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Thanks ;)

Verify    (cos(x + y)) / (cos(x) + sin(y)) = (cos(x) - sin(y)) / (cos(y - x))

step 1

\(\frac{cos(x + y)}{cos(x) + sin(y)} \stackrel{?}{=} \frac{cos(x) - sin(y)}{cos(y - x)}\)

step 2

\([cos(x + y)][cos(y - x) ]\stackrel{?}{=} [cosx - siny][cosx+ siny]\\\)

Step 3

cos is an even funtion so cos(y-x)=cos (-(y-x))

\([cos(x + y)][cos(x - y) ]\stackrel{?}{=} [cosx - siny][cosx+ siny]\\ \)

step4

\(let A=x+y\;\;\;B=x-y\\ LHS=cosAcosB\\ LHS=[(cosAcosB+sinAsinB)+(cosAcosB-sinAsinB)]*0.5\\ LHS=[(cos(A+B))+(cos(A-B))]*0.5\\ LHS=[(cos(x+y+x-y))+(cos(x+y-(x-y)))]*0.5\\ LHS=[(cos(2x))+(cos(2y))]*0.5\\ LHS=\frac{cos(2x)+cos(2y)}{2}\)

step 5

so we have

\(\frac{cos(2x)+cos(2y)}{2}\stackrel{?}{=} [cosx - siny][cosx+ siny]\\ \frac{cos(2x)+cos(2y)}{2}\stackrel{?}{=} [cos^2x - sin^2y]\\ \)

step 6

As an aside:

\(cos(2y)=cos^2y-sin^2y\\cos(2y)=1-2sin^2y\\sin^2y=\frac{1-cos(2y)}{2}\\ and\;\; likewise\;\;\\ cos^2x=\frac{1+cos2x}{2} \)

step 7

\(\frac{cos(2x)+cos(2y)}{2}\stackrel{?}{=} [cosx - siny][cosx+ siny]\\ \frac{cos(2x)+cos(2y)}{2}\stackrel{?}{=} [cos^2x -sin^2y]\\ likewise\\ \frac{cos(2x)+cos(2y)}{2}\stackrel{?}{=} [\frac{cos(2x)+1}{2} - \frac{1-cos(2y)}{2}]\\ \frac{cos(2x)+cos(2y)}{2}\stackrel{?}{=} \frac{cos(2x)+cos(2y)}{2} \\ \)

Step 8

Left hand side and right hand side are identical so the equality is verified.

What have you thought about so far?

What will the 2 numbers have to be if their sum is odd?  etc

Talk about your questions. This will help you figure out at least some of it by yourslef and that is how you will lean.

Please no one answer over the top of me.

Let the point be  ( x , (x^2 - 3) / sqrt(2) )

D = sqrt (  x^2  + (x^2  - 3)^2 /2 )         take the derivative and set to 0

D' =   (2x + 2x(x^2 - 3) ) /  sqrt (  x^2  + (x^2  - 3)^2 /2 )    = 0

2x  + 2x^3 - 6x    = 0

2x^3 - 4x  = 0

2x ( x^2 - 2) = 0

x = 0,  x = sqrt (2)     or x = -sqrt (2)

When x = 0

D =  sqrt (0 + 9/2)  =  sqrt (9/2)  

When x = ±sqrt (2)

D = sqrt ( 2 + 1/2)  =  sqrt (5/2)  =  sqrt (10)/2

a = 10    and b = 2

So

a + b  =  12

All these are equivalent:
1 - 10 nPr 4 - 8 nPr 4             =3,360 
2 - [10 nPr 4]/2 + [8 nPr 4]/2 =3,360
3 - 9 nPr 4 =3,024 + 3024/9   =3,360

First, add the number of children and adults: 298 people

298/42 about 7.095, but we have to round up because we can't have the part of a coach. Thus, the answer is 8 coaches.

Thank you very much! Do you have any idea on how to approach the other two problems and solve them?

(272 + 26) / 42  ≈ 7.095 coaches =  8 coaches

y =    8x^3  + 36x  + 54x^(-1) + 27x^(-3)        take the derivative....set to 0

y '  =  24x^2 + 36 - 54x^(-2) - 81x^(-4)  = 0   

y ' = 3 ( 8x + 12 - 18x^(-2) - 27x^(-4)  )  =  0

8x^2 + 12 - 18x^(-2) - 27x^(-4) = 0

8x^6 + 12x^4 - 18x^2 - 27 = 0   factor

4x^4 ( 2x^2 + 3)  -  9(2x^2 + 3)  = 0

(4x^4 - 9) (2x^2 + 3) = 0      the second factor gives a non-real answer when set to 0

So

4x^4 - 9 = 0

4x^4 = 9

x^4 = 9/4

x =  (9/4)^(1/4)  =  [(3/2)^2] ^(1/4)  = √[3/2]

Subbing this back into the original function

8[√[3/2] ]^3 + 36(√[3/2] ) + 54 / (√[3/2]) + 27 / [ √[3/2] ]^3   =

8 (3/2)(√[3/2] )  + 18√6   +     54 ( √[ 3/2] ) / (3/2)   +  27 /  [ (3/2) √[3/2] )

12√[3/2]   + 18√6 +  + 36  √[3/2]  +   18 / √[3/2]

6√6  + 18√6  + 18√6   +  6√6  =

48√6

\(E[profit] = P[win] (\text{profit if win} )+ P[lose](\text{"profit" if lose})= \\ (0.2)((0.1)400000-2000)+ (1-0.2)(-2000) = \$6000 \)

How did you get that?

