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Adding \(2\pi\) to the degree will result in the same thing, since radius is not changing so another possible polar coordinate to express \((3, \frac{3\pi}{4})\) is \((3, \frac{11\pi}{4})\).

If we have a negative degree than, it essentially flips the side so:

\((-3, \frac{7\pi}{4})\) and \((-3, \frac{-\pi}{4})\) are all equal. These are our three.

Just figured it out! Each concentric circle has a radius of 1, so the x coordinate for \(A_1\) would be 2, and the y coordinate is the angle \(\theta\) based off the unit circle so in this case for \(A_1\) it would be \(\frac{\pi}{6}\).

For the function f(x), you can have x be any real number other than negative numbers because the square root of a negative number is imaginary. We can see in the denominator that x can be negatives because it gets squared which makes it positive, but that doesn't change the domain because the top would still be imaginary. That means the smallest possible x value to be inputted into the function is 0.

This makes the domain [0, inf).

How many ways can you write two different letters chosen from 
A B C D E F G H I J K L M N O P 
so that they are in alphabetical order, and the first letter is a vowel?  

two different letters  

in alphabetical order  

the first letter is a vowel  

If the first letter is A, then there are 15 possibilities.  (Second letter can be anything past A.)  

If the first letter is E, then there are 11 possibilities.  (Second letter can be anything past E.)  

If the first letter is I, then there are 7 possibilities.     (Second letter can be anything past I.)  

If the first letter is O, then there is 1 possibility.         (Second letter can be only P.)  

Add them up.                                34 possibilities.   

_.

Just solve for each variable from the bottom-up

a = 6

b = -22/5

And so on

Borrowing from proyaop's answer

14 + 2sqrt (41)  ≈ 26.8

This is  the larger root  and all  integer x's > than this will make the polynomial > 0

So.....the greatest integer that makes  this le to 0   is 26

a + 2b = 3

5a + (2+ m)b = k 

This system will have infinite solutions when the equations are the  same

Multiply the first equation through by 5

5a + 10b  = 15

5a + (2 + m)b  =  k

This will  have infinite solutions when   m = 8  and  k  = 15

******

4a + 6b = 5

        (2 - m) b  =  k

This only has infinite solutions if  both equations  are the same......not possible

There are \(6^6\) total ways to color the cube. There are 12 ways to choose 2 adjecent sides in a cube(\({6\choose 3}-3\)) and 6 ways to color them. There are also \(6^4 \) ways to color the rest of the sides but we must take into account symmetry of a cube. There are 24 possible ways to orient a certain coloring of a cube(\(6\cdot4\) 6 ways to choose the base and 4 ways to choose a side) So the total number of ways to get a coloring with atleast one adjacent side is:
\(12\times6\times6^4\div24=\frac{6^5}{2}\)
Therefore:
\(\Large P=\frac{\frac{6^5}{2}}{6^6}=\boxed{\frac{1}{12}}\) Im pretty sure its right

Observing, we have 3 overarching fractions in which their denominators cannot be 0, and thus if we set their denominators equal to 0, we can find the 3 values of x that are not in the domain of f(x) (domain means all the values of x that "work/are allowed" for the function):

1) \(1 + {2\over{1 + {3\over{x}}}}=0\)

Solve for x: \(-1 = {2\over{1 + {3\over{x}}}}\); \(-1 - {3\over{x}}=2\); x = -1

2) However, what if \(1 + {3\over{x}} = 0\)? Solving for x, you would get x = -3

3) Lastly, x = 0 is quite obvious, because then 3/0 would be undefined.

Their sum is -1 + (-3) + 0 = -4.

         To not be in the domain of a function, then x must be a value in which the denominator is equal to 0 and thus the expression is undefined. x^2 + 7 can never be 0 for a real value of x, because x^2 is nonnegative. x^3 - x^4 = x^3(1 - x), which this expression is 0 when x = 0 and 1. Additionally, x^2 - 3x + 2 = (x - 2)(x - 1) is 0 for when x = 2 and 1.

         Altogether, we have x = 0, 1, or 2 that will result in one of the fractions having a denominator of 0... Since 1/0 is undefined, x = 0, 1, and 2 are not part of the domain of f.

         Hence, their sum 0 + 1 + 2 = 3, which is your final answer.

If x is 0, then you would get undefined when calculating 1/x, or 1/(x^2), or 1/(x^3), with x = 0 being the only value not in the domain of f(x). Thus, there is ONE real number that is not in the domain of the function above.

\(f(x) = \frac{2x + 3}{x} - 7 + \frac{1}{x - 2} + \frac{x^2}{x + 4}\)

Put this all over a common denominator to get a rational function:

\({(2x + 3)(x - 2)(x + 4)\over(x)(x-2)(x+4)}-{7(x)(x-2)(x+4)\over(x)(x-2)(x+4)}+{1(x)(x+4)\over(x)(x-2)(x+4)}+{x^3(x-2)\over(x)(x-2)(x+4)}\)

Expanding:

\({2x^3+7x^2-10x-24\over{x^3+2x^2-8x}}-{7x^3+14x^2-56x\over{x^3+2x^2-8x}}+{x^2+4x\over{x^3+2x^2-8x}}+{x^4-2x^3\over{x^3+2x^2-8x}}\)

Combine numerators:

\(x^4-7x^3-6x^2+50x-24\over{x^3+2x^2-8x}\)

Clearly, after graphing or observing the degree of the numerator, f will have a range of (-inf, inf).

Substitution: f(k,-3) = 2k - 3(-3)^2 + 7 - 4(k)(-3) + 8(k)^2 = = 2k - 3*9 + 7 + 12k + 8k^2 = 8k^2 + 14k - 20 = -10

8k^2 + 14k - 10 = 0; divide by 2: 4k^2 + 7k - 5 = 0

Use quadratic formula: \(k = {-b \pm \sqrt{b^2-4ac} \over 2a}\); substituting, you have \(k = {-7\pm\sqrt{49+80}\over8}={-7\pm\sqrt{129}\over8}\), two possible values.

Opening parenthesis for only addition never hurts: \(1 - 3i + 2-i+2+i+1+3i\)

One can treat \(i=\sqrt{-1}\) as a variable, and combine like terms normally: \(1+2+2+1-3i+3i-i+i=6\)

Combine like terms: \({15 - 16i\over{-3+4i}}\)

Rationalize the denominator by multiplying by conjugate to get difference of squares which removes the imaginary term:

\({(15 - 16i)(-3 - 4i)\over{(-3 + 4i)(-3-4i)}}={-45+48i-60i+64i^2\over{9-16i^2}}={19-12i\over25}={19\over25}-{12\over25}i\)

Combine like terms: \(x^2 - 4x +48 = 0\)

Quadratic formula: \(x = {4\pm\sqrt{16-4*48}\over2}={4\pm4i\sqrt{11}\over2}=2\pm2\sqrt{11}i\)

So, your answers are \(2 - 2\sqrt{11}i\), \(2 + 2\sqrt{11}i\)

From the two equations you have given in the system there is no k and m to get infinite solutions (a,b) could you double check your equations?