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Ohhh I thought there might be a better way to do it...  

Thanks, hectictar....

The intersection point of the three medians is called the centroid

There's an easy way to figure this......

Median   =   ( sum of the x coordinates of the vertices / 3 , sum of y coordinates of the vertices / 3 )  =

[  (4 + 2 - 4 ) / 3  ,  ( 8 - 6 + 4 )  / 3  ]    =

[ 2 /3  , 2  ]

m∠BCA  +  m∠BCD   =   180

m∠BCA   =   180  -  m∠BCD

m∠BCA   =   180  -  (27x - 3)

the sum of the angles in a triangle  =  180°

6x  +  (18x + 12)  +  m∠BCA   =   180

6x  +  (18x + 12)  +  (180 - (27x - 3))   =   180

6x  +  18x - 27x + 12 + 180 + 3   =   180

-3x   =   -15

x   =   5

m∠BCD   =   (27x - 3)°

m∠BCD   =   (27(5) - 3)°

m∠BCD   =   132°

We know that

the shortest side on the photo * the scale factor  =  the shortest side on land

3 * s  =  120

s  =  120/3  =  40

So each side on land is  40  times the corresponding side in the photo.

the longest side  =  40 * 8  =  320  m

the middle side   =  40 * 7  =  280  m

the midpoint of  AB   =   \((\frac{4+2}{2},\frac{8-6}{2})\)   =   (3, 1)

the midpoint of  AC   =   \((\frac{4-4}{2},\frac{8+4}{2})\)   =   (0, 6)

the midpoint of  BC   =   \((\frac{2-4}{2},\frac{-6+4}{2})\)   =   (-1, -1)

a line that passes through  (3, 1)  and  (-4, 4)  has the equation   \(y=-\frac37x+\frac{16}{7}\)

a line that passes through  (0, 6)  and  (2, -6)  has the equation   \(y=-6x+6\)

a line that passes through  (-1, -1)  and  (4, 8)  has the equation  \(y=\frac95x+\frac45\)

The first two lines intersect when

\(-\frac37x+\frac{16}{7}=-6x+6 \\ -\frac37x+6x=6-\frac{16}{7}\\ \frac{39}{7}x=\frac{26}{7} \\x=\frac{2}{3}\)

And    \(y=-6(\frac23)+6=2\)

The first two lines intersect at the point  \((\frac23,2)\) . So if  \((\frac23,2)\)  is on the third line, then we know that all three lines pass through  \((\frac23,2)\)  .

\(y=\frac95(\frac23)+\frac45=\frac65+\frac45=\frac{10}{5}=2\)

C  =  2*radius  + 273.92       (1)

C  =  2 * 3.14 * radius   =  6.28* radius   (2)

Set  (1)  = (2)    and we have that

2r + 273.92  =    6.28r        subtract  2r  from both sides

273.92  =  4.28r      divide both sides by 4.28 

273.92 / 4.28  =   r   =  64 ft

6a^2  ( 4a  - 5)  =

6a^2 * 4a  -   6a^2 * 5   =

24a^3   -  30a^2

The compound interest earned is just the final amount - original amount  =

2176.82   -  1200     =  $976.82

f (f(x) )   =  

c  [ cx / (2x + 3) ]

________________         =     x

2 [ cx / (2x + 3)] + 3

[c^2x] / (2x + 3) ]

___________________     =    x

[ 2cx + 6x + 9] / (2x + 3)

c^2x    =  [2cx + 6x + 9 ] * x

c^2x  = 2cx^2 + 6x^2 + 9x

(2c + 6)x^2 + (9 - c^2)x  =  0

For this to be 0  for all  real x {except x = -3/2}   we need both

2c + 6  = 0       and    9 - c^2  = 0

So  using the second    c  = 3   or  c = -3

But when  c  =  3,   2c + 6  =  12

But  when  c   = -3      both  2c + 6  and 9 - c^2    will = 0

So....  c  = - 3 

Using the Law of Cosines

BC^2 = BP^2  + PC^2  - 2(BP)(PC)cos (60)

BC^2  =  2^2 + 3^2  - 2(2)(3)(1/2)