[10 nPr 4] - [8 nPr 4] = 3,360

Set the circles equal

(x - 3)^2 + (y - 4)^2  =  (x + 4)^2 + (y - 3)^2

x^2 -  6x + 9   + y^2  - 8y + 16  =  x^2 + 8x + 16 + y^2 -6y + 9

-6x - 8y =   8x - 6y

-2y = 14x

y = -7x

Sub this into (x - 3)^2 + (y - 4)^2 = 25

(x - 3)^2 + ( -7x - 4)^2 = 25

x^2 - 6x + 9 + 49x^2 + 56x + 16 = 25

50x^2 + 50x + 25 = 25

x ( x + 1) = 0

x = 0     or x = -1

So....the intersection of the circles will occur at   (0,0) and (-1, 7)

Find the intersection of    (x - 3)^2 + ( y - 4)^2 = 25   and   y = - 6x

(x - 3)^2 + ( -6x - 4) = 25

x^2 - 6x + 9 + 36x^2 + 48x + 16 = 25

37x^2 + 42x  = 0

x (37x + 42) = 0 

x = 0   or x = -42/37

Putting these into y = -6x

(0, 0)  and (-42/37, 252/37)

And find the intersection of   (x +4)^2 + ( y - 3)^2 = 25   and   y = - 6x

(x + 4)^2 + (-6x - 3)^2 = 25

x^2 + 8x + 16 + 36x^2 + 36x = 9 = 25

37x^2 + 44x = 0

x (37x + 44) = 0

x = 0   or x = -44/37

Putting these into y = -6x

(0, 0)   and ( -44/37, 264/37)

Note that (0,0 )   will satisfy all three

But only  (-1, 7) ( -42/37, 252/37)  and (-44/37, 264/37)  will satisfy  two of the three

F.C. =13.54

Thank you for the heads up!! I greatly appreciate it!! With your help this is what I come up with...

P(defective) = 1-P(defective)

E(profit) = P(defective) · $350 - P(defective) · $700

P(defective) = 1 - 0.025 = 0.975

E(profit) = 0.975 · $350 - 0.975 · $700

                           $341.25 - $68.25 = $273

E(profit) = $273

Either box can be selected first

If the first box is the first one selected

(3/7)(2/6)  =   6/42  = 1/7

If the second box is selected first

(4/7) (3/6) = 12/42  =  2/7

So...the combined probabilities =   1/7 + 2/7  = 3 / 7

Let the fuel consumption before modification = FC

So

FC ( 1.27) =  17.2      divide both sides by 1.27

FC ≈   13.54

Call the side of the square, S

So...the area of the square = S^2

The area of the small triangle at the bottom left  = S* [ sqrt (4^2 - S^2) ] /2 / 2      (1)

The area of the small triangle at the top right = S* [ sqrt (5^2 - S^2)] /2        (2)

And the area of the small triangle at the top left =   [ S  - sqrt (4^2 - S^2)] [ S - sqrt (5^ - S^2) ] / 2     (3)

And the area of the right triangle = 6     (4)

So

(1) + (2) + (3) + (4)  =  S^2

S* [ sqrt (16 - S^2 ] /2  + S* [ sqrt (25 - S^2)] / 2   + [ S  - sqrt (16 - S^2)] [ S - sqrt (25 - S^2) ] / 2 + 6 = S^2 

S* [ sqrt (16 - S^2 ]   + S* [ sqrt (25 - S^2)] + [ S  - sqrt (16 - S^2)] [ S - sqrt (25 - S^2) ]  + 12 = 2 S^2 

S* [ sqrt (16 - S^2 ]   + S* [ sqrt (25 - S^2)] + [ S  - sqrt (16 - S^2)] [ S - sqrt (25 - S^2) ]  + 12 = 2 S^2 

S* [ sqrt ( 16 - S^2) ] + S*[ sqrt (25 - S^2) ]  + S^2  - S*[ sqrt (16 - S^2)] - S* [ sqrt (25 - S^2)] +

sqrt [ ( 16 - S^2)  (25 - S^2) ] + 12  = 2S^2

sqrt [ (16 - S^2) (25 - S^2)] =  S^2 - 12       square both sides

(16 - S^2) ( 25 - S^2)  =  S^4 - 24S^2 + 144

400 - 41S^2 + S^4  = S^4 - 24S^2 + 144

17S^2  - 256  = 0

17S^2 = 256

S^2  =   256 / 17