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f(3)  = 4

So

(2/5)x + 3  = 3

x = 0

So

y  = 4f ( 3)  - 7   =

4 (4)  - 7   = 

9

(0, 9)  

is not one-to-one

[0, pi]

period  = pi

y intercept  = 0

Plot points   ( -pi/4, -1) , (0,0) and ( pi/4  ,1)

Simplify as

x^2 + (4m - 2)x + (m + 6)  =  0

We have one root when  the discriminant  =  0

(4m - 2)^2 - 4 (1) ( m + 6)  = 0

16m^2 - 16m + 4 - 4m  - 24   = 0

16m^2 -20m - 20  =  0 

4m^2 - 5m - 5  = 0 

m =   [ 5 + sqrt [ 25 + 80 ] ] / 8  =    [ 5 + sqrt 105 ] / 8

Simplify as

x^2 + 12x + y^2 + 2y  =  10        complete the square on x, y

x^2 + 12x + 36  + y^2 + 2y + 1 =  10 + 36 + 1

(x + 6)^2  + (y + 1)^2  =  47

47 = r^2

Area of this  circle =   pi r^2  =  pi * 47  =   47 pi

Period  = the distance between  successive "peaks"  on the curve

x value at first peak  =  1

x value at second peak  = 5

Period =  5   -1   =   4

a)  3(8/10)  + 8(2/10)  =  [ 24 + 16 ] / 10  =   40 / 10   = 4

b) 3(8/11) + 8 (3/11)  =  [ 24 + 24 ]  / 11 =   48/11

c) 3(8/12) + 8(4/12)  = [ 24 + 32 ]  / 12 = 58/12 =  29/6 

d)  3 ( 8 / (10+n))  + 8 ( 2 + n) / (10 + n)  = 6

24 / (10+n)  + [ 16 + 8n ] / (10+n)  = 6

24 + 16 + 8n   =  6(10+ n)

40 + 8n  = 60 + 6n

2n  = 20

n = 10

e)   3 (8  / (10 + n)) + 8 ( 2 + n)  /(10+n)  =  7

24 + 16 + 8n  =  7(10 + n)

40 + 8n  = 70 + 7n

n  =  30

x^2 -10x + y^2 + 10y  =  -c

Complete the square on x,y

x^2 -10x + 25  + y^2 + 10y + 25 =  -c + 25 + 25

(x + 5)^2  + ( y -5)^2   =  50  - c

We want

50 - c   =  1^2

50 - c  =  1

c  = 50 - 1 =   49

First one

Altitude YW =  sqrt [ XZ^2  - (YZ /2)^2 ]  = sqrt  [12^2  - (6sqrt 3)^2  ] = sqrt [ 144  - 108] = sqrt [ 36] = 6

sin (YZW  )  =  YW / YZ  = 6 / 12   = 1/2

Draw  WN  where N is the tangent point of  the semi-circle to YZ

radius of  semi-circle / sin (YZW) = WZ / sin (WNZ)

r  / (1/2)  = (6sqrt 3) / sin 90°

r = (1/2) (6sqrt 3)  =  3sqrt 3  = sqrt (27)

Area of semi-circle   = (1/2)pi r^2   = (1/2)pi (27)  =  13.5 pi

a^2 - 9   = (a + 3) (a - 3)

3a^2 - 3a  =  3a ( a  - 1)

GCF  =  1

sin (arcos (-sqrt (3) / 2)) =  sin (5/6 pi)  =  1/2

arctan ( sin (-pi/2))  =  arctan (-1)  = 7pi/4  =  -pi/4

1/2 - pi/4  =

[ 2  - pi ] /  4

Second one

Law of Cosines

AC^2  = AB^2 + BC^2  - 2(AB * BC) cos ABC

30^2  = 25^2 + 25^2  - 2(25 * 25) cos ABC

[ 30^2 - 25^2 -25^2 ] /  (-2 * 25 * 25)   = cos ABC

cos ABC =  7/25

sin ABC   = sqrt [ 25^2  - 7^2 ] / 25 =  sqrt [ 576 ] / 25  =   24 / 25

First one

Let the center  of the smaller circle  = M   and the larger one, N

Draw MA and NB

Extend   NM  and BA  to meet at  C

Let MC   =  3 + x

We have similar triangles  CMA and CNB

So

CM / MA = CN / NB

(3 + x) / 3   = (13 + x) / 7

7(3 + x) = 3(13 + x)

21 + 7x  = 39 + 3x

4x   = 18

x = 18/4 = 9/2  =  4.5

And we  can use similar triangles again  to  find TP

CM / MA  =  CT / TP

(3 + 4.5) / 3 = (4.5 + 6)  / TP

TP  = 3(4.5 + 6) / (3 + 4.5)

TP  = 3 (10.5) / ( 7.5)  =  4.2

Set the functions equal

x^3 + x^2  - 3x + 5  = x^3 + 2x^2 - 8x  + 11     simplify as

x^2 -5x + 6  =  0      factor as

(x -3) ( x -2)  = 0

x - 3 = 0       x  - 2   =   0

x = 3                x  = 2 

When x  = 2     y =  2^3 + 2^2 -3(2) + 5 =  11

When x =  3     y =  3^3 + 3^2 - 3(3) + 5  = 32

The intersection pts   are  ( 2, 11)   (3, 32)

Find all points (x, y) that are 5 units away from the point (2, 7) and that lie on the line 2x + y = 13

Helo ABJelly!

