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2. 

Area of Sector  = pi *r^2 *  (100  / 360)

250 pi  =  pi * r^2 * (5/18)

r^2  = 250 (18 / 5)

r^2  =  50 * 18

r^2  = 100 * 9

r^2 = 900

r  = 30

1.

Let O be the center of the semi-circle

Draw CE  perp  to AB

Connect OC  = r

OE = 8.5

EB = r - 8.5

CB = 3

Using the Pythagorean Theorem

Height of triangle OCE   = Height of triangle ECB

CE^2  =  CE^2

r^2  - (OE)^2  = (CB)^2  - ( EB)^2

r^2 - (8.5)^2  =  (3)^2 - (r - 8.5)^2

r^2  - 72.25  =  9 - r^2 + 17r - 72.25

r^2 - 72.25  =  -r^2 + 17r - 63.25

2r^2 - 17r -9  = 0

(2r + 1) ( r -  9)  =  0

r =  9 

(2 -n)(2-n) + (2n -4) (n + 1)  =  (n + 1) ( 2 - n)

n^2 -4n + 4  + 2n^2 - 2n -4  = -n^2 + n + 2

3n^2 - 6n  = -n^2 + n + 2

4n^2 -7n   - 2  = 0

(4n + 1) ( n - 2)  = 0

Set each factor to 0 and  solve

n = -1/4     and  n =  2  (reject this answer....it makes a denominayor =  0 )

Simplify

-3y =  5x + 12

x intercept                             y intercept

0 = 5x + 12                            -3y = 12

-12 = 5x                                    y =  -4

x = -5/12

Distance between  x int and  y  int = 

sqrt [  (-5/12)^2 + (-4)^2 ] = 

sqrt [ 25/144  + 16 ]  =

sqrt [ 2329 / 144  ]   =

sqrt [ 2329  ] /  12   ≈   4.02

Write the equation of this line in the form Ax + By = C

\(m=\dfrac{y_p-y_q}{x_p-x_q}=\dfrac{2-7}{-8-4}=\dfrac{5}{12}\\ y=m(x-x_q)+y_q\\ y=\frac{5}{12}(x-4)-7\\ y=\frac{5}{12}x-\frac{5}{3}-7\\ -\frac{5}{12}x+y=-\frac{5}{3}-7\\ \dfrac{-5x+12y}{12}=\dfrac{-20-84}{12}\\ \dfrac{5x-12y}{12}=\dfrac{20+84}{12}\)

\(Ax+By=C\\ \color{blue}5x-12y=104\)

 !

A(-2,-4), B(-3,12), C(4,y) find y.

\(m= \dfrac{y_B-y_A}{x_B-x_A}=\dfrac{12-(-4)}{-3-(-2)}=-16\\ y=m(x-x_A)+y_A=-16(x-(-2))+(-4)=-16x-32-4\\ y=-16x-36\\ y_C=-16x_C-36=-16\cdot 4-36\\ \color{blue}y_C=-100\)

 !

6x + (3/5)y

xy = 2/5

y = 2 /(5x)

6x + (3/5)(2) / (5x)

6x  + (6/25)x^(-1)           take the derivative and set to 0

6 - (6/25)x^(-2)  =  0    multiply

6x^2  = (6/25)

x^2 =  1/25

x =   1/5

Take the scond derivative to see that this is a  min

 2(6/25)(.2)^(-3)  =  60   which is > 0 ....so this is a  min

Min  at (.2, 2)    =  6(.2) + (3/5)(2)  =  1.2 + 1.2  =  2.4

(-4,0)  and (0,2)  are on the line

Slope   = [ 2 - 0  ] /  [ 0 -  - 4]  =  2/ 4   =1/2

Equation  

y = (1/2)x + 2          multiply through by 2

2y = x + 4

x - 2y =  -4

OK...i'll try to think about it a litle more

Simplify  as

x^2 - 21x + y^2 + 13y  =  25     complate the square on x, y

x^2 -21x + 441/4   + y ^2 + 13y + 169/4 =   25 + 441/4 + 169/4

(x - 21/2)^2  + (  y +13/2)^2 =  355/2

This is a circle centered at  ( 21/2 , - 13/2)  with a radius of sqrt [ 355 / 2 ] 

