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Good job, CC!!!

Focus  =  (-8, -3)

Directrix  x = - 2

This parabola opens to the left because the focus is to the left of the directrix

For this reason....we can throw out answers B and D.....these parabolas open up/down

The  vertex is  [ (-8 - 2)/2, -3)  =  ( -10/2, -3)  =  (-5, -3)

We can stop here

"C"   has  a vertex of (-5, -3)   and the presence of the " - "  in front of the "12" indicates that this one opens to the left

Expanding  (a + bi)⁴  =  a⁴ + 4a³bi + 6a²b²i² + 4ab³i³ + b⁴i⁴  =  a⁴ + 4a³bi - 6a²b² - 4ab³i + b⁴

Since all the i-terms must cancel for the answer to be a real number:   4a³bi - 4ab³i - 24i  must have value 0.

   --->   4abi(a² - b²) - 24i  =  0

   --->   ab(a² - b²) - 6  =  0                 

   --->   ab(a² - b²) = 6 

Therefore, using factors of 6:  either  ab = 1  and a² - b² = 6  which is impossible because a = 1 and b = 1 and then a² - b² isn't 6

                                                      or  ab = 2  and a² - b² = 3  which means a = 2 and b = 1  -- this works!

                                                      or  ab = 3  and a² - b² = 2  which means a = 3 and b = 2 which is impossible

                                                      or  ab = 6  and  a² - b² = 1  which means a = 6 and b = 1 which is impossible

Now that you know that a = 2 and b = 1, place these back into the original formula   (a + bi)⁴ - 24i  to find the value of c.

Sorry! I meant to get back with you on this question... I believe the answer is D. 5 + 4i

I believe you have to switch the sign of the imaginary part (the part with the i).

a + bi is a - bi–– technically how you evaluated yours 

the conjugate... 5 - 4i is 5+4i 

Am I correct? 

jkhnjk.lk./,..

Odd

Remember that the odd function has the property that....

f(-x)  = -f(x)

So

sin(-150)  =  -sin(150)  =  - 1/2    [ or, -0.5 ]

Here's my best shot, K

P(win both races) = (.25)(.35)  = .0875

P(win one of the two races) =  P(win 1st, lose second) + P(lose first, win second) =

(.25)(.65)  = .1625   +  (.75)(.35)  = .2625

So (.1625 + .2625)  = .425

P(lose both) = (.75)(.65) = .4875

Expected value   =

100000(.0875)  + 60000(.425) + 15000(.4875)  -  25000  =

$16,562.50  ⇒  D

Sorry, I accidently labeled it trig because this question is part of my trig test.

Don't see any trig, here.....

(C)   cosine is even, sine  is odd

Therefore

cos(-B)  = cos(B)    and

sin(-B)  =  -sin(B)

\(\text{In general an arithmetic sequence has the explicit form }\\ s_n = n d + s_0\\ \text{It's recursive form is }\\ s_n = s_{n-1}+d\\ \text{In both cases }s_0 \text{ is specified}\)

\(fine_5=2.25=5d+fine_0\\ fine_{21}=6.25=21d+fine_0\\ fine_{21}-fine_5 = 4= 16d\\ d=\dfrac 1 4\\ 2.25=\dfrac 9 4 = \dfrac 5 4 + fine_0\\ fine_0=1\)

\(a) ~fine_n = fine_{n-1}+0.25,~fine_0=1\\ b)~fine_n = (0.25)n + 1\)

\(c) ~\text{Hopefully clearly he'll choose the explicit formula. Think about it.}\)

\(d)~fine_{60}=60(0.25)+1 = 16\)

Let  CO  =  x

△COB  is a  30° - 60° - 90°  triangle. So     BC  =  2x

△COA  is a  45° - 45° - 90°  triangle. So     AC  =  √2 x

And  x  is positive, so     √2 x  <  2x .  So     AC  <  BC

AC  =   √3

√2 x  =  √3

x  =  √3 / √2       Multiply the numerator and denominator by  √2

x  =  √6 / 2

BC  =  2x

BC  =  2( √6 / 2 )

BC  =  √6

By the Pythagorean theorem,

AC² + BC²  =  AB²

(√3)²  +  (√6)²  =  AB²

3  +  6  =  AB²

9  =  AB²           AB is a length, so take the positive sqrt of both sides.

3  =  AB

thank you all, sorry, was afk

a)  Here's  the graph, GM :   https://www.desmos.com/calculator/et1lsbetqj

b) Since a log of 0 isn't defined....the graph has a vertical asymptote of x  = 3

c) The function is "down" as x approaches  3 from the right  and "up" as x approaches infinity

I'd like someone with a very good understanding of complex numbers to look at this question and answer.

Maybe Heureka or Alan or Rom or Tiggsy.

Since angle CAO is greater than angle CBO, then AO < BO

And since CO = CO.....then AC < BC

So....AC is the shorter leg of  triangle ABC

And since CO is perpendicular to the plane....then triangles AOC and BOC are right

Then

sin 45  =  OC / sqrt(3)

1/sqrt(2) * sqrt (3)  =  OC  =  sqrt (3/2)

And

sin (30)  = OC  / BC

(1/2) = sqrt (3/2) / BC

BC  = 2sqrt(3/2)

So.....AC^2  = 3   and BC^2  = 6

AB  =  sqrt [ AC^2 + BC^2 ]   =    sqrt [ 3  + 6 ]  =  sqrt (9) =  3

Thanks Chris :)

\(\text{Given the removable singularity at }x=1 \text{, and the vertical asymptote at }x=-1\\ \text{and the fact that }q(x) \text{ is quadratic}\\ q(x) =q_0 (x-1)(x+1)\\ \text{because of the removable singularity we also know that }p(x) \text{ has }(x-1) \text{ as a factor}\)

\(q(2)=3\\ q_0 (1)(3)=3\\ q_0=1\\ q(x) = (x+1)(x-1)=x^2-1\)

\(\text{As }\dfrac{p(x)}{q(x)} \text{ has a horizontal asymptote of }0\\ \text{the degree of }p(x) \text{ must be less than that of }q(x)\\ \text{so }p(x)=p_0(x-1)\\ p(2)=1\\ p_0=1\\ p(x)=x-1\)

\(p(x)+q(x) = (x-1)+(x^2-1) = x^2 + x - 2\)

Guest,

Don't you have please or thankyou in your vocabulary? 

The last one

x^10 + 7x^9 + 14x^8 + 1729x^7 - 1379x^6  = 0

x^6 ( x^4 + 7x^3 + 14x^2 + 1729x - 1379)  = 0        [ we know that x = 0 is a solution ]

x^4 + 7x^3  + 14x^2 + 1729x - 1379  = 0

Using the Rational Zeroes Theorem...the possible factors of 1379   are    ± [1, 7, 197, 1379]

However.....none of these are roots

The graph here shows that there is only one positive real solution :

https://www.desmos.com/calculator/u9lxz5cq5q

c(x) = log₆ x

2) The graph has no y-intercept.

Yep they're the same!