\(f_1(x)=\pm\sqrt{5^2-(x-2)^2}+7\\ f_2(x)=-2x+13\\ f_1(x)=f_2(x)\\ \pm\sqrt{5^2-(x-2)^2}+7=-2x+13\\ \sqrt{5^2-(x-2)^2}\ ^2=(-2x+6)^2\\ 5^2-(x-2)^2=4x^2-24x+36\)

\(25-x^2+4x-4=4x^2-24x+36\\ 5x^2-28x+15=0\\ x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ \color{blue}x\in \{\frac{3}{5},5\}\\ \color{blue}f_1(x)=11.8\\ \color{blue}f_2(x)=3\)

The points we are looking for are (0.6,11.8) and (5,3).

 !

There is color A and Color B.

Either there is:

3 sides color A 0 sides color B

3 sides color B 0 sides color A

2 sides color A 1 side color B

1 side color B 2 sides color A

Each of these has exactly one way of occuring, because they can always be rotated back to the same position

So there are 4 colorings. 

Simplify as

2x^2 + 16x + 30 < 3x^2 + 19x + 6

Rearrange as

x^2 + 3x - 24 > 0     (1)

Express as an equality

x^2 + 3x - 24 =  0   

x^2 + 3x  + 9/4   =  24 + 9/4

(x + 3/2)^2   = 105/4        take  both roots

x + 3/2  = sqrt (105) /2                                  x + 3/2   =   -sqrt(105)/2

x = [ -3 + sqrt (105) ] / 2   ≈  3.6                      x  = [ -3 -sqrt (105) ]  / 2 ≈  -6.6

Our answer  comes from  either  ( ≈ -6.6 , ≈ 3.6)     or  (-inf, ≈ -6.6) U ( ≈ 3.6 , inf)

Testing x =0 in (1)  makes it false

So

The solution comes  from  (-inf, [-3-sqrt (105) ] / 2)  U [ -3 + sqrt (105) ]  / 2  ,  inf)

z = a  + bi

2  - 3i(a + bi)    = 3 + 2(a + bi)

2 - 3ai - 3bi^2  =  3 + 2a + 2bi

2 - 3ai + 3b  =  3 + 2a + 2bi

-3ai =  2bi

-3a = 2b

a = (-2/3)b

2 - 3(-2/3)b i  + 3b  = 3 + 2(-2/3)b + 2bi

2 + 3b  = 3 - (4/3)b

(13/3)b =  1

b = 3/13

a = (-2/3)(3/13)  = -2/13

z = -2/13  + (3/13)i

Like so:

Use \(\lim_{n\rightarrow\infty}Summation\)

I'll leave you to write the whole expression in LaTeX.

To solve this problem, consider extending $\overline{BC}$ and $\overline{AB}$ until they intersect at point $G$, forming an equilateral triangle $ABCG$.

Now, $\angle ACB = \angle ACB = \angle GCB = 60^\circ$ because all three angles of equilateral triangle $DEF$ are $60^\circ$.

Triangle $GBC$ is isosceles with $GB = GC$ because $G$ is the intersection of the equilateral triangle's sides extended. Thus, $\angle GBC = \angle GCB = 60^\circ$.

Now, $\angle BGC = 60^\circ + 60^\circ = 120^\circ$. Since $\angle ACB = 60^\circ$, the exterior angle at $B$ of triangle $ABC$ is $120^\circ - 60^\circ = 60^\circ$.

Therefore, $\triangle ABC$ is an equilateral triangle, and $AE = \boxed{4}$.

This formula 

\(\cos(x)=\frac{a\cdot b}{||a||\,||b||}\). Just to clarify, The \(\cdot\) between the a and b is not a multiplication sign, it is a dot product.

This formula is very useful for calculating the angle between 2 vectors.

To prove: draw a vector, connecting vector  \(\vec{a}\) and \(\vec{b}\), remembering vector subtraction, this vector is \(\vec{a-b}\).

We see a cosine, so we hope to use the law of cosines to help us relate side lengths. However, we don't know the side lengths of these vectors. However, we have a very important symbol called the norm, \(||~||\), and this the distance to the endpoint to the origin. Remember, vectors aren't defined with position, so norm gives an effective way to represent length. Here is a graph:

Set \(\theta = \angle AOB\)

Therefore by the law of cosines, we have:

\(\cos(\theta) = \frac{{||a||}^{2}+{||b||}^{2}-{||a-b||}^{2}}{2||a||\,||b||}\).

