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3x-2 = 1/2y

y = 6x-4          this is an equation of a line (slope , m = 6   and b , y-intercept = -4)....ANY point on this LINE solves the equation, thus there are INFINITE possibilities.

Nice, EP  !!!!

Also note that

1*2  = 2

2*3  = 6

3*4  = 12

So....it appears that the number of dots in the nth figure  = (n) (n + 1)

(10) (10 + 1)  =  10 * 11  =   110

1.

n^3  = ( n - 1)^3 +( n - 2)^3 + ( n - 3)^3

Simplify

n^3  =  3n^3 -18n^2 + 42n - 36     rearrange as

2n^3 - 18n^2 + 42n - 36  = 0        divide throught by 2

n^3 - 9n^2 + 21n - 18  =  0    [ split up the 9n^2 term as 6n^2 + 3n^2 ]

n^3 - 6n^2 + 3n^2 +21n - 18  = 0     this factors as

n^2 ( n - 6) + 3(n^2 + 7n - 6)  = 0

n^2(n - 6)  + 3 (n - 6) ( n + 1) = 0

(n - 6 ) [ n^2 + 3(n + 1) )  = 0

(n - 6) [ n^2 + 3n + 3]  = 0

The second factor has no real solutions when set to 0  [ the discriminant is negative ]

So

n - 6  =  0

n = 6

1)  y = 1/3 x + 1

2)  y = 2x-2              are the equations of the lines

Multiply 2) by -1 and add to 1) to get

x=9/5

sub in to either  1) or 2)    to find y = 8/5

Multiply together to get     9/5 * 8/5 = 72/25

This is just solving a system of equations...

1    y =- 3x+5

2   y = 2x-6            multiply by -1

    -y = -2x+6          and add to equation 1

     0 =-5 x +11         x = 11/5      sub this in to either 1   or 2   to find y = -8/5     (11/5, -8/5 )

\($\frac{\sqrt{2} + a + \sqrt{b}}{c}$ \)

     1   [ 1 - ( √2 - √3) ]                         1 - √2 + √3                  1 - √2 + √3             1 - √2 + √3

__________________________  =  ______________  =   _____________ =  ________  =

[1 + (√2 - √3)]   [ 1 - ( √2 - √3) ]            1  -  (√2 - √3)^2         1 - [ 2 - 2√6 +3]         2√6 - 4

1 - √2 + √3

_________  =

2(√6 - 2)

[1 - √2 + √3 ]  ( √6 + 2)         [ √6 - √12 + √18 + 2 - 2√2 + 2√3 ]

__________________   =  ____________________________   =

2(√6 - 2)  (√6 + 2)                    2 ( 6 - 4)

[ √6 - 2√3 + 3√2 + 2 - 2√2 + 2√3 ]

__________________________  =      

   2 (6 - 4)

√2 + 2 + √6

__________

        4

So

a + b + c  =    2 + 6 + 4   =   12

1^2 +1

2^2 +2

3^2 +3

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10^2 + 10 = 110 dots

a)  r  =      2

          __________       

            5 +  7cos θ

Standard Form  is

r  =     ep

      ___________

       1 + e cos θ

We  need   a "1" as a constant in the denominator to put this into SF.....divide every term by 5

r  =       2/5

       ____________

        1 + (7/5)cos θ

The eccentricity, e,  =   7/5

If

0 ≤ e < 1      we have an ellipse

e = 1     we have a parbola

1 < e  we have a hyperbola

So.....this is a hyperbola

(b)   If this is an ellipse....we must have that

0 ≤ e < 1

0 ≤ b / 5 < 1

0 ≤ b < 5

The lines y=5x+3$, $y=-2x-25$ and $y=3x+k$ intersect at the same point. What is the value of $k$?

y = 5x + 3         y  = -2x - 25          y = 3x + k

Set the first two equations equal for y.....solve for x

5x + 3  =  - 2x - 25

7x  =  - 28

x = -4

So  y =  5(-4) + 3  =  -17

So......the intersection point is   (-4, -17)

So

y = 3x + k       implies that

-17  = 3(-4) + k

-17= -12 + k

k = -5

Here's a graph : https://www.desmos.com/calculator/vgpkw03igb

What is the value of a$ if the lines $2y - 2a = 6x$ and $y + 1 = (a + 6)x$ are parallel?

Put these in slope intercept form.......

2y - 2a = 6x          y + 1  = (a + 6)x

y - a  = 3x             y  =  (a + 6)x - 1

y = 3x + a

If these lines are parallel  then  the slopes  are equal which implies that

3 =  a + 6      subtract 6 from both sides

-3  = a

See the graph here : https://www.desmos.com/calculator/nk5b6mfkdu

The product of the LCM and GCD of two numbers is the same as the produce of the two numbers.

