jfan17

since i haven't seen anyone answer yet, i'll try pitching my 2 cents.

we have that z+1/z = 1

if we square our given expression, we get:

z^2 + 2 + 1/z^2 = 1

z^2 + 1/z^2 = -1

if we square this again, we get:

z^4 + 2 + 1/z^4 = 1

z^4 + 1/z^4 = -1

as you might be able to deduce, this will equal -1 however many times we square our expression, because the 2ab term will always equal 2, and the RHS(right hand side) will always equal (-1)^2 = 1, so subtract the two, and you get -1.

z^2 + 1/z^2 = -1

z^4 + 1/z^4 = -1

z^8 + 1/z^8 = -1

Now we have a problem because we are asked to find z^10 + 1/z^10 which does not entail these powers of 2. We can try experimenting to get the z^10 term, and the first thing that comes to mind is multiplying by z^2 + 1/z^2

(z^2+1/z^2)(z^8+1/z^8) = z^10 + 1/z^10 + z^6 + 1/z^6 = 1

now we need the value of z^6 + 1/z^6 to get the value of our desired expression.

To do so, we can repeat these steps with z^2+1/z^2 and z^4 + 1/z^4

(z^2 + 1/z^2)(z^4+1/z^4) = z^6 + 1/z^6 + z^2 + 1/z^2 = 1

this is good, because we have the value of z^2 + 1/z^2 = -1

z^6 + 1/z^6 - 1 = 1

z^6 + 1/z^6 = 2

now we go back to our expression with the z^10 term and substitute the value of z^6 + 1/z^6 to get;

z^10 + 1/z^10 + 2 = 1

z^10 + 1/z^10 = -1

By triangle inequality theorem, any two given sides, a, b, of a triangle, must be greater than or equal to the third side

a+b >= c

Just check all of the 4 cases for the answer, it will be the outlier that does not satsify the above condition. In other words, when you add two of the sides, it is smaller than the third side.

Sum of cubes factorization, which can be generalized to all odd powers of n:

a^3 + b^3 = (a+b)(a^2-ab+b^2)

replace a and b with c and 1/c respectively in this case to get:

c^3 + (1/c)^3 = (c+1/c)(c^2-1+1/c^2)

since we are already told c^2 + 1/c^2 = 14, we can substitute to get:

c^3 + (1/c)^3 = (c+1/c)(14-1) = (c+1/c)(13)

now all we need to find is the value of c+1/c

realize that:

(c+1/c)^2 = c^2 + 1/c^2 + 2 = 14 + 2 = 16 

this is because we are given that c^2 + 1/c^2 = 14, so just substitute

(c+1/c) = sqrt(16) = +-4

Since we want the largest value possible of the expression, we can assume it's positive, because if it's negative, then -4(13) would likewise be negative. Our answer is then:

4(13) = 52

vieta's formulas would really help in this situation! With vietas, you don't need to find the specific roots a,b of the quadratic, but rather, you can derive and find the elementary symmetric sums of the roots (a + b + c, ab + bc + ac, abc as an example for a third degree poly.)

By vieta's, the roots a,b of the given quadratic are such that:

a+b = -4/2 = -2

ab = -5/2

Now that we have these two values, just expand the numerator as:

((a+b)-ab) = (-2+5/2) = 1/2 for the numerator

and rewrite the denominator as:
(a+b)^2 - 2ab = (-2)^2 +5 = 4 + 5 = 9

our answer is then:

(1/2)/9 = 1/18

you are given the vertex of the quadratic, so that should be telling you that you can use vertex form of the quadratic.

the vertex form of g(x) can be written as

a(x-h)^2 + k where the vertex is (h,k)

g(x) = a(x+5)^2

since it has a y intercept of 5, the point (0,-5) should lie on this graph as well

-5 = a(5)^2 = 25a 

a = -1/5

now we have our function:

g(x) = -1/5(x+5)^2

g(1) = -36/5

You should check out de moivre's theorem

I'm not sure about the approach you took, but have you tried multiplying both sides by the denominator($9\sin{\theta}-\cos{\theta}$) from the beginning? Maybe that would get you somewhere, since you know that $r * \cos{\theta} = x$ and $r * \sin{\theta} = y$

Not sure if you're asking for the solution, but if you are, then you can keep reading I guess.