Largest value of x =  21/2  + sqrt [ 355 / 2 ]  =   21/2 + sqrt [ 710 ]  / 2

yeah i dont think thats right cuz i tried it and it said it was wrong

also it doesnt have to line up on the same line they started on

Here's what I get

Planet Y  travels  360 / 450 =  4/5  as fast as Planet X

So.....Planet X  has traveled 360°  in 360 days

And Planet Y has traveled (4/5) * 360°  = 288°  in 360 days

So  there is a 72°  difference between them

So it takes   360 / 72   = 5   360 day periods before they are on the same line again

Planet  Z  travels  360 / 540  = 2/3  as fast as Planet X

Planet  Z travels  (2/3) 360  = 240°  when Planet X travels 360°

So....there is a 120° difference  between them

So...it  takes   360 / 120 =  3   360 day periods   before they are in the same  line again

5 * 3 =    15  360 day periods =  5400 days

solved it

360

i think the answer is 420 but its wrong

Take the derivative with respect to  and  set to 0

2x -16 = 0

x = 8

Take the derivative with respect to y and  set to 0

2y + 16 = 0 

y = -8                            

The coldest temp is

  (8)^2 + (-8)^2-16(8) + 16(-8) + 26  =  -102

Simplify as

3a + 2b  = 2

3a + (2 - m)b = k

For this to have infinite solutions, the equations must be the same

This implies that  m   =  0     and   k    =  2

There really isn't a direct explanation for this, but I'll try my best. Say 0.x repeating is a decimal, and you want to convert it to a fraction. First, name that value n and name 10 times that value 10n. Equation below:

 n = 0.xx...

10n = x.xx...

{Minus n from both sides}

 9n = x

 n = x/9

 \(Compute\ a^3b + ab^3\)

\(2x^2 - 8x + 7 = x^2 - 4x + 13\\ x^2-4x-6=0\\ x=2\pm\sqrt{4+6}\\ x\in \{2-\sqrt{10}\ ,\ 2+\sqrt{10}\}=\{a,b\}\)

\(a^3b + ab^3\\ =(2-\sqrt{10})^3(2+\sqrt{10})+(2-\sqrt{10})(2+\sqrt{10})^3\\ \color{blue}=-168\)

The only possible degree of p+q is 11.

No matter how you cancel out, you can't cancel out the coefficient of the ^11 term because the highest degree of q is 7. Because degree is defined by the highest exponent the degree is always 11.

The degree of the polynomial p + q depends on the terms of both p and q. There are two main scenarios to consider:

No common factors: If p and q have no common factors (i.e., no variable raised to the same power in both polynomials), then the degree of p + q will simply be the larger of the two individual degrees.

In this case, p has a degree of 11 and q has a degree of 7. So, the degree of p + q would be 11.

Common factors: If p and q share some common factors (i.e., a variable raised to the same power appears in both), then the degree of p + q can be lower than the highest individual degree.

However, the degree can only be lowered by at most 1. This happens when the terms with the common factors cancel out when added together.

In this scenario, the possible degree of p + q would be between 10 (1 less than the highest degree) and 11.

Therefore, depending on the specific form of p and q, the possible degrees of the polynomial p + q are:

11: If there are no common factors between p and q.

10: If there is a common factor that cancels out upon addition.

There's a really good inequality for this, Cauchy-Schwarz inequality.

\((a_1^2+a_2^2+\dots+a_n^2)(b_1^2+b_2^2+\dots+b_n^2)\ge(a_1b_1+a_1b_2+\dots+a_nb_n)^2\) , where a and b are a sequence of real numbers.

Applying this, we want to get xy on the right side so we do:

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

\(18*\frac{4}{3}\ge(x+y)^2\)

\(x+y\le\sqrt{24}\).

The maximum value is \(\sqrt{24}\).

(Yes the equality condition can be satisfied).