BC^2 = 13  - 6

BC  = √7

Using the Law of Sines

sin BPC / BC  =  sin BCP / BP

√3/[2√7]  =  sin BCP / 2

sin BCP  = √(3/7)

arcsin [ √(3/7)  ]  = BCP ≈ 40.89°

And BCP  and  ACQ  are complementary....so ACQ ≈ 49.106°

Which means that CAQ  =  180 -  60  - 49.106  ≈ 70.893°

So.... using the Law of Sines again

AC / sin AQC   = CQ / sin CAQ

AC =  √3/ sin (70.893) ≈ 1.833

And using the Law of Sines once more

AC / sin AQC  = AQ / sin ( ACQ)

AQ  =  sin (ACQ) * AC / sin (AQC)

AQ =  sin (49.106) * 1.833 / sin (60)    ≈ 1.5996  =   1.6

That's great, I wish I could actually finish all mines. 

y  =  4x - 19

2x + y  =  95

Substitute  4x - 19  for  y  into the second equation.

2x + (4x - 19)  =  95

2x + 4x - 19  =  95

2x + 4x  =  95 + 19

6x  =  114

x  =  114/6   =   19

Let's call the second longest side  x  .

Now we can see that the figure is a rectangle with  10  square feet removed from it. So...

area  =  (9 ft)(x ft)  -  10 ft²

area  =  9x ft²  -  10 ft²

53 ft²  =  9x ft²  -  10 ft²

53  =  9x - 10

63  =  9x

x  =  7

And the perimeter, in feet  =  9 + 7 + 9 + 7  =  32

shaded area   =   (area of rectangle) - 2(area of triangle)

area of rectangle   =   length * width   =   (55 + 55) * 76   =   110 * 76   =   8360   cm²

area of triangle   =   (1/2) * base * height   =   (1/2) * 55 * 36   =   990   cm²

shaded area   =   8360 cm²  - 2(990) cm²   =   8360 cm²  -  1980 cm²   =   6380 cm²

The sphere and the wire will have the same volume.

volume of sphere  =  4/3 * pi * 3³  =  4/3 * pi * 27  =  36pi

volume of wire  =  area of cross section * length  =  pi * (0.4/2)² * length  =  0.04pi * length

volume of sphere  =  volume of wire

36pi  =  0.04pi * length                       Divide both sides of this equation by  0.04pi .

900  =  length

So the wire's length is  900 cm .

plz answer .. 

Just ask me if you have any questions

Hahaha

Awesome!

a)  f(x)  =  \(\frac95\)x - 4          This has an inverse because it is just a linear equation.

y  =  \(\frac95\)x - 4        To find the inverse, solve this equation for  x , so add  4  to both sides.   

y + 4  =  \(\frac95\)x       Multiply both sides by  \(\frac59\) .

\(\frac59\)(y + 4)  =  x     So the inverse function is...

f^-1(x)  =  \(\frac59\)(x + 4)      And to find  f^-1(20) , plug in  20  for  x  into this function.

f^-1(20)  =  \(\frac59\)(20 + 4)   =   \(\frac59\)(24)   =   \(\frac{40}3\)

b)  g(x)  =  4x² + 8x + 13

g(x)  does not have an inverse function because it would have two different  y  values for an  x  value, and for an equaton to qualify as a function, there can only be one  y  value for every  x  value.

c)  h(x)  =  \(\frac1{\sqrt{x}}\)  for  x > 0        Yes this has an inverse.

y  =  \(\frac1{\sqrt{x}}\)             To find the inverse, solve this equation for  x .

y\(\sqrt{x}\)  =  1

\(\sqrt{x}\)  =  \(\frac1{y}\)                 Square both sides.

x  =  \(\frac1{y^2}\)                    So the inverse function is..

f^-1(x)  =  \(\frac1{x^2}\)   for   x > 0

f^-1(4)  =  \(\frac{1}{4^2}\)   =   \(\frac{1}{16}\)

that looks neat. i think ill give it a try

MBND  is a parallelogram.....

The height is  1    and the width is 1/2.....so its area is   1 * (1/2)  = 1/2  units^2

And the shaded area  is (1/2)  of this   =  1/4 units^2

(e) How many of these students like Haydn and exactly one of the other two?

E isnt 55. (55 is wrong)

(f) How many of these students like no more than two of these composers?

 (-inf, 1] is the answer

c=-3