CPhill

y  = x^3         

Sequentially....we have

1 - Vertcal stretch by a factor of 3  ⇒  y =  3x^3

2 - Shift 4 units to the left  ⇒ y =   3(x + 4)^3

3 - Shift 3 units down ⇒  y  =  3(x + 4)^3  -  3

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P = kv^3           we need to find the variation constant, k

500 = k (10)^3

500 = k * 1000   divide both sides by 1000

(1/2)  = k

So   at   v  = 12 m/s , we have that

P  = (1/2) (12)^3

P = (1/2) (1728)

P  = 864 kW

(d - 4b)³  =

C(3,0) *d³  -  C(3, 1)* d² * (4b)  +  C(3, 2)* d * (4b)²   -  C(3,3) * (4b)³  =

d³    -  3* d²* 4b   +  3 * d * 16b²    -  64b³ =

d³   -  12d²b  + 48db²  - 64b³

Thanks to Geno, Melody, Heureka and Guest  !!!

Here's an algebraic solution.......

As the Guest pointed out.....this will be a 4th degree polynomial in the form

ax^4 + bx^3 + cx^2 + dx + e

Since (0, -3)  is on the graph, then  e  =  - 3

We can first solve for a and c

So we have that

a(-2)^4 + b(-2)^3 + c(-2)^2 + d(-2)  - 3  = 29

a(2)^4  + b(2)^3  + c(2)^2  + d(2)   - 3   = 29

Adding these, we have that

32a  + 8c  -  6   =  58

32a + 8c  =  64

4a + c   =  8        (1)

And  we have that

a(1)^4 + b(1)^3 + c(1)^2 + d(1)  - 3  = -1

a(-1)^4  + b(-1)^3  + c(-1)^2  + d(-1)   - 3   = -1

Adding these we get that

2a  + 2c   =  4

a + c   =  2      (2)

Subtract   (2)  from (1)

3a  =  6   ⇒   a   =  2

And    2 + c =  2    ⇒   c  = 0

And we have that

2 (1)^4  + b(1)^3  + d(1)  - 3  =  -1

2(-1)^4  + b(-1)^3  + d(-1) - 3  = - 1

Subtract these

2b +  2d  =  0

b + d  = 0       (3)

And

2(2)^4 + b(2)^3 + d(2)  - 3  = 29

2(-2)^4 + b(-2)^3 + d(-2)  - 3  = 29

Subtract these

16b + 4d  = 0

4b + d   = 0           (4)

Subtract (3)  from (4)

3b = 0

b  = 0

And    b + d   = 0  ⇒   0 + d = 0  ⇒   d  = 0

So......the function is

2x^4  - 3

We need to solve this for A, the amount originally invested....

439.23  =  A(1.10)^4       divide both sides by (1.10)^4

439.23 / (1.10)^4  =  A   =  $300 

There will be 8 compounding periods....since the compounding is quarterly, the effective interest rate  is (.075/4)....so we have

A  =  P ( 1 + i/4)^(n * t)     

Where  P = 1000, i  = .075, n = 4 and t  = 2      ......so.....

A  = 1000 ( 1 + .075/4)^8  =  $1160

Cedric has deposited $12,000 into an account that pays 5% interest compounded annually. Daniel has deposited $12,000 into an account that pays 7% simple annual interest. In 15 years Cedric and Daniel compare their respective balances. To the nearest dollar, what is the positive difference between their balances?

Cedric will have   12000 * ( 1/05)^15  ≈ $ 24947.14

Daniel will have   12000 * (.07) * (15)  = $ 12600

Difference in balances  =  24947 - 12600   =  $ 12347

(a)   M and B  =  14

       B and H   =  14

       M and H  =  20         Total  48

(b)  M  =  9

      B   =  15

      H   =  5             Total   29

(c)   2      [ the number outside the circles ]

(d)   9

(e)  No more than  2   =   Total - Those who like 3 of them  =  63 - 8   =  55

Good job, OBT......!!!

These symbols often confuse me, too.....

The second one says that AC is congruent to PR....however....the triangles are similar, not congruent, so this is false....for the same reason, the fourth one is also false....the triangles would have to be congruent for BC = QR

The "formula" to figure the new points is as follows :

New point  =  ( Vertex of triangle - Dilation Center)*Scale Factor + Dilation Center

So we have

[ ( -3 , 7) - ( 5, -5)] *14  + (5, -5)  =

[ (-8, 12) ] *14  + ( 5, -5)  =

(-112, 168) + (5, -5) =

(-107, 163)

[ ( -7 , -5) - ( 5, -5)] *14  + (5, -5)  =

[ (-12, 0) ] *14 + (5, -5) =

(-168, 0) + (5, -5) =

(-163, -5)

[ ( 9 , 3 ) - ( 5, -5)] *14  + (5, -5)  =

[ (4, 8) ] *14 + (5, -5)  =

(56, 112) + (5, - 5)  =

(61, 107 )

Here's the pre and post images :