proyaop

By pythagorean theorem, with diagonal EF, we know triangle EFG has side length EG = \sqrt{64 - 36} = 2\sqrt{7}

Now you can do trig: secE = 1/cosE. cosE = adj/hyp, so secE = hyp/adj. hypotenuse = 8, adjacent = 2\sqrt{7}

So, secE = 8/2\sqrt{7} = 4/\sqrt{7} = \(4\sqrt{7}/7\) after rationalizing the denominator.

cosF = adjacent/hypotenuse = 6/8 = 3/4

tanF = sinF/cosF (remember this). We know cosF is 3/4, and for sinF = opp/hyp, where opposite = 2\sqrt{7}, hypotenuse = 8.

2\sqrt{7}/8 = \sqrt{7}/4 thus, sinF/cosF = (\sqrt{7}/4)/(3/4) = \(\sqrt{7}/3\)

Hey xXValeXx- The reference angle is the acute angle made with the x-axis and the line that makes the angle. 

So, for -23pi/6, you can add 12pi/6 = 2pi to the value to keep it the same but a "neater" number. -23pi/6 + 2pi + 2pi = pi/6.

Next, for 7pi/6, this angle is pi/6 greater than pi, and pi is 180 degrees, thus the point (-1,0) on the unit circle. So again, pi/6 is the acute angle made with the x-axis.

Lastly, for -11pi/12, after you make it positive by adding 2pi, you get 13pi/12, which makes a pi/12 angle with the x-axis in quadrant 3.

Thus, your answers are pi/6, pi/6, and pi/12, respectively.

Order of operations:

First do parenthesis

Then do exponents

Then do multiplication (and if there is a negative sign treat it as -1 * (_))

Then do division

Then do addition

Then do subtraction

Abbreviation being PEMDAS... good luck...

Because each variable has a positive or negative value, we have to account for both possible values, creating two "cases", or options, and we test each one to get a different solution/ordered pair. If you don't understand, you should try searching "casework" problems, but this problem is just getting used to algebraic expansion. Remember difference of cubes!!!

(c)

Now that we know a + b = p/n 2sqrt(3), and that a - b = 4, we can use system of equations to get the individual values of each variable.

Case 1: a + b = 2sqrt(3)

a + b = 2sqrt(3)

a - b = 4

Adding both equations, 2a = 4 + 2sqrt(3), a = 2 + sqrt(3).

Plugging in 2 + sqrt(3) for the top equation, 2 + sqrt(3) - 2sqrt(3) = -b, so b = sqrt(3) - 2.

So for case 1, \(a = 2 + \sqrt{3}\), and \(b = -2 + \sqrt{3}\)

Case 2: a + b = -2sqrt(3)

a + b = -2sqrt(3)

a - b = 4

Adding both equations, 2a = 4 - 2sqrt(3), a = 2 - sqrt(3)

Plugging in 2 - sqrt(3) for the top equation, 2 - sqrt(3) + 2sqrt(3) = -b, so b = -2 - sqrt(3).

So for case 2, \(a = 2 - \sqrt{3}\), and \(b = -2 - \sqrt{3}\)

Hence, we have:

a = 2 + sqrt(3)     =>        b = -2 + sqrt(3)

                           or

a = 2 - sqrt(3)      =>        b = -2 - sqrt(3)

(b)

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

Earlier, we manipulated a^2 + ab + b^2 into (a - b)^2 + 3ab, however you can also manipulate it to (a + b)^2 - ab (do the expansion and see that it works!)

So... a^3 - b^3 = (a - b)[(a + b)^2 - ab] = 52

However, from (a), we learned that a - b = 4, and that ab = -1, substituting in, we have a^3 - b^3 = 4[(a + b)^2 + 1] = 52.

Dividing by 4, we have (a + b)^2 + 1 = 13, a + b = positive/negative square root of (12), which sqrt(12) = 2sqrt(3), so positive negative 2sqrt(3).

Hence, \(a + b = \pm2\sqrt{3}\)

(a)

By difference of cubes, a^3 - b^3 = (a - b)(a^2 + ab + b^2)

Substituting and manipulating, we have a^3 - b^3 = 4[(a - b)^2 + 3ab]

Substituting again, we have a^3 - b^3 = 4(16 + 3ab) = 64 + 12ab = 52

Subtracting and dividing by 12, we have 12ab = 52 - 64 = -12, ab = -1.

Hence, ab = -1

You can use distributive property => f(x) = -12x + 4

Domain is legal inputs for x, so domain is "all real numbers"

Range is legal outputs of f(x) (or values of f(x)), so range is "all real numbers"

\(x = {1\over2 + {1\over x - 3}}\), multiply by (x - 3)/(x - 3)

\(x = {x - 3\over 2x - 6 + 1}\)

2x^2 - 5x = x - 3

2x^2 - 6x + 3 = 0

Quadratic formula!!!

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

\(x - {1\over x} = \pm1\)