Using the method mentioned above:

$9r\sin{\theta} - r\cos{\theta} = 7$

Using our substitutions, we rewrite this as:

$9y - x = 7$

From here on out, it gets pretty intuitive. Dividing and rearranging to get our desired form, we get:

$9y = x+7$

$y = x/9 + 7/9$

or 

$y = \frac19 x + \frac79$

Hope this helped!

If you multiply all three of our given expressions together, you get:

$(abc)^2 = (1/2 * 8 * 128)abc$

Dividing by abc on both sides, we then get:
$abc = 64 * 8 = 512$

If we look back to our first expression $bc = a/2$, we can multiply a on both sides to get a known value(abc).

This gives us:

$abc = a^2/2$

$512 = a^2/2$

$a^2=1024$

Realize that 1024 = 2¹⁰, so the square root of it is simply 2⁵ = 32

$a = 2^5 = 32$

Repeat the same steps for our second equation by multiplying by b on both sides, and we get:
$abc = 8b^2$

$512 = 8b^2$

$64 = b^2$

$b = 8$

$abc = 512$

Since a = 32 and b = 8

That only leaves us with c, which is then:

$512/(32*8) = 2$

a = 32

b = 8

c = 2

a+b+c = 32 + 8 + 2 = 42

Have you tried the sum of cubes formula? Try expressing the left side term with the sum of cubes factorization(the 5 term and 1).

Realize that if the value of x is less than 1, that causes the numerator to be the square root of an imaginary(since 0-1 would be -1), which is clearly non-real. The smallest possible integer value for x then becomes 1.

in general, a line given two vectors can be parametrized as :

a + t*d 

where:

a = starting vector

t = any real number

d = the direction vector

In this case, we have (1,1,2) and (2,3,4)

the direction vector can be found through subtracting the two. We get:

(2,3,4) - (1,1,2) = (1,2,2)

with our starting point being (1,1,2)

We then get the line equation as:

(1,1,2) + t(1,2,2)

this then turns into:

(1,1,2) + (t,2t,2t)

Adding the two, we get:
(t+1,2t+1,2t+2)

a = 1

b = 1

c = 2

d = 1

e = 2

f = 2

c/a = 2/1 = 2

e/a = 2/1 = 2

the ordered pair (c/a, e/a) is then (2,2). Please correct me if I'm wrong, precalc has never been one of my strong suits. 

By Thales' Theorem: angle BAC = 90 degrees(it subtends an arc of 180 degrees)

Because we know that, we know that triangle BAC is a right triangle with angle measurements of 45, 45, 90

Next, realize that the hypotenuse of said triangle is the diameter BC, which has length 16( which is 8 *2).

Now that we have our hypotenuse, realize that the ratio of a 45-45-90 triangle is 

x, x, $x\sqrt{2}$

That means that our leg of our right triangle is equal to:

$16\over{\sqrt{2}}$= $16\sqrt{2}\over2$ by rationalizing the denominator(multiplying it by $\sqrt{2}\over\sqrt{2}$)

This equals 

$8\sqrt{2}$

Triangle area formula is 1/2 * b * h

Since the base and height are the same in this scenario, we have that the area of triangle BAC is:

$(8\sqrt{2})^2/2 = 64*2/2 = 64$

The area of the entire circle is 64pi, so half of it is 32pi. Our area of the shaded region is then:

$32\pi-64$

The probability is then this part divided by the entire area of the dart board, which is 64pi, so our answer is:

$32\pi-64\over64\pi$