answer:To find f^-1(x), we start with the equation y = 7x - 2, where y represents the output of f(x). To invert the function, we solve for x in terms of y. First, we add 2 to both sides, getting y + 2 = 7x. Then, we divide both sides by 7, resulting in x = (y + 2)/7. Since we're finding the inverse function, we swap x and y to express the inverse in terms of x, giving f^-1(x) = (x + 2)/7. For g^-1(x), we start with y = x^2 - 4x - 1. To invert this function, we need to solve the quadratic equation for x in terms of y. First, we rearrange the equation to x^2 - 4x - (y + 1) = 0. We can then use the quadratic formula: x = (-b ± sqrt(b^2 - 4ac)) / 2a, where a = 1, b = -4, and c = -(y + 1). Applying the formula, we get x = (4 ± sqrt(16 + 4(y + 1))) / 2. Simplifying, we get x = (4 ± sqrt(20 + 4y)) / 2, which can be further simplified to x = 2 ± sqrt(5 + y). Since the inverse function should be expressed in terms of x, we swap y and x, giving g^-1(x) = 2 ± sqrt(5 + x). However, note that g(x) is not one-to-one, so its inverse is not a function in the classical sense. We can consider the two possible branches of the inverse separately, if needed. Given the inverses, we can find (g^-1 o f^-1)(x). We first apply f^-1(x) = (x + 2)/7, then apply g^-1 to the result. Substituting into g^-1, we get (g^-1 o f^-1)(x) = 2 ± sqrt(5 + (x + 2)/7). Simplifying, we have (g^-1 o f^-1)(x) = 2 ± sqrt((35 + x + 2)/7) = 2 ± sqrt((37 + x)/7). To find (f^-1 o g^-1)(x), we first apply g^-1(x) = 2 ± sqrt(5 + x), then apply f^-1 to the result. Substituting into f^-1, we get (f^-1 o g^-1)(x) = ((2 ± sqrt(5 + x)) + 2)/7. Simplifying, we have (f^-1 o g^-1)(x) = (4 ± sqrt(5 + x))/7. Note that the process of finding inverses and compositions highlights the importance of understanding function properties, such as injectivity and surjectivity. In this case, g(x) is not one-to-one, which complicates the process of finding its inverse.

answer:To prove the Angle Bisector Theorem, we can use similar triangles and the properties of angle bisectors. Let's draw a diagram of the triangle ABC and the angle bisectors of angles A and B, intersecting at point I. First, we notice that the angle bisector of angle A divides the side c into two segments, c1 and c2. We can also draw a line from point I to the vertex C, forming two triangles, ACI and BCI. Now, consider the triangles ACI and BCI. Since the angle bisector of angle A intersects side c at point I, we know that the angle CAI is equal to the angle IAB (by definition of angle bisector). Similarly, the angle bisector of angle B intersects side c at point I, so the angle CBI is equal to the angle IBA. Using the fact that the sum of the angles in a triangle is 180 degrees, we can write: ∠CAI + ∠ACI + ∠CIA = 180° (in triangle ACI) ∠CBI + ∠BCI + ∠CIB = 180° (in triangle BCI) Since ∠CAI = ∠IAB and ∠CBI = ∠IBA, we can substitute these angles into the equations above: ∠IAB + ∠ACI + ∠CIA = 180° ∠IBA + ∠BCI + ∠CIB = 180° Now, notice that ∠CIA and ∠CIB are vertical angles, so they are equal. Also, ∠IAB + ∠IBA = ∠A + ∠B (by definition of angle sum in a triangle). Therefore, we can rewrite the equations as: ∠A + ∠B + ∠ACI = 180° ∠A + ∠B + ∠BCI = 180° This implies that ∠ACI = ∠BCI, which means that triangles ACI and BCI are similar (by the AA similarity theorem). As a result, we can write: c1 / a = c2 / b or, equivalently: c1 / c2 = a / b which proves the Angle Bisector Theorem. The Angle Bisector Theorem has many applications in geometry and trigonometry, and it's a fundamental result that helps us understand the properties of triangles and angle bisectors. It's also worth noting that this theorem can be generalized to polygons with more than three sides, leading to interesting and useful results in geometry and other areas of mathematics.

answer:Thank you, sir. I've been fortunate enough to call this place home for nigh on 40 years now. Started out as a young lad, learnin' the trade from me old dad, who was the gardener here before me. I've seen a lot of changes over the years, but the gardens have always been me pride and joy. There's somethin' special about watchin' a seedling you planted grow into a beautiful bloom, and knowin' it's bringin' joy to the family and their guests. I've had the pleasure of workin' under three generations of the family now, and each one's had their own ideas and tastes, but they've all shared a love for these gardens. (pauses to inspect his clipping work) Excuse me, just makin' sure this hedge is even... can't have it lookin' sloppy, now can I?

answer:Now where were we, sir? Ah yes, the gardens... I was wonderin', would you like a tour? I could show you some of me favorite spots, and tell you a bit about the history of the estate?