To relate norm to dot product, we have another very important formula:

\(v\cdot v ={||v||}^{2}\). (dot product again). PS, notice the resemblance to the formula \(z*\overline{z}={|z|}^{2}\), for complex number z.

Using this formula, 

\(\cos(\theta) = \frac{a\cdot a+b\cdot b-(a-b)\cdot(a-b)}{2||a||\,||b||}\).

Dot products have a special property, because they are commutative, and distributive.

Therefore, \(\cos(\theta) = \frac{a\cdot a+b\cdot b-(a\cdot a- 2a\cdot b + b\cdot b)}{2||a||\,||b||}\)

Simplifying:

\(\cos(\theta) = \frac{2a \cdot b}{2||a|| \, ||b||}=\frac{a\cdot b}{||a||\,||b||}\).

This gives our desired formula.

Last one

CO  = r

PO = 20

PC  =sqrt [ 20^2 + r^2 ]

AP  = AB  - PB  =  2r - [ r + 20 ] =  r - 20

PQ = 7

Product of intersecting chord segments

PQ * PC  = AP * PB

7 * sqrt (20^2 + r^2)  =  (r-20)(r + 20)

7* sqrt [ 400 + r^2 ]  = r^2 - 400           square both side

49 (400 + r^2)  =  r^4 - 800r^2 + 160000

19600 +  49r^2  = r^4 - 800r^2 + 160000

r^4  - 849r^2  + 140400  =  0

This produces two positive solutions for r

r  =15   (reject)  PO is already greater than 15

r = 4sqrt (39)   

r^2  =  16 * 39   =    624

Second one

There may be a way better way to  solve this !!!

Using the Law of Cosines extensively

-cos DCB   =   cos DAB

Draw DB

DB^2  = 10^2 + 3^2  - 2(10*3) cos (DCB)

DB^2  = 6^2  + 2^2  + 2(6*2) cos (DCB)

100 + 9  - 60(cosDCB)  =  36 + 4 + 24cos(DCB)

69 = (24 + 60)cos (DCB)

69/84  = cos (DCB) = 23/28

sinDCB  = sqrt [ 28^2 - 23^2 ] / 28 = sqrt (255)/28 = sin DAB =  sin BCP

cosADC  = -cosABC

Draw AC

AC^2  =  10^2 + 6^2  - 2(6 * 10)cos ADC

AC^2  =  2^2 + 3^2   + 2(2*3)cos ADC

100 + 36  - 120cosADC  = 4 + 9 + 12cosADC

136 - 120cosADC  = 13 + 12cos ADC

123  =  132 cos ADC

cos ADC = 123/132  = 41/44 

sin ADC =  sqrt (44^2 - 41^2) / 44 =  sqrt (255) / 44  = sin ABC = sin PBC

sin PBC / sin BCP  =   PC / BP =  [ sqrt (255)/ 44 ] / [ sqrt (255) / 28]

PC/BP  = 28/44  = 7/11

PC = (7/11)BP

BP^2  = BC^2  + PC^2  - 2(BC * PC)cosBCP

BP^2  = 3^2 + [ (7/11)BP]^2  - 2 [ 3 * (7/11)BP ] cos DAB

BP^2  = 9 + [ (7/11) BP ]^2  - (42/11)BP (-cosDCB)

BP^2 = 9 + [ 49/121]BP^2  + (42/11)(23/28) BP

Let BP  =  x

x^2 - (49/121)x^2 - 69/22x  - 9 =  0

(72/121)x^2  - (69/22)x - 9  =  0

Solving this  for x  produces  x  = 22 / 3   =  BP

Set the distance to d, and the speed of the plane as x.

When there is no wind using the formula, distance/rate = speed.

\(\frac{2d}{x}=\frac{27}{4}\).

When there is wind on our war there we are 50 mph faster, but on our way back we are 50 mph slower, so

\(\frac{d}{x+50}+\frac{d}{x-50}=\frac{15}{2}\).

\(\begin{cases}\frac{2d}{x}=\frac{27}{4}\\\frac{d}{x+50}+\frac{d}{x-50}=\frac{15}{2}\end{cases}\)

\(\begin{cases}d=\frac{27}{8}x \\ \frac{d}{x+50}+\frac{d}{x-50}=\frac{15}{2} \end{cases}\)

\(\frac{27x}{4x+200}+\frac{27x}{4x-200}=15\)

We get our positive answer is \(x=50\sqrt{10}\), note the negative answer is \(-50\sqrt{10}\), which is the same because it just means the wind is blowing in the opposite direction.

Therefore,  \(\bf{d = 50\sqrt{10}*\frac{27}{8}=\frac{675}{4}\sqrt{10}}\).