13

p(Child and Female)=0.20

Thanks, guest....that's easier than my approach  !!!!

A(w)  =   -w (w  - 80)  =  -w^2  + 80w

The width that maximizes the area can  be found  by :

-80 / [2* -1)  = -80/-2  = 40

And the maximum area  is

-40 (40 - 80)  =  -40 * -40  = 1600 m^2

[ This isn't surprising....a square always maximizes the area for a given perimeter....so....really....all we need to do is to divide the perimeter by 4  and square this  result....so we would get  (160 /4)^2  =

(40)^2  =   1600 m ^2   }

Ty

I believe the second one is P(A ∪ B) :)

 The perimeter of the rectangle:

    2L + 2 W = 160

   A = WL

    L + W = 80

  A = W ( 80 -w) = -W 2 + 80 W 

    The quadratic: - W2 +80 W 

     f( W)=  - W2 + 80 W  has maximum at Vertex:

                W = -80/-2 = 40 yard

                 L = 80 - 40 = 40

       Maximum area obtained by square with side 40 Yard

         The area : 40 * 40 = 1600 sq yds^2

Probabilty is  :

( Number Part Time )  /  ( Number35 - 49 )   =

(179) / ( 581 + 179 + 104 )  =

179 / 864  ≈   20.7 %

Don't know the second one, RP.....maybe someone else does

How about this method:

Area of the 1/4 circle =4Pi...........................................................(1)

Area of the segment={[pi/2 - sin(pi/2)] /2 x 4^2} / 2 =2Pi - 4........(2)

Area of rectangle MCND =4 x 1 =4

Shaded area in this rectangle =4 - (2Pi - 4) =8 - 2Pi....................(3)

(1) - (2) - (3) =4Pi - (2Pi - 4) - (8 - 2Pi) =4Pi - 4

These are all the same type of problem

Area  = (1/2) (product of known sides) * (sine  of included angle)

So...the first one is

Area  = (1/2) (11.1 * 7 ) sin ( 84°)  =  (1/2)(77.7) sin (84°)  = 38.6 in^2

Knowing this....can you do the other two    ???

P ( C l D )  =   P (D and C) / P (D )  = .16 / .80  =  .20  =  20%

The first is odd....it is symmetric to the origin

The last two are even....they are symmetric to the y axis

We have  21  marbles...we are drawing with replacement

P(red, red )  =  6/21 * 6/21  =  2/7 * 2/7  =   4/49    true  

P (blue, yellow)  = 4/21 * 9/21  =  4/21 * 3/7 = 4/49 false

P (yellow, red )  =  9/21 * 6/21  =  3/7 * 2/7  =  6/49 false

P(green, green )  = 2/21 * 2/21     =  4/441  false

P (yellow, green )  =  9/21 * 2/21  =  3/7 * 2/21  =  6/147  = 2/49  true

P (red, blue )  = 6/21 * 4/21  =  2/7 * 4/21  =  8 / 147  true

BTW.....we don't have an "amount" of marbles....we only have a "number" of marbles  !!!!

Second  one

P (G/ S)  =     P(S and G) / P(S)  =  .32 / .50   =  .64   = 64%