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Tysm! Can you help me with another question as well??

\( PA \cdot PB \)

Let the center of  both be (0,0)

Let the ellipse have its major axis along x....its equation is

x^2 / 12^2   + y^2 / (6.5)^2   = 1

The  foci  of the ellipse  =  (+/-  sqrt ( 12^2 - 6.5^5] , 0)  = ( +/- sqrt (407) / 2) , 0)

For  the hyperbola  the equation is

x^2/a^2  - y^2/b^2  =1

a^2 = 2.5^2  

c =  sqrt (407)/2

b^2  =  c^2  - a^2 =  407/4  -  2.5^2  =  191/2

The equation of the hyperbola is

x^2/ 2.5^2  - y^2/ (191/2)  =1

This is  a little difficult to solve  but here's a graphical solution using GeoGebra

PA = 9.5

PB =14.5

PA * PB  =  9.5 * 14.5 =  137.75

The function  can be written as

y = Asin (Bx)  +  C

A= the amplitude

B = 2pi / period =  2pi /pi  =  2

C = the vertical shift  =   -2

So

The function is

y = 3sin (2x) - 2

\(a_1=1 \)

\(a_n=(-5)a_{n-1}\)

Let the point be (x, y)

You want the minimum of

\(\sqrt{x^2+y^2}\)

and 

\(y = \frac{1}{\sqrt{3}} (x^2 - 7 + 2x)\)

Substitute y in, then use differentiation to find the minimum value of x

The value is about 1.82843 at x = 1.82843

Anything from 0-11 is possible (0 is only possible when p = 11, otherwise p wont be prime)

but the general form 11n+a where 0<=a<=10 should satisfy any prime number

You can easily factor into

\(x(9x^3+z^2-12)+21\)

So the remainder is 21

\(f(x) = 2x^2 - 13x + 20 - 5x^2 + 19x + 7.\\ f(x)=-3x^2+3x+27\\ x^2-x-9=0\\ x_0\in \{\frac{1}{2}\pm \sqrt{(\frac{1}{2})^2+9}\}\\ x_0\in \{\frac{1}{2}+\sqrt{9.25},\frac{1}{2}-\sqrt{9.25}\}\\ \color{blue}x_0\in \{3.541, -2.541\}\)

The x-coordinate(s) of all point(s) where the parabola y = f(x) intersects the line y = 0

are \(\color{blue}x_0\in \{3.541, -2.541\}.\)

 !

f(x)  = -3x^2  -6x + 27

Set to 0  to find the roots

-3x^2  -6x + 27 = 0

3x^2 + 6x  -27 = 0

3 (x^2 + 2x - 9) = 0

x^2 + 2x - 9 = 0

x^2  + 2x  =  9    complete the square on x

x^2  + 2x + 1 =  9 + 1

(x + 1)^2    = 10

x + 1  =  sqrt 10          or       x +  1  =  -sqrt 10

x = sqrt 10  -1           or x =  -sqrt 10  - 1

Negative on

(inf , -sqrt (10 )  - 1 )   U  (sqrt (10) -1, inf )

a(1)^2 + b(1) + c =  -24

a(4)^2 + b(4) + c = 10

a(3)^2 + b(3) + c = 60         simplify these

a +  b +  c =  -24           (1)

16a + 4b + c = 10          (2)

9a + 3b + c = 60            (3)

Subtract (2) - (1)  and (3) -(1)

15a + 3b  = 34     (4)

8a + 2b  =  84  →  4a + b  = 42  →  b = 42 - 4a       (5)

Sub (5)  into (4)

15a + 3(42 - 4a)  =  34

15a + 126 -12a = 34

3a  = -92

a = -92/3

b = 42 - 4(-92/3)  = 494/3

a + b + c =  -24

-92/3 + 494/3 + c = -24

c = -24 + 92/3  - 494/3   = -158

f(-1)  = (-92/3)(-1)^2  + (494/3)(-1)  - 158  =   -1060 / 3 

200        <------- (2)(100)     <-------- (i+1-i+1)(100)  <-----\(((i+1)-(i-1))(100)\)

Compute    

(2 + 3i) + (2 - 3i) + (-2 + 3i) + (-2 - 3i).     

given                                                           (2 + 3i) + (2 – 3i) + (–2 + 3i) + (–2 – 3i)    

without the parentheses                               2 + 3i + 2 – 3i – 2 + 3i – 2 – 3i    

rearranged for ease of recognition             +2 +2 –2 –2   +3i +3i –3i –3i    

looks like everything cancels out                 zero    

_.    

The answer is [3,infty).

Find the length of the shortest altitude in this triangle.

general circle equation

\((x – x_P)² + (y – y_P)² = r²\)

\(y^2=10^2-x^2\\ y^2=9^2-(15-x)^2\\ y^2=81-225+30x-x^2\\ 100-x^2=-144+30x-x^2\\30x=244\\ x_C=8.1\overline{33}\\ \color{blue}y_C=h_C=5.818\)

\(y^2=15^2-x^2\\ y^2=10^2-(9-x)^2\\ y^2=100-x^2+18x-81\\ 225-x^2=19+18x-x^2\\ 18x=206\\ x_A=11.\overline{44}\\ y_A=h_A=9.697\)

\(y^2=9^2-x^2\\ y^2=15^2-(10-x)^2\\ y^2=225-100+20x-x^2\\ 81-x^2=125+20x-x^2\\20x=-44\\ x_B=-2.2\\ y_B=h_B=8.726\)

\(\color{blue}The\ length\ of\ the\ shortest\ altitude\ in\ this\ triangle\ is\ h_C=5.818\)

 !

Give an example of a quadratic function that has zeroes at x = 2 and x = -2, and that takes the value 0 when x = 2.   

zeros at +2 & –2 gives us (x – 2)(x + 2)    

then multiplied out, the quadratic is x² – 4    

_.    

The answer is 18 degrees.

Using Herons Formula,

sqrt(s(s-AB)(s-BC)(s-CA)) = sqrt(17(17-15)(17-9)(17-10)) = sqrt(17 * 2 * 8 * 7) = sqrt(1904) = 4√119

Then, you solve for the altitudes:

BC_a = (2 * A) / BC = (2 * 4√119) / 9 = (8√119) / 9

CA_a = (2 * A) / CA = (2 * 4√119) / 10 = (8√119) / 10 = (4√119) / 5

AB_a = (2 * A) / AB = (2 * 4√119) / 15 = (8√119) / 15

The shortest one is (8√119) / 15.