(This makes sense- when you create factor trees to find the LCM of two numbers you take the common factors between the two trees and multiply them. When finding the GCD you take all the unique factors and one of the doubles.)

3720*8 = 29760 so 120 times the other number is 29760

29760/120 = \(\boxed{248}\) is the other number

If x + 2y= 4 and xy = -8, what is the value of x^2 + 4y^2?

x + 2y = 4        square both sides

(x + 2y)^2  = 16

x^2 + 4xy + 4y^2  = 16          [xy = -8 ]

x^2 + 4(-8) + 4y^2  = 16

x^2 + 4y^2 - 32  =  16

x^2 + 4y^2  = 48

10 )  Using the same idea as in (7)....we have that

(sqrt(2)/3)^2 + y^2  = 1

(2/9) + y^2  = 1

y^2  =  1 - 2/9

y^2  = 7/9

In Q$....y is negative....so we need the negative square root

y = -sqrt(7)/3

So

tan  θ  =  y/x  = -sqrt(7)/ 3          -sqrt (7)

                        _________  =  ________    rationalize the denominator  and we get

                         sqrt(2) / 3          sqrt (2)

-sqrt (7) sqrt (2)             -sqrt (14)

_____________    =     _________

sqrt(2)  sqrt (2)                  2

Did you proof-read your question? Missing some numbers?

7)   Since we have a unit circle, the form is  x^2 + y^2  = 1

So....we  have that

x^2 + (0.6)^2  = 1

x^2 + .36  = 1

x^2  = 1 -.36

x^2  = .64

Since we are in Q2, x is negative so we need the negative value of the square root

x = -sqrt (.64)   =  -.0.8

The tangent  = y / x  =    0.6/ -0.8   =  - 60/80  =  -3/4

5)  (-3, 2)

The cosecant  =  r/y       we need to find r  

r = sqrt [ (-3)^2 + (2)^2 ]  =   sqrt [ 9 + 4 ]  =   sqrt (13)

So    csc  θ   =  √13 / 2

6)  (5, -8)

Again we need r  =    sqrt [ (5)^2 + (-8)^2 ]  =  sqrt [ 25 + 64]  = sqrt (89)

So   csc  P   =  √89 / -8   =  - √89 / 8

x^2 + 16x +c

If this is a square of a binomial..we only have one real solution...so...the discriminant must = 0.....therefore

16^2  - 4(1)c = 0

256  =  4c

64  =  c

The binomial  is (x + 8)^2

\(\text{The cheat way is to remember the vertex of $a x^2 + bx + c$ occurs at $x=-\dfrac{b}{2a}$}\\~\\ \text{we see here that $x=-\dfrac{-2}{2}=1,~f(1) = 1-2-35=-36$}\\ v=(1,-36)\\~\\ \text{A better way (imo) is to complete the square}\\~\\ x^2-2x-35 = x^2 - 2x + 1-1-35 = (x-1)^2 - 36\\~\\ \text{and we can read the vertex right off as $(1,-36)$}\)

Copy the answer from your question about Umbriel's orbit !

\(h(t) = -16t^2 -24t+160 = 8(-2t^2-3t+20)\\~\\ h(t) = 0 \Rightarrow (2t^2+3t-20)=0\\~\\ t \dfrac{-3\pm \sqrt{9+160}}{4} = \dfrac{-3\pm 13}{4} = \dfrac 5 2,-4\\~\\ \text{Clearly $\dfrac 5 2s = 2.5s$ is the solution we want}\)

\((x+2)^2 + (y+7)^2 = 81 = 9^2\\~\\ \text{you can read the center right off as $(-2,-7)$ and the radius as $9$}\)

thank you!!!

Orbital Period = Kepler's Third Law 
P =2*pi*sqrt(a^3/(G*M)), where a =semi-major axis.
P =2*pi*sqrt(2.66E8^3 / (8.68E25*6.67428E-11))
P = 358130 - seconds
P =358,130 / 3,600 =99.48 - hours

Yolanda made 4 liters of punch for a birthday party. If she served 3,250 milliliters of punch, how many liters of punch were left  

There are a thousand milliliters in a liter, so the 4 liters Yolanda made is the same as 4,000 milliliters.

Subtract the 3,250 milliliters she served from the 4,000 milliliters she made, and she has 750 milliliters left. 

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can someone please help me i dont understand what i got wrong